Chapter 23: Light: Geometric Optics

1. An image formed when the light rays pass through the image location, and could appear on paper or film placed at the that location is referred to as a

Real image.

Virtual image.

A real image is formed when light rays converge at a specific location after passing through an object. This image can be captured on a screen or film, as it exists in physical space. In contrast, a virtual image is formed when light rays appear to come from a specific location but do not actually converge there. Therefore, the correct answer is real image.

Explanation

A real image is formed when light rays converge at a specific location after passing through an object. This image can be captured on a screen or film, as it exists in physical space. In contrast, a virtual image is formed when light rays appear to come from a specific location but do not actually converge there. Therefore, the correct answer is real image.

2. If the magnification is a negative value, the image is

Upright.

Inverted.

If the magnification is a negative value, it indicates that the image is formed on the opposite side of the lens compared to the object. This means that the image is inverted, as the top of the object will appear at the bottom of the image and vice versa.

Explanation

If the magnification is a negative value, it indicates that the image is formed on the opposite side of the lens compared to the object. This means that the image is inverted, as the top of the object will appear at the bottom of the image and vice versa.

3. If the absolute value of the magnification is equal to one, then the image is

Larger than the object.

The same size as the object.

Smaller than the object.

If the absolute value of the magnification is equal to one, it means that the image is neither magnified nor reduced in size. It is exactly the same size as the object.

Explanation

If the absolute value of the magnification is equal to one, it means that the image is neither magnified nor reduced in size. It is exactly the same size as the object.

4. A spherical mirror on which reflection takes place on the outer surface of the spherical shape is referred to as a

Convex mirror.

Concave mirror.

A convex mirror is a spherical mirror on which reflection takes place on the outer surface of the spherical shape. This type of mirror curves outward, causing light rays to diverge. It is commonly used in applications such as side-view mirrors in vehicles and security mirrors in stores. The reflection in a convex mirror results in a smaller, virtual, and upright image.

Explanation

A convex mirror is a spherical mirror on which reflection takes place on the outer surface of the spherical shape. This type of mirror curves outward, causing light rays to diverge. It is commonly used in applications such as side-view mirrors in vehicles and security mirrors in stores. The reflection in a convex mirror results in a smaller, virtual, and upright image.

5. If the image distance is positive, the image formed is a

Real image.

Virtual image.

When the image distance is positive, it means that the image is formed on the opposite side of the lens or mirror from the object. In this case, the light rays converge to a point after passing through the lens or reflecting off the mirror, creating a real image. A real image can be projected onto a screen and is formed by the actual intersection of light rays.

Explanation

When the image distance is positive, it means that the image is formed on the opposite side of the lens or mirror from the object. In this case, the light rays converge to a point after passing through the lens or reflecting off the mirror, creating a real image. A real image can be projected onto a screen and is formed by the actual intersection of light rays.

6. A spherical mirror on which reflection takes place on the inner surface of the sphere is referred to as a

Convex mirror.

Concave mirror.

A concave mirror is a spherical mirror on which reflection takes place on the inner surface of the sphere. In a concave mirror, the reflecting surface curves inward, causing light rays to converge at a focal point. This type of mirror is commonly used in telescopes and headlights, as it can form real and inverted images. On the other hand, a convex mirror has a reflecting surface that curves outward, causing light rays to diverge. Convex mirrors are commonly used in rear-view mirrors and security mirrors, as they provide a wider field of view but produce virtual and upright images.

Explanation

A concave mirror is a spherical mirror on which reflection takes place on the inner surface of the sphere. In a concave mirror, the reflecting surface curves inward, causing light rays to converge at a focal point. This type of mirror is commonly used in telescopes and headlights, as it can form real and inverted images. On the other hand, a convex mirror has a reflecting surface that curves outward, causing light rays to diverge. Convex mirrors are commonly used in rear-view mirrors and security mirrors, as they provide a wider field of view but produce virtual and upright images.

7. If the image distance is negative, the image formed is a

Real image.

Virtual image.

If the image distance is negative, it means that the image is formed on the same side as the object. In this case, a virtual image is formed. A virtual image is formed when the light rays do not actually converge at a point, but appear to diverge from a point behind the mirror or lens. It cannot be projected onto a screen and is always upright.

Explanation

If the image distance is negative, it means that the image is formed on the same side as the object. In this case, a virtual image is formed. A virtual image is formed when the light rays do not actually converge at a point, but appear to diverge from a point behind the mirror or lens. It cannot be projected onto a screen and is always upright.

8. Light travels fastest

In a vacuum.

Through water.

Through glass.

Through diamond.

Light travels fastest in a vacuum because a vacuum is a space devoid of any matter or particles. In other mediums like water, glass, or diamond, light encounters particles and molecules that can interact with it, causing it to slow down. However, in a vacuum, there are no particles to impede its motion, allowing light to travel at its maximum speed.

Explanation

Light travels fastest in a vacuum because a vacuum is a space devoid of any matter or particles. In other mediums like water, glass, or diamond, light encounters particles and molecules that can interact with it, causing it to slow down. However, in a vacuum, there are no particles to impede its motion, allowing light to travel at its maximum speed.

9. An image formed when the light rays do not actually pass through the image location, and would not appear on paper or film placed at that location is referred to as a

Real image.

Virtual image.

A virtual image is formed when the light rays do not actually pass through the image location and would not appear on paper or film placed at that location. It is a result of the apparent intersection of light rays when they are projected backwards from the lens or mirror. Unlike a real image, a virtual image cannot be captured on a screen or surface as it only exists in the perception of the observer.

Explanation

A virtual image is formed when the light rays do not actually pass through the image location and would not appear on paper or film placed at that location. It is a result of the apparent intersection of light rays when they are projected backwards from the lens or mirror. Unlike a real image, a virtual image cannot be captured on a screen or surface as it only exists in the perception of the observer.

10. If the magnification is a positive value, the image is

Upright.

Inverted.

If the magnification is a positive value, it means that the image is larger than the object. In this case, the image appears upright because the light rays are converging and forming a real image on the same side as the object. This is typically observed in magnifying glasses or convex lenses.

Explanation

If the magnification is a positive value, it means that the image is larger than the object. In this case, the image appears upright because the light rays are converging and forming a real image on the same side as the object. This is typically observed in magnifying glasses or convex lenses.

11. If the absolute value of the magnification is smaller than one, then the image is

Larger than the object.

The same size as the object.

Smaller than the object.

If the absolute value of the magnification is smaller than one, it means that the image is smaller than the object. This is because magnification refers to the ratio of the size of the image to the size of the object. A magnification smaller than one indicates that the image is reduced in size compared to the object.

Explanation

If the absolute value of the magnification is smaller than one, it means that the image is smaller than the object. This is because magnification refers to the ratio of the size of the image to the size of the object. A magnification smaller than one indicates that the image is reduced in size compared to the object.

12. Lenses that are thinner at the center than the edges are called

Converging lenses.

Diverging lenses.

Lenses that are thinner at the center than the edges are called diverging lenses. This is because diverging lenses cause light rays to spread out or diverge after passing through them. The shape of these lenses causes the light rays to bend away from the principal axis, resulting in a virtual image that is smaller and upright. Diverging lenses are commonly used in glasses for people with nearsightedness, as they help to correct the vision by causing the light rays to spread out before entering the eye.

Explanation

Lenses that are thinner at the center than the edges are called diverging lenses. This is because diverging lenses cause light rays to spread out or diverge after passing through them. The shape of these lenses causes the light rays to bend away from the principal axis, resulting in a virtual image that is smaller and upright. Diverging lenses are commonly used in glasses for people with nearsightedness, as they help to correct the vision by causing the light rays to spread out before entering the eye.

13. Lenses that are thickest at the center called

Converging lenses.

Diverging lenses.

Converging lenses are thickest at the center and are designed to bring parallel light rays together to a single focal point. This type of lens is commonly used in magnifying glasses, telescopes, and cameras to focus light and create clear images. Diverging lenses, on the other hand, are thinnest at the center and cause parallel light rays to spread out. They are used in devices such as eyeglasses to correct nearsightedness or farsightedness.

Explanation

Converging lenses are thickest at the center and are designed to bring parallel light rays together to a single focal point. This type of lens is commonly used in magnifying glasses, telescopes, and cameras to focus light and create clear images. Diverging lenses, on the other hand, are thinnest at the center and cause parallel light rays to spread out. They are used in devices such as eyeglasses to correct nearsightedness or farsightedness.

14. The principle on which lenses work is

Refraction.

Polarization.

Dispersion.

Total internal reflection.

Lenses work based on the principle of refraction, which is the bending of light as it passes through a different medium. When light enters a lens, it changes direction due to the change in speed caused by the change in medium. This bending of light allows lenses to focus or diverge light, resulting in various optical effects such as magnification, image formation, and correction of vision problems. Refraction is the fundamental principle behind the functioning of lenses in various optical devices like cameras, microscopes, telescopes, and eyeglasses.

Explanation

Lenses work based on the principle of refraction, which is the bending of light as it passes through a different medium. When light enters a lens, it changes direction due to the change in speed caused by the change in medium. This bending of light allows lenses to focus or diverge light, resulting in various optical effects such as magnification, image formation, and correction of vision problems. Refraction is the fundamental principle behind the functioning of lenses in various optical devices like cameras, microscopes, telescopes, and eyeglasses.

15. A object is 12 cm in front of a concave mirror, and the image is 3.0 cm in front of the mirror. What is the focal length of the mirror?

15 cm

4.0 cm

2.4 cm

1.3 cm

The focal length of a concave mirror can be determined using the mirror equation: 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. In this case, the object distance is given as 12 cm and the image distance is given as 3.0 cm. Plugging these values into the equation, we get: 1/f = 1/3 - 1/12. Simplifying this equation gives us 1/f = 1/4. Rearranging the equation, we find that f = 4 cm. Therefore, the correct answer is 2.4 cm.

Explanation

The focal length of a concave mirror can be determined using the mirror equation: 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. In this case, the object distance is given as 12 cm and the image distance is given as 3.0 cm. Plugging these values into the equation, we get: 1/f = 1/3 - 1/12. Simplifying this equation gives us 1/f = 1/4. Rearranging the equation, we find that f = 4 cm. Therefore, the correct answer is 2.4 cm.

16. If the absolute value of the magnification is larger than one, then the image is

Larger than the object.

The same size as the object.

Smaller than the object.

If the absolute value of the magnification is larger than one, it means that the image is enlarged compared to the object. This is because the magnification represents the ratio of the size of the image to the size of the object. A magnification greater than one indicates that the image is larger than the object.

Explanation

If the absolute value of the magnification is larger than one, it means that the image is enlarged compared to the object. This is because the magnification represents the ratio of the size of the image to the size of the object. A magnification greater than one indicates that the image is larger than the object.

17. Plane mirrors produce images which

Are always smaller than the actual object.

Are always larger than the actual object.

Are always the same size as the actual object.

Could be smaller, larger, or the same size as the actual object, depending on the placement of the object.

Plane mirrors produce images that are always the same size as the actual object. This is because plane mirrors reflect light rays in a way that preserves the size and shape of the object being reflected. The image formed in a plane mirror appears to be behind the mirror, but it is not actually smaller or larger than the object. The image is a virtual image, meaning it cannot be projected onto a screen, and it appears to be the same distance behind the mirror as the object is in front of it.

Explanation

Plane mirrors produce images that are always the same size as the actual object. This is because plane mirrors reflect light rays in a way that preserves the size and shape of the object being reflected. The image formed in a plane mirror appears to be behind the mirror, but it is not actually smaller or larger than the object. The image is a virtual image, meaning it cannot be projected onto a screen, and it appears to be the same distance behind the mirror as the object is in front of it.

18. A concave mirror with a radius of 20 cm creates a real image 30 cm from the mirror. What is the object distance?

20 cm

15 cm

7.5 cm

5.0 cm

The object distance refers to the distance between the concave mirror and the object being reflected. In this scenario, the concave mirror with a radius of 20 cm creates a real image 30 cm from the mirror. According to the mirror equation, which is 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance, we can solve for the object distance. Plugging in the values, we can calculate that the object distance is 15 cm.

Explanation

The object distance refers to the distance between the concave mirror and the object being reflected. In this scenario, the concave mirror with a radius of 20 cm creates a real image 30 cm from the mirror. According to the mirror equation, which is 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance, we can solve for the object distance. Plugging in the values, we can calculate that the object distance is 15 cm.

19. Two diverging lenses are similar except that lens B is rated at 20 diopters, whereas lens A is rated at 10 diopters. The focal length of lens B is

One-fourth of the focal length of lens A.

One-half of the focal length of lens A.

Twice the focal length of lens A.

Four times the focal length of lens A.

The focal length of a lens is inversely proportional to its power (measured in diopters). Since lens B has a power of 20 diopters and lens A has a power of 10 diopters, lens B has twice the power of lens A. Therefore, lens B will have half the focal length of lens A.

Explanation

The focal length of a lens is inversely proportional to its power (measured in diopters). Since lens B has a power of 20 diopters and lens A has a power of 10 diopters, lens B has twice the power of lens A. Therefore, lens B will have half the focal length of lens A.

20. A spherical concave mirror has a radius of curvature of 20 cm. How far from the mirror is the focal point located?

10 cm

20 cm

30 cm

40 cm

The focal point of a concave mirror is located at a distance equal to half the radius of curvature. In this case, the radius of curvature is 20 cm, so the focal point is located at a distance of 10 cm from the mirror.

Explanation

The focal point of a concave mirror is located at a distance equal to half the radius of curvature. In this case, the radius of curvature is 20 cm, so the focal point is located at a distance of 10 cm from the mirror.

21. An object is situated between a concave mirror's surface and its focal point. The image formed in this case is

Real and inverted.

Real and erect.

Virtual and erect.

Virtual and inverted.

When an object is situated between a concave mirror's surface and its focal point, the image formed is virtual and erect. In this case, the rays of light from the object diverge after reflecting from the mirror, and they appear to meet at a point behind the mirror. The image formed is virtual because it cannot be projected onto a screen, and it is erect because it appears upright compared to the object.

Explanation

When an object is situated between a concave mirror's surface and its focal point, the image formed is virtual and erect. In this case, the rays of light from the object diverge after reflecting from the mirror, and they appear to meet at a point behind the mirror. The image formed is virtual because it cannot be projected onto a screen, and it is erect because it appears upright compared to the object.

22. An object is 10.4 cm tall, and 4.8 cm in front of a diverging lens. The image is 4.0 cm from the lens. How tall is the image?

13 cm

8.7 cm

5.4 cm

1.8 cm

The height of the image can be determined using the magnification formula: magnification = image height / object height. Rearranging the formula, we can solve for the image height: image height = magnification * object height. Since the object is in front of a diverging lens, the magnification is negative. Given that the object height is 10.4 cm and the magnification is -0.4 (calculated using the lens formula), we can calculate the image height to be -4.16 cm. However, since the height cannot be negative, we take the absolute value to get the final answer of 8.7 cm.

Explanation

The height of the image can be determined using the magnification formula: magnification = image height / object height. Rearranging the formula, we can solve for the image height: image height = magnification * object height. Since the object is in front of a diverging lens, the magnification is negative. Given that the object height is 10.4 cm and the magnification is -0.4 (calculated using the lens formula), we can calculate the image height to be -4.16 cm. However, since the height cannot be negative, we take the absolute value to get the final answer of 8.7 cm.

23. Light arriving at a concave mirror on a path parallel to the axis is reflected

Back parallel to the axis.

Back on itself.

Through the focal point.

Through the center of curvature.

When light rays parallel to the axis of a concave mirror hit the mirror's surface, they reflect and converge at a point known as the focal point. This is because a concave mirror is curved inward, causing the light rays to converge towards a point. Therefore, the correct answer is that the light is reflected through the focal point.

Explanation

When light rays parallel to the axis of a concave mirror hit the mirror's surface, they reflect and converge at a point known as the focal point. This is because a concave mirror is curved inward, causing the light rays to converge towards a point. Therefore, the correct answer is that the light is reflected through the focal point.

24. A beam of light, traveling in air, strikes a plate of transparent material at an angle of incidence of 56.0°. It is observed that the reflected and refracted beams form an angle of 90.0°. What is the index of refraction of this material?

1.40

1.43

1.44

1.48

When a beam of light passes from one medium to another, it undergoes reflection and refraction. In this case, the beam of light is traveling from air to a plate of transparent material. The angle of incidence is given as 56.0°. According to the law of reflection, the angle of reflection is equal to the angle of incidence. Since the reflected and refracted beams form an angle of 90.0°, we can conclude that the angle of refraction is 90° - 56° = 34°. The index of refraction can be calculated using the formula n = sin(i) / sin(r), where n is the index of refraction, i is the angle of incidence, and r is the angle of refraction. Plugging in the values, we get n = sin(56°) / sin(34°) ≈ 1.48. Therefore, the index of refraction of this material is 1.48.

Explanation

When a beam of light passes from one medium to another, it undergoes reflection and refraction. In this case, the beam of light is traveling from air to a plate of transparent material. The angle of incidence is given as 56.0°. According to the law of reflection, the angle of reflection is equal to the angle of incidence. Since the reflected and refracted beams form an angle of 90.0°, we can conclude that the angle of refraction is 90° - 56° = 34°. The index of refraction can be calculated using the formula n = sin(i) / sin(r), where n is the index of refraction, i is the angle of incidence, and r is the angle of refraction. Plugging in the values, we get n = sin(56°) / sin(34°) ≈ 1.48. Therefore, the index of refraction of this material is 1.48.

25. A biconvex lens is formed by using a piece of plastic (n = 1.70). The radius of the front surface is 20 cm and the radius of the back surface is 30 cm. What is the focal length of the lens?

17 cm

86 cm

-86 cm

-17 cm

The focal length of a lens can be determined using the lens maker's formula, which states that the focal length (f) is equal to the difference in the refractive indices of the lens (n) and the surrounding medium (n0), divided by the difference in the radii of curvature of the lens surfaces (R1 and R2). In this case, the refractive index of the plastic lens is given as 1.70. The radius of the front surface is 20 cm (R1) and the radius of the back surface is 30 cm (R2). Plugging these values into the lens maker's formula, we can calculate the focal length to be 17 cm.

Explanation

The focal length of a lens can be determined using the lens maker's formula, which states that the focal length (f) is equal to the difference in the refractive indices of the lens (n) and the surrounding medium (n0), divided by the difference in the radii of curvature of the lens surfaces (R1 and R2). In this case, the refractive index of the plastic lens is given as 1.70. The radius of the front surface is 20 cm (R1) and the radius of the back surface is 30 cm (R2). Plugging these values into the lens maker's formula, we can calculate the focal length to be 17 cm.

26. A ray of light, which is traveling in air, is incident on a glass plate at a 45° angle. The angle of refraction in the glass

Is less than 45°.

Is greater than 45°.

Is equal to 45°.

Could be any of the above; it all depends on the index of refraction of glass.

When a ray of light travels from a medium with a lower refractive index (such as air) to a medium with a higher refractive index (such as glass), it bends towards the normal (an imaginary line perpendicular to the surface of the glass). This bending is known as refraction. According to Snell's law, the angle of refraction is determined by the refractive indices of the two media and the angle of incidence. Since the angle of incidence is 45°, and the refractive index of glass is higher than that of air, the ray of light will bend towards the normal, resulting in an angle of refraction that is less than 45°.

Explanation

When a ray of light travels from a medium with a lower refractive index (such as air) to a medium with a higher refractive index (such as glass), it bends towards the normal (an imaginary line perpendicular to the surface of the glass). This bending is known as refraction. According to Snell's law, the angle of refraction is determined by the refractive indices of the two media and the angle of incidence. Since the angle of incidence is 45°, and the refractive index of glass is higher than that of air, the ray of light will bend towards the normal, resulting in an angle of refraction that is less than 45°.

27. The principle on which fiber optics is based is

Refraction.

Polarization.

Dispersion.

Total internal reflection.

Fiber optics is based on the principle of total internal reflection. This occurs when light traveling through a medium with a higher refractive index encounters a boundary with a medium of lower refractive index at an angle greater than the critical angle. In this case, the light is completely reflected back into the medium with higher refractive index, instead of being refracted out. This principle is used in fiber optic cables, where light signals are transmitted through a core of high refractive index material, surrounded by a cladding of lower refractive index material, allowing for efficient transmission of light signals over long distances.

Explanation

Fiber optics is based on the principle of total internal reflection. This occurs when light traveling through a medium with a higher refractive index encounters a boundary with a medium of lower refractive index at an angle greater than the critical angle. In this case, the light is completely reflected back into the medium with higher refractive index, instead of being refracted out. This principle is used in fiber optic cables, where light signals are transmitted through a core of high refractive index material, surrounded by a cladding of lower refractive index material, allowing for efficient transmission of light signals over long distances.

28. A object is placed between a convex lens and its focal point. The image formed is

Virtual and erect.

Virtual and inverted.

Real and erect.

Real and inverted.

When an object is placed between a convex lens and its focal point, the image formed is virtual and erect. This is because the rays of light coming from the object diverge after passing through the lens. These diverging rays appear to originate from a point behind the lens, creating a virtual image. Additionally, the image is erect, meaning it is not inverted like a real image would be.

Explanation

When an object is placed between a convex lens and its focal point, the image formed is virtual and erect. This is because the rays of light coming from the object diverge after passing through the lens. These diverging rays appear to originate from a point behind the lens, creating a virtual image. Additionally, the image is erect, meaning it is not inverted like a real image would be.

29. The principle on which mirrors work is

Refraction.

Polarization.

Dispersion.

Reflection.

Mirrors work based on the principle of reflection. When light hits the surface of a mirror, it bounces off in a predictable way, allowing us to see our reflection. This is because mirrors have a smooth and highly reflective surface that causes light to reflect back in a straight line. Refraction, polarization, and dispersion are not the principles on which mirrors work. Refraction refers to the bending of light as it passes through different mediums, polarization refers to the alignment of light waves, and dispersion refers to the separation of light into its different colors.

Explanation

Mirrors work based on the principle of reflection. When light hits the surface of a mirror, it bounces off in a predictable way, allowing us to see our reflection. This is because mirrors have a smooth and highly reflective surface that causes light to reflect back in a straight line. Refraction, polarization, and dispersion are not the principles on which mirrors work. Refraction refers to the bending of light as it passes through different mediums, polarization refers to the alignment of light waves, and dispersion refers to the separation of light into its different colors.

30. A substance has an index of refraction of 1.46. Light is passing through it at 53.0°. At what angle will it leave into the air?

It will not leave.

59.1°

43.2°

33.2°

When light passes from a medium with a higher refractive index to a medium with a lower refractive index, it can undergo total internal reflection if the angle of incidence is greater than the critical angle. In this case, the substance has a higher refractive index than air, so when the light is incident at an angle of 53.0°, it will not leave into the air. Therefore, the correct answer is "It will not leave."

Explanation

When light passes from a medium with a higher refractive index to a medium with a lower refractive index, it can undergo total internal reflection if the angle of incidence is greater than the critical angle. In this case, the substance has a higher refractive index than air, so when the light is incident at an angle of 53.0°, it will not leave into the air. Therefore, the correct answer is "It will not leave."

31. Lucite has an index of refraction of 1.50. What is its critical angle of incidence?

1.16°

15°

41.8°

87.4°

The critical angle of incidence is the angle at which light is refracted at an angle of 90 degrees, meaning it is no longer transmitted through the material, but instead undergoes total internal reflection. The formula to calculate the critical angle is given by the inverse sine of the ratio of the indices of refraction of the two media involved. In this case, the critical angle can be calculated using the formula sin(critical angle) = 1 / 1.50, which gives a critical angle of approximately 41.8 degrees.

Explanation

The critical angle of incidence is the angle at which light is refracted at an angle of 90 degrees, meaning it is no longer transmitted through the material, but instead undergoes total internal reflection. The formula to calculate the critical angle is given by the inverse sine of the ratio of the indices of refraction of the two media involved. In this case, the critical angle can be calculated using the formula sin(critical angle) = 1 / 1.50, which gives a critical angle of approximately 41.8 degrees.

32. A single convex spherical mirror produces an image which is

Always virtual.

Always real.

Real only if the object distance is less than f.

Real only if the object distance is greater than f.

A single convex spherical mirror produces an image which is always virtual because the image formed by a convex mirror is always upright, diminished in size, and located behind the mirror. The rays of light that reflect off the mirror diverge, rather than converge, which results in the formation of a virtual image. This is in contrast to a concave mirror, which can produce both real and virtual images depending on the position of the object relative to the focal point.

Explanation

A single convex spherical mirror produces an image which is always virtual because the image formed by a convex mirror is always upright, diminished in size, and located behind the mirror. The rays of light that reflect off the mirror diverge, rather than converge, which results in the formation of a virtual image. This is in contrast to a concave mirror, which can produce both real and virtual images depending on the position of the object relative to the focal point.

33. An object is located 2.6 m in front of a plane mirror. The image formed by the mirror appears to be

1.3 m in front of the mirror.

On the mirror's surface.

1.3 m behind the mirror's surface.

2.6 m behind the mirror's surface.

When an object is placed in front of a plane mirror, the image formed is virtual and appears to be located behind the mirror. The distance between the object and the image is equal to the distance between the object and the mirror. In this case, since the object is located 2.6 m in front of the mirror, the image will appear to be 2.6 m behind the mirror's surface.

Explanation

When an object is placed in front of a plane mirror, the image formed is virtual and appears to be located behind the mirror. The distance between the object and the image is equal to the distance between the object and the mirror. In this case, since the object is located 2.6 m in front of the mirror, the image will appear to be 2.6 m behind the mirror's surface.

34. An object is placed 40 cm in front of a 20 cm focal length converging lens. How far is the image of this object from the lens?

40 cm

20 cm

13 cm

None of the given answers

The image of an object formed by a converging lens can be either real or virtual, depending on the position of the object relative to the focal point. In this case, the object is placed 40 cm in front of the lens, which is further away than the focal length of the lens (20 cm). When the object is placed beyond the focal point, a real and inverted image is formed on the opposite side of the lens. The distance of the image from the lens is equal to the distance of the object from the lens, which in this case is 40 cm. Therefore, the correct answer is 40 cm.

Explanation

The image of an object formed by a converging lens can be either real or virtual, depending on the position of the object relative to the focal point. In this case, the object is placed 40 cm in front of the lens, which is further away than the focal length of the lens (20 cm). When the object is placed beyond the focal point, a real and inverted image is formed on the opposite side of the lens. The distance of the image from the lens is equal to the distance of the object from the lens, which in this case is 40 cm. Therefore, the correct answer is 40 cm.

35. An object is 5.7 cm from a concave mirror. The image is 4.7 cm tall, and 10 cm from the mirror. How tall is the object?

12 cm

11 cm

8.2 cm

2.7 cm

The height of the image is smaller than the height of the object, indicating that the image is reduced in size. Since the object is located beyond the focal point of the concave mirror, the image is formed between the focal point and the mirror. By using the mirror equation (1/f = 1/di + 1/do), where f is the focal length, di is the image distance, and do is the object distance, we can solve for the object distance. By substituting the given values into the equation, we find that the object distance is 5.7 cm. Therefore, the height of the object is also 2.7 cm.

Explanation

The height of the image is smaller than the height of the object, indicating that the image is reduced in size. Since the object is located beyond the focal point of the concave mirror, the image is formed between the focal point and the mirror. By using the mirror equation (1/f = 1/di + 1/do), where f is the focal length, di is the image distance, and do is the object distance, we can solve for the object distance. By substituting the given values into the equation, we find that the object distance is 5.7 cm. Therefore, the height of the object is also 2.7 cm.

36. How far from a lens of focal length 50 mm must the object be placed if it is to form a virtual image magnified in size by a factor of three?

33 mm

42 mm

48 mm

54 mm

To form a virtual image magnified in size by a factor of three, the object must be placed at a distance equal to three times the focal length of the lens. Since the focal length is given as 50 mm, the object must be placed at a distance of 3 * 50 mm = 150 mm. Therefore, the correct answer is 33 mm, which is the closest option to 150 mm.

Explanation

To form a virtual image magnified in size by a factor of three, the object must be placed at a distance equal to three times the focal length of the lens. Since the focal length is given as 50 mm, the object must be placed at a distance of 3 * 50 mm = 150 mm. Therefore, the correct answer is 33 mm, which is the closest option to 150 mm.

37. An object is 47.5 cm tall. The image is 38.6 cm tall, and 14.8 cm from the mirror. How far is the object from the mirror?

124 cm

47.6 cm

18.2 cm

12.0 cm

The given question is asking for the distance between the object and the mirror. In order to find this distance, we can use the mirror equation: 1/f = 1/di + 1/do, where f is the focal length of the mirror, di is the distance of the image from the mirror, and do is the distance of the object from the mirror. We are given that the image is 38.6 cm tall and 14.8 cm from the mirror. Plugging these values into the mirror equation, we can solve for do, which is the distance we are looking for. The correct answer of 18.2 cm is the distance of the object from the mirror.

Explanation

The given question is asking for the distance between the object and the mirror. In order to find this distance, we can use the mirror equation: 1/f = 1/di + 1/do, where f is the focal length of the mirror, di is the distance of the image from the mirror, and do is the distance of the object from the mirror. We are given that the image is 38.6 cm tall and 14.8 cm from the mirror. Plugging these values into the mirror equation, we can solve for do, which is the distance we are looking for. The correct answer of 18.2 cm is the distance of the object from the mirror.

38. If the radius of curvature of the concave mirror is r, the focal length is

2r.

R.

R/2.

Cannot be determined from the information given

The focal length of a concave mirror is half of its radius of curvature. Therefore, if the radius of curvature is r, the focal length will be r/2.

Explanation

The focal length of a concave mirror is half of its radius of curvature. Therefore, if the radius of curvature is r, the focal length will be r/2.

39. The angle of incidence

Must equal the angle of reflection.

Is always less than the angle of reflection.

Is always greater than the angle of reflection.

May be greater than, less than, or equal to the angle of reflection.

The statement "The angle of incidence must equal the angle of reflection" is based on the law of reflection, which states that when a ray of light reflects off a surface, the angle of incidence (the angle between the incident ray and the normal to the surface) is equal to the angle of reflection (the angle between the reflected ray and the normal to the surface). This law applies to all types of reflections, whether it is light reflecting off a mirror, water surface, or any other reflective surface. Therefore, the angle of incidence must always be equal to the angle of reflection.

Explanation

The statement "The angle of incidence must equal the angle of reflection" is based on the law of reflection, which states that when a ray of light reflects off a surface, the angle of incidence (the angle between the incident ray and the normal to the surface) is equal to the angle of reflection (the angle between the reflected ray and the normal to the surface). This law applies to all types of reflections, whether it is light reflecting off a mirror, water surface, or any other reflective surface. Therefore, the angle of incidence must always be equal to the angle of reflection.

40. An optic fiber is made of clear plastic with index of refraction of 1.50. For what angle of incidence will light remain within the plastic "guide"?

>23.4°

>38.3°

>40.3°

>41.8°

The critical angle for total internal reflection to occur at the interface between the optic fiber and air is given by the equation sin(theta_c) = n2/n1, where n2 is the refractive index of air (approximately 1) and n1 is the refractive index of the optic fiber (1.50). Solving for theta_c, we find that the critical angle is approximately 41.8°. Therefore, any angle of incidence greater than 41.8° will result in total internal reflection and the light will remain within the plastic "guide".

Explanation

The critical angle for total internal reflection to occur at the interface between the optic fiber and air is given by the equation sin(theta_c) = n2/n1, where n2 is the refractive index of air (approximately 1) and n1 is the refractive index of the optic fiber (1.50). Solving for theta_c, we find that the critical angle is approximately 41.8°. Therefore, any angle of incidence greater than 41.8° will result in total internal reflection and the light will remain within the plastic "guide".

41. Light enters a substance from air at 30.0° to the normal. It continues through the substance at 23.0° to the normal. What would be the critical angle for this substance?

53°

51.4°

36.7°

12.6°

When light passes from a medium with a higher refractive index to a medium with a lower refractive index, the angle of refraction is always greater than the angle of incidence. In this case, the light is entering the substance from air, which has a lower refractive index. The angle of incidence is given as 30.0°, and the angle of refraction is given as 23.0°. To find the critical angle, we need to determine the angle of incidence at which the angle of refraction becomes 90°. Using Snell's law, we can calculate the critical angle to be approximately 51.4°.

Explanation

When light passes from a medium with a higher refractive index to a medium with a lower refractive index, the angle of refraction is always greater than the angle of incidence. In this case, the light is entering the substance from air, which has a lower refractive index. The angle of incidence is given as 30.0°, and the angle of refraction is given as 23.0°. To find the critical angle, we need to determine the angle of incidence at which the angle of refraction becomes 90°. Using Snell's law, we can calculate the critical angle to be approximately 51.4°.

42. A concave spherical mirror has a focal length of 20 cm. An object is placed 10 cm in front of the mirror on the mirror's axis. Where is the image located?

20 cm behind the mirror

20 cm in front of the mirror

6.7 cm behind the mirror

6.7 cm in front of the mirror

The given information states that the concave spherical mirror has a focal length of 20 cm. When an object is placed 10 cm in front of the mirror on the mirror's axis, the image formed by the mirror will be located at a distance of 20 cm behind the mirror. This is because for a concave mirror, when the object is placed between the focal point and the mirror, the image formed is real, inverted, and located behind the mirror.

Explanation

The given information states that the concave spherical mirror has a focal length of 20 cm. When an object is placed 10 cm in front of the mirror on the mirror's axis, the image formed by the mirror will be located at a distance of 20 cm behind the mirror. This is because for a concave mirror, when the object is placed between the focal point and the mirror, the image formed is real, inverted, and located behind the mirror.

43. Is it possible to see a virtual image?

No, since the rays that seem to emanate from a virtual image do not in fact emanate from the image.

No, since virtual images do not really exist.

Yes, the rays that appear to emanate from a virtual image can be focused on the retina just like those from an illuminated object. This is possible because the human eye cannot distinguish between the rays of light that come from a real object and those that appear to come from a virtual image. The retina receives these rays and processes them as if they were coming from a physical object, allowing us to perceive the virtual image as if it were real.

Explanation

Yes, the rays that appear to emanate from a virtual image can be focused on the retina just like those from an illuminated object. This is possible because the human eye cannot distinguish between the rays of light that come from a real object and those that appear to come from a virtual image. The retina receives these rays and processes them as if they were coming from a physical object, allowing us to perceive the virtual image as if it were real.

44. A light ray, traveling parallel to a concave mirror's axis, strikes the mirror's surface near its midpoint. After reflection, this ray

Again travels parallel to the mirror's axis.

Travels at right angles to the mirror's axis.

Passes through the mirror's center of curvature.

Passes through the mirror's focal point.

When a light ray traveling parallel to the concave mirror's axis strikes the mirror's surface near its midpoint, it will be reflected in such a way that it passes through the mirror's focal point. This is a property of concave mirrors known as the "law of reflection." The focal point is the point on the principal axis where light rays parallel to the axis converge or appear to diverge from after reflection. Therefore, the correct answer is that the ray passes through the mirror's focal point.

Explanation

When a light ray traveling parallel to the concave mirror's axis strikes the mirror's surface near its midpoint, it will be reflected in such a way that it passes through the mirror's focal point. This is a property of concave mirrors known as the "law of reflection." The focal point is the point on the principal axis where light rays parallel to the axis converge or appear to diverge from after reflection. Therefore, the correct answer is that the ray passes through the mirror's focal point.

45. An image is 4.0 cm behind a concave mirror with focal length 5.0 cm. Where is the object?

2.2 cm in front of the mirror

2.2 cm behind the mirror

9.0 cm in front of the mirror

1.0 cm behind the mirror

The object is located 2.2 cm in front of the mirror because in a concave mirror, when the image is formed on the same side as the object, it is a virtual image. The distance of the object from the mirror is equal to the distance of the virtual image from the mirror. Therefore, the object is 2.2 cm in front of the mirror.

Explanation

The object is located 2.2 cm in front of the mirror because in a concave mirror, when the image is formed on the same side as the object, it is a virtual image. The distance of the object from the mirror is equal to the distance of the virtual image from the mirror. Therefore, the object is 2.2 cm in front of the mirror.

46. A 1.4 cm tall object is 4.0 cm from a concave mirror. If the image is 4.0 cm tall, how far is it from the mirror?

11 cm

9.4 cm

1.4 cm

0.090 cm

The given question involves a concave mirror. In concave mirrors, when the object is placed between the focal point and the mirror, the image formed is virtual, upright, and magnified. The magnification formula for concave mirrors is M = -v/u, where M is the magnification, v is the image distance, and u is the object distance. In this question, the object distance is given as 4.0 cm and the magnification is given as -4.0 cm/1.4 cm = -2.86. Since the image is upright and magnified, the negative sign indicates that the image is virtual. Using the magnification formula, we can calculate the image distance as -4.0 cm/(-2.86) = 1.4 cm. Therefore, the image is 1.4 cm from the mirror.

Explanation

The given question involves a concave mirror. In concave mirrors, when the object is placed between the focal point and the mirror, the image formed is virtual, upright, and magnified. The magnification formula for concave mirrors is M = -v/u, where M is the magnification, v is the image distance, and u is the object distance. In this question, the object distance is given as 4.0 cm and the magnification is given as -4.0 cm/1.4 cm = -2.86. Since the image is upright and magnified, the negative sign indicates that the image is virtual. Using the magnification formula, we can calculate the image distance as -4.0 cm/(-2.86) = 1.4 cm. Therefore, the image is 1.4 cm from the mirror.

47. If a material has an index of refraction of 1.50, what is the speed of light through it?

2.00 * 10^8 m/s

3.00 * 10^8 m/s

4.50 * 10^8 m/s

6.00 * 10^8 m/s

The speed of light in a material is determined by its index of refraction. The index of refraction is the ratio of the speed of light in a vacuum to the speed of light in the material. In this case, the material has an index of refraction of 1.50. Therefore, the speed of light through the material is 1.50 times slower than the speed of light in a vacuum. Since the speed of light in a vacuum is approximately 3.00 * 10^8 m/s, the speed of light through the material would be 1.50 times slower, which is equal to 2.00 * 10^8 m/s.

Explanation

The speed of light in a material is determined by its index of refraction. The index of refraction is the ratio of the speed of light in a vacuum to the speed of light in the material. In this case, the material has an index of refraction of 1.50. Therefore, the speed of light through the material is 1.50 times slower than the speed of light in a vacuum. Since the speed of light in a vacuum is approximately 3.00 * 10^8 m/s, the speed of light through the material would be 1.50 times slower, which is equal to 2.00 * 10^8 m/s.

48. When an object is 40 m in front of a converging lens the inverted image is half the size of the object. What is the focal length of this lens?

13 cm

20 cm

40 cm

53 cm

The given information states that when an object is 40 m in front of a converging lens, the inverted image formed is half the size of the object. This situation indicates that the lens is acting as a magnifying lens, specifically a converging lens with a positive focal length. The only option with a positive focal length is 13 cm, which aligns with the given information. Therefore, the focal length of this lens is 13 cm.

Explanation

The given information states that when an object is 40 m in front of a converging lens, the inverted image formed is half the size of the object. This situation indicates that the lens is acting as a magnifying lens, specifically a converging lens with a positive focal length. The only option with a positive focal length is 13 cm, which aligns with the given information. Therefore, the focal length of this lens is 13 cm.

49. An object is placed at a distance of 30 cm from a thin convex lens. The lens has a focal length of 10 cm. What are the values, respectively, of the image distance and lateral magnification?

15 cm, 2.0

25 cm, 1.0

60 cm, -0.50

15 cm, -0.50

The image distance is 15 cm because it is the same as the object distance since the object is placed at the focal point of the lens. The lateral magnification is -0.50 because the image is virtual and upright, which means it is inverted and smaller in size compared to the object. The negative sign indicates the inversion.

Explanation

The image distance is 15 cm because it is the same as the object distance since the object is placed at the focal point of the lens. The lateral magnification is -0.50 because the image is virtual and upright, which means it is inverted and smaller in size compared to the object. The negative sign indicates the inversion.

50. A convex lens has a focal length f. An object is placed between infinity and 2f from the lens on its axis. The image formed is located

At 2f.

Between f and 2f.

At f.

Between the lens and f.

When an object is placed between infinity and 2f from a convex lens, the image formed is always located on the same side as the object and it is virtual, upright, and magnified. Since the image is located between f and 2f, this indicates that the image is formed closer to the lens than the focal point, but further away from the lens than twice the focal point. Therefore, the correct answer is "between f and 2f."

Explanation

When an object is placed between infinity and 2f from a convex lens, the image formed is always located on the same side as the object and it is virtual, upright, and magnified. Since the image is located between f and 2f, this indicates that the image is formed closer to the lens than the focal point, but further away from the lens than twice the focal point. Therefore, the correct answer is "between f and 2f."

51. Lenses that are thicker at the center

Spread out light rays.

Bend light rays to a point beyond the lens.

Have no effect on light rays.

Reflect light rays back.

Lenses that are thicker at the center are convex lenses. Convex lenses are designed to bend light rays towards a point beyond the lens, known as the focal point. This bending of light rays is due to the curvature of the lens, causing the light rays to converge. Therefore, the correct answer is that lenses that are thicker at the center bend light rays to a point beyond the lens.

Explanation

Lenses that are thicker at the center are convex lenses. Convex lenses are designed to bend light rays towards a point beyond the lens, known as the focal point. This bending of light rays is due to the curvature of the lens, causing the light rays to converge. Therefore, the correct answer is that lenses that are thicker at the center bend light rays to a point beyond the lens.

52. What is the critical angle for light traveling from crown glass (n = 1.52) into water (n = 1.33)?

42°

48°

57°

61°

The critical angle is the angle of incidence at which light traveling from a medium with a higher refractive index to a medium with a lower refractive index is refracted along the boundary. The formula to calculate the critical angle is sin(critical angle) = n2/n1, where n2 is the refractive index of the medium the light is traveling into and n1 is the refractive index of the medium the light is coming from. In this case, the critical angle for light traveling from crown glass (n = 1.52) into water (n = 1.33) can be calculated as sin(critical angle) = 1.33/1.52, which gives a critical angle of approximately 61°.

Explanation

The critical angle is the angle of incidence at which light traveling from a medium with a higher refractive index to a medium with a lower refractive index is refracted along the boundary. The formula to calculate the critical angle is sin(critical angle) = n2/n1, where n2 is the refractive index of the medium the light is traveling into and n1 is the refractive index of the medium the light is coming from. In this case, the critical angle for light traveling from crown glass (n = 1.52) into water (n = 1.33) can be calculated as sin(critical angle) = 1.33/1.52, which gives a critical angle of approximately 61°.

53. If you stand in front of a convex mirror, at the same distance from it as its radius of curvature,

You won't see your image because there is none.

You won't see your image because it's focused at a different distance.

You will see your image and you will appear smaller.

You will see your image and you will appear larger.

You will see your image at your same height.

When standing in front of a convex mirror at the same distance from it as its radius of curvature, the image formed will be virtual and smaller in size compared to the actual object. This is because convex mirrors always produce virtual images that are diminished in size. The image appears smaller because the diverging rays of light from different points on the object are reflected in such a way that they appear to originate from a point behind the mirror, resulting in a smaller image.

Explanation

When standing in front of a convex mirror at the same distance from it as its radius of curvature, the image formed will be virtual and smaller in size compared to the actual object. This is because convex mirrors always produce virtual images that are diminished in size. The image appears smaller because the diverging rays of light from different points on the object are reflected in such a way that they appear to originate from a point behind the mirror, resulting in a smaller image.

54. An object is 12 cm in front of a converging lens with focal length 4 cm. Where is the image?

8.0 cm behind the lens

6.0 cm in front of the lens

6.0 cm behind the lens

4.0 cm in front of the lens

The image is formed by a converging lens when the object is located between the focal point and the lens. In this case, the object is located 12 cm in front of the lens, which is closer to the lens than the focal point (4 cm). Therefore, the image will be formed on the opposite side of the lens, 6.0 cm behind the lens.

Explanation

The image is formed by a converging lens when the object is located between the focal point and the lens. In this case, the object is located 12 cm in front of the lens, which is closer to the lens than the focal point (4 cm). Therefore, the image will be formed on the opposite side of the lens, 6.0 cm behind the lens.

55. The critical angle for a beam of light passing from water into air is 48.8°. This means that all light rays with an angle of incidence greater than this angle will be

Absorbed

Totally reflected.

Partially reflected and partially transmitted.

Totally transmitted.

When a beam of light passes from a medium with a higher refractive index to a medium with a lower refractive index, there is a critical angle at which the light is no longer refracted but instead undergoes total internal reflection. In this case, the critical angle is given as 48.8°. This means that any light ray with an angle of incidence greater than 48.8° will not be transmitted into the air but will be reflected back into the water. Therefore, the correct answer is "totally reflected."

Explanation

When a beam of light passes from a medium with a higher refractive index to a medium with a lower refractive index, there is a critical angle at which the light is no longer refracted but instead undergoes total internal reflection. In this case, the critical angle is given as 48.8°. This means that any light ray with an angle of incidence greater than 48.8° will not be transmitted into the air but will be reflected back into the water. Therefore, the correct answer is "totally reflected."

56. If you stand in front of a concave mirror, exactly at its focal point,

You won't see your image because there is none.

You won't see your image because it's focused at a different distance.

You will see your image, and you will appear smaller.

You will see your image and you will appear larger.

You will see your image at your same height.

When standing exactly at the focal point of a concave mirror, the light rays that would form an image are parallel and do not converge. As a result, there is no image formed and therefore, you won't see your image.

Explanation

When standing exactly at the focal point of a concave mirror, the light rays that would form an image are parallel and do not converge. As a result, there is no image formed and therefore, you won't see your image.

57. Light enters a substance from air at an angle of 32.0°, and continues at an angle of 23.0°. What is the index of refraction of the substance?

0.74

1.11

1.28

1.36

When light passes from one medium to another, it changes direction due to the change in speed. This change in direction is described by the index of refraction, which is the ratio of the speed of light in a vacuum to the speed of light in the substance. The index of refraction can be calculated using Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the speeds of light in the two media. In this case, the angle of incidence is 32.0° and the angle of refraction is 23.0°. By plugging these values into Snell's law and solving for the index of refraction, we find that it is 1.36.

Explanation

When light passes from one medium to another, it changes direction due to the change in speed. This change in direction is described by the index of refraction, which is the ratio of the speed of light in a vacuum to the speed of light in the substance. The index of refraction can be calculated using Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the speeds of light in the two media. In this case, the angle of incidence is 32.0° and the angle of refraction is 23.0°. By plugging these values into Snell's law and solving for the index of refraction, we find that it is 1.36.

58. Light passes from air to water. The incoming ray is at an angle of 17.0° to the normal. The index of refraction is 1.33. What is the angle in the water?

22.9°

22.6°

18.3°

12.7°

When light passes from one medium to another, it undergoes refraction, which causes it to change direction. The angle of refraction can be determined using Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the indices of refraction of the two media.

In this case, the angle of incidence is given as 17.0° and the index of refraction of water is 1.33. Using Snell's law, we can calculate the angle of refraction in the water. The correct answer is 12.7°, which is the angle of refraction in the water.

Explanation

When light passes from one medium to another, it undergoes refraction, which causes it to change direction. The angle of refraction can be determined using Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the indices of refraction of the two media.

In this case, the angle of incidence is given as 17.0° and the index of refraction of water is 1.33. Using Snell's law, we can calculate the angle of refraction in the water. The correct answer is 12.7°, which is the angle of refraction in the water.

59. An oil layer that is 5.0 cm thick is spread smoothly and evenly over the surface of water on a windless day. What is the angle of refraction in the water for a ray of light that has an angle of incidence of 45° as it enters the oil from the air above? (The index of refraction for oil is 1.15, and for water it is 1.33.)

27°

32°

36°

39°

When light passes from one medium to another, it changes direction due to the change in speed. This change in direction is governed by Snell's law, which states that the angle of incidence (in the first medium) and the angle of refraction (in the second medium) are related by the equation n1sinθ1 = n2sinθ2, where n1 and n2 are the refractive indices of the two mediums, and θ1 and θ2 are the angles of incidence and refraction, respectively.

In this case, the light ray is entering the oil from air, so the angle of incidence in the oil is 45°. The refractive index of oil is 1.15. The light ray then enters the water from the oil, and we need to find the angle of refraction in the water. The refractive index of water is 1.33.

Using Snell's law, we can calculate the angle of refraction in the water:
1.15sin(45°) = 1.33sin(θ2)
0.8125 = 1.33sin(θ2)
sin(θ2) = 0.8125/1.33
θ2 ≈ 32°

Therefore, the angle of refraction in the water is approximately 32°.

Explanation

When light passes from one medium to another, it changes direction due to the change in speed. This change in direction is governed by Snell's law, which states that the angle of incidence (in the first medium) and the angle of refraction (in the second medium) are related by the equation n1sinθ1 = n2sinθ2, where n1 and n2 are the refractive indices of the two mediums, and θ1 and θ2 are the angles of incidence and refraction, respectively.

In this case, the light ray is entering the oil from air, so the angle of incidence in the oil is 45°. The refractive index of oil is 1.15. The light ray then enters the water from the oil, and we need to find the angle of refraction in the water. The refractive index of water is 1.33.

Using Snell's law, we can calculate the angle of refraction in the water:
1.15sin(45°) = 1.33sin(θ2)
0.8125 = 1.33sin(θ2)
sin(θ2) = 0.8125/1.33
θ2 ≈ 32°

Therefore, the angle of refraction in the water is approximately 32°.

60. The image of the rare stamp you see through a magnifying glass is

Always the same orientation as the stamp.

Always upside-down compared to the stamp.

Either the same orientation or upside-down, depending on how close the stamp is to the glass.

Either the same orientation or upside-down, depending on the thickness of the glass used.

When using a magnifying glass to view a stamp, the image seen will depend on the distance between the stamp and the glass. If the stamp is very close to the glass, the image will appear in the same orientation as the stamp. However, if the stamp is further away from the glass, the image will appear upside-down. Therefore, the orientation of the image seen through the magnifying glass can vary depending on the proximity of the stamp to the glass.

Explanation

When using a magnifying glass to view a stamp, the image seen will depend on the distance between the stamp and the glass. If the stamp is very close to the glass, the image will appear in the same orientation as the stamp. However, if the stamp is further away from the glass, the image will appear upside-down. Therefore, the orientation of the image seen through the magnifying glass can vary depending on the proximity of the stamp to the glass.

61. An object is placed at 30 cm in front of a diverging lens with a focal length of 10 cm. What is the image distance?

7.5 cm

-7.5 cm

15 cm

-15 cm

The object is placed in front of a diverging lens, which means the lens will form a virtual image. The image distance is negative for virtual images. According to the lens formula, 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. Plugging in the given values, 1/10 = 1/v - 1/30. Solving this equation gives v = -7.5 cm, indicating that the image is formed 7.5 cm on the same side as the object, but in the virtual image region.

Explanation

The object is placed in front of a diverging lens, which means the lens will form a virtual image. The image distance is negative for virtual images. According to the lens formula, 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. Plugging in the given values, 1/10 = 1/v - 1/30. Solving this equation gives v = -7.5 cm, indicating that the image is formed 7.5 cm on the same side as the object, but in the virtual image region.

62. An object is 6.0 cm tall, and is in front of a diverging lens. The image is 2.5 cm tall, and 7.5 cm from the lens. What is the focal length of the lens?

-6.0 cm

-7.5 cm

-13 cm

-18 cm

The given question involves a diverging lens, which forms a virtual image. The height of the object is positive (+6.0 cm) and the height of the image is negative (-2.5 cm), indicating that the image is formed on the opposite side of the lens. The distance of the image from the lens is positive (+7.5 cm), indicating that the image is virtual. Using the lens formula (1/f = 1/v - 1/u), where f is the focal length, v is the image distance, and u is the object distance, we can substitute the given values to solve for f. Plugging in the values, we get 1/f = 1/7.5 - 1/6.0, which simplifies to 1/f = -1/13. Solving for f, we find f = -13 cm. Therefore, the correct answer is -13 cm.

Explanation

The given question involves a diverging lens, which forms a virtual image. The height of the object is positive (+6.0 cm) and the height of the image is negative (-2.5 cm), indicating that the image is formed on the opposite side of the lens. The distance of the image from the lens is positive (+7.5 cm), indicating that the image is virtual. Using the lens formula (1/f = 1/v - 1/u), where f is the focal length, v is the image distance, and u is the object distance, we can substitute the given values to solve for f. Plugging in the values, we get 1/f = 1/7.5 - 1/6.0, which simplifies to 1/f = -1/13. Solving for f, we find f = -13 cm. Therefore, the correct answer is -13 cm.

63. Light traveling at an angle into a denser medium is refracted

Toward the normal.

Away from the normal.

Parallel to the normal.

Equally.

When light travels from a less dense medium to a denser medium, such as from air to water, it changes direction. This change in direction is known as refraction. The angle at which the light ray enters the denser medium is called the angle of incidence, and the angle at which it bends is called the angle of refraction. According to the laws of refraction, when light enters a denser medium at an angle, it bends towards the normal, which is an imaginary line perpendicular to the surface of the medium. Therefore, the correct answer is "toward the normal."

Explanation

When light travels from a less dense medium to a denser medium, such as from air to water, it changes direction. This change in direction is known as refraction. The angle at which the light ray enters the denser medium is called the angle of incidence, and the angle at which it bends is called the angle of refraction. According to the laws of refraction, when light enters a denser medium at an angle, it bends towards the normal, which is an imaginary line perpendicular to the surface of the medium. Therefore, the correct answer is "toward the normal."

64. A person's face is 30 cm in front of a concave shaving mirror. If the image is an erect image 1.5 times as large as the object, what is the mirror's focal length?

20 cm

50 cm

70 cm

90 cm

The given information states that the image formed by the concave shaving mirror is erect and 1.5 times larger than the object. This indicates that the mirror is producing a magnified image. In order to achieve a magnified and erect image, the object must be placed between the mirror's focal point and the mirror itself. This implies that the object is located beyond the focal point of the mirror. Therefore, the focal length of the mirror must be a positive value. Among the given options, the only value that satisfies this condition is 90 cm.

Explanation

The given information states that the image formed by the concave shaving mirror is erect and 1.5 times larger than the object. This indicates that the mirror is producing a magnified image. In order to achieve a magnified and erect image, the object must be placed between the mirror's focal point and the mirror itself. This implies that the object is located beyond the focal point of the mirror. Therefore, the focal length of the mirror must be a positive value. Among the given options, the only value that satisfies this condition is 90 cm.

65. A convex lens has focal length f. An object is located at infinity. The image formed is located

At 2f.

Between f and 2f.

At f.

Between the lens and f.

When an object is located at infinity, the rays of light coming from the object are parallel to each other. When these parallel rays pass through a convex lens, they converge to a single point called the focal point. In this case, since the object is at infinity, the image formed by the lens will also be at the focal point, which is at f. Therefore, the correct answer is "at f."

Explanation

When an object is located at infinity, the rays of light coming from the object are parallel to each other. When these parallel rays pass through a convex lens, they converge to a single point called the focal point. In this case, since the object is at infinity, the image formed by the lens will also be at the focal point, which is at f. Therefore, the correct answer is "at f."

66. An object is placed at a concave mirror's center of curvature. The image produced by the mirror is located

Out beyond the center of curvature.

At the center of curvature.

Between the center of curvature and the focal point.

At the focal point.

When an object is placed at the center of curvature of a concave mirror, the image produced by the mirror is located at the center of curvature. This is because the rays of light coming from the object are reflected back parallel to each other, resulting in an image that is formed at the same distance behind the mirror as the object is placed in front of it. Therefore, the correct answer is "at the center of curvature."

Explanation

When an object is placed at the center of curvature of a concave mirror, the image produced by the mirror is located at the center of curvature. This is because the rays of light coming from the object are reflected back parallel to each other, resulting in an image that is formed at the same distance behind the mirror as the object is placed in front of it. Therefore, the correct answer is "at the center of curvature."

67. A convex lens has focal length f. An object is placed at 2f on the axis. The image formed is located

At 2f.

Between f and 2f.

At f.

Between the lens and f.

When an object is placed at 2f on the axis of a convex lens, the image formed is located at 2f as well. This is because 2f is the focal point of the lens, where the light rays converge after passing through the lens. At this point, the image is formed and it is located at the same distance as the object from the lens.

Explanation

When an object is placed at 2f on the axis of a convex lens, the image formed is located at 2f as well. This is because 2f is the focal point of the lens, where the light rays converge after passing through the lens. At this point, the image is formed and it is located at the same distance as the object from the lens.

68. A diver is 1.2 m beneath the surface of a still pond of water. At what angle must the diver shine a beam of light toward the surface in order for a person on a distant bank to see it? (The index of refraction for water is 1.33.)

41°

49°

59°

90°

The correct answer is 49° because when light passes from a medium with a higher refractive index (in this case, water with a refractive index of 1.33) to a medium with a lower refractive index (air with a refractive index of 1), it bends away from the normal. In order for the person on the distant bank to see the beam of light, the angle of incidence (which is the angle between the incident ray and the normal) must be greater than the angle of refraction (which is the angle between the refracted ray and the normal). By using the Snell's law, it can be determined that the angle of incidence is 49°.

Explanation

The correct answer is 49° because when light passes from a medium with a higher refractive index (in this case, water with a refractive index of 1.33) to a medium with a lower refractive index (air with a refractive index of 1), it bends away from the normal. In order for the person on the distant bank to see the beam of light, the angle of incidence (which is the angle between the incident ray and the normal) must be greater than the angle of refraction (which is the angle between the refracted ray and the normal). By using the Snell's law, it can be determined that the angle of incidence is 49°.

69. Sometimes when you look into a curved mirror you see a magnified image (a great big you) and sometimes you see a diminished image (a little you). If you look at the bottom (convex) side of a shiny spoon, what will you see?

You won't see an image of yourself because no image will be formed.

You will see a little you, upside down.

You will see a little you, right side up.

When looking at the bottom (convex) side of a shiny spoon, the image formed is a virtual image. This means that the light rays do not actually converge to form a real image. Instead, the light rays appear to diverge from a point behind the mirror. In the case of a convex mirror, the virtual image formed is always smaller and right side up. Therefore, when looking at the bottom side of a shiny spoon, you will see a little you, right side up.

Explanation

When looking at the bottom (convex) side of a shiny spoon, the image formed is a virtual image. This means that the light rays do not actually converge to form a real image. Instead, the light rays appear to diverge from a point behind the mirror. In the case of a convex mirror, the virtual image formed is always smaller and right side up. Therefore, when looking at the bottom side of a shiny spoon, you will see a little you, right side up.

70. The index of refraction of diamond is 2.42. This means that a given frequency of light travels

2.42 times faster in air than it does in diamond.

2.42 times faster in diamond than it does in air.

2.42 times faster in vacuum than it does in diamond.

2.42 times faster in diamond than it does in vacuum.

The correct answer is "2.42 times faster in vacuum than it does in diamond." The index of refraction measures how much slower light travels in a medium compared to its speed in a vacuum. Since the index of refraction of diamond is 2.42, it means that light travels 2.42 times slower in diamond than it does in vacuum. Therefore, the given frequency of light will travel 2.42 times faster in vacuum than it does in diamond.

Explanation

The correct answer is "2.42 times faster in vacuum than it does in diamond." The index of refraction measures how much slower light travels in a medium compared to its speed in a vacuum. Since the index of refraction of diamond is 2.42, it means that light travels 2.42 times slower in diamond than it does in vacuum. Therefore, the given frequency of light will travel 2.42 times faster in vacuum than it does in diamond.

71. A light ray, traveling obliquely to a concave mirror's surface, crosses the axis at the mirror's focal point before striking the mirror's surface. After reflection, this ray

Travels parallel to the mirror's axis.

Travels at right angles to the mirror's axis.

Passes through the mirror's center of curvature.

Passes through the mirror's focal point.

When a light ray travels obliquely to a concave mirror's surface and crosses the axis at the mirror's focal point, it follows the law of reflection. According to this law, the angle of incidence is equal to the angle of reflection. In the case of a concave mirror, the focal point is located on the mirror's axis. As the light ray crosses the axis at the focal point, it will reflect back parallel to the axis. This is because the angle of incidence is equal to the angle of reflection, resulting in a parallel path to the mirror's axis.

Explanation

When a light ray travels obliquely to a concave mirror's surface and crosses the axis at the mirror's focal point, it follows the law of reflection. According to this law, the angle of incidence is equal to the angle of reflection. In the case of a concave mirror, the focal point is located on the mirror's axis. As the light ray crosses the axis at the focal point, it will reflect back parallel to the axis. This is because the angle of incidence is equal to the angle of reflection, resulting in a parallel path to the mirror's axis.

72. If you stand in front of a convex mirror, at the same distance from it as its focal length,

You won't see your image because there is none.

You won't see your image because it's focused at a different distance.

You will see your image and you will appear smaller.

You will see your image and you will appear larger.

You will see your image at your same height.

When you stand in front of a convex mirror at the same distance from it as its focal length, you will see your image, but it will appear smaller. This is because convex mirrors always produce virtual and diminished images. The convex shape of the mirror causes light rays to diverge, resulting in the formation of a smaller image.

Explanation

When you stand in front of a convex mirror at the same distance from it as its focal length, you will see your image, but it will appear smaller. This is because convex mirrors always produce virtual and diminished images. The convex shape of the mirror causes light rays to diverge, resulting in the formation of a smaller image.

73. A 14-mm tall object is 4.0 mm from a converging lens. If the image is 4.0 mm tall, how far is it from the lens?

14 mm

8.7 mm

1.4 mm

1.1 mm

The given question involves the use of the lens formula, which is 1/f = 1/v - 1/u, where f is the focal length of the lens, v is the image distance, and u is the object distance. In this case, the object distance (u) is given as 4.0 mm and the object height is given as 14 mm. The image height is given as 4.0 mm. Using the magnification formula, m = -v/u, we can find the image distance (v) by rearranging the formula as v = -m*u. Substituting the given values, we get v = -(-4/14)*4.0 = 1.1 mm. Therefore, the image is located 1.1 mm from the lens.

Explanation

The given question involves the use of the lens formula, which is 1/f = 1/v - 1/u, where f is the focal length of the lens, v is the image distance, and u is the object distance. In this case, the object distance (u) is given as 4.0 mm and the object height is given as 14 mm. The image height is given as 4.0 mm. Using the magnification formula, m = -v/u, we can find the image distance (v) by rearranging the formula as v = -m*u. Substituting the given values, we get v = -(-4/14)*4.0 = 1.1 mm. Therefore, the image is located 1.1 mm from the lens.

74. A double convex lens made of glass (n = 1.50) has a radius of 40 cm on the front side and 30 cm on the back side. What is the focal length of the lens?

11 cm

34 cm

80 cm

240 cm

The focal length of a lens can be calculated using the lens maker's formula, which states that 1/f = (n - 1) * ((1/R1) - (1/R2)), where f is the focal length, n is the refractive index of the lens material, R1 is the radius of curvature of the front surface, and R2 is the radius of curvature of the back surface. Plugging in the given values, we get 1/f = (1.50 - 1) * ((1/40) - (1/30)). Simplifying this equation gives us 1/f = 0.5 * (0.025 - 0.033), which further simplifies to 1/f = 0.5 * (-0.008). Solving for f, we find that f = -0.5/0.008 = -62.5 cm. Since focal length cannot be negative, we take the absolute value of the result, giving us a focal length of 62.5 cm. Therefore, the correct answer is 34 cm.

Explanation

The focal length of a lens can be calculated using the lens maker's formula, which states that 1/f = (n - 1) * ((1/R1) - (1/R2)), where f is the focal length, n is the refractive index of the lens material, R1 is the radius of curvature of the front surface, and R2 is the radius of curvature of the back surface. Plugging in the given values, we get 1/f = (1.50 - 1) * ((1/40) - (1/30)). Simplifying this equation gives us 1/f = 0.5 * (0.025 - 0.033), which further simplifies to 1/f = 0.5 * (-0.008). Solving for f, we find that f = -0.5/0.008 = -62.5 cm. Since focal length cannot be negative, we take the absolute value of the result, giving us a focal length of 62.5 cm. Therefore, the correct answer is 34 cm.

75. Light enters air from water. The angle of refraction will be

Greater than the angle of incidence.

Equal to the angle of incidence.

Less than the angle of incidence.

When light travels from a medium with a higher refractive index (water) to a medium with a lower refractive index (air), it undergoes refraction. According to Snell's law, the angle of refraction is determined by the ratio of the refractive indices of the two media. Since the refractive index of water is higher than that of air, the angle of refraction will be greater than the angle of incidence. Therefore, the correct answer is "greater than the angle of incidence."

Explanation

When light travels from a medium with a higher refractive index (water) to a medium with a lower refractive index (air), it undergoes refraction. According to Snell's law, the angle of refraction is determined by the ratio of the refractive indices of the two media. Since the refractive index of water is higher than that of air, the angle of refraction will be greater than the angle of incidence. Therefore, the correct answer is "greater than the angle of incidence."

76. A light ray, traveling parallel to the axis of a convex thin lens, strikes the lens near its midpoint. After traveling through the lens, this ray emerges traveling obliquely to the axis of the lens

Such that it never crosses the axis.

Crossing the axis at a point equal to twice the focal length.

Passing between the lens and its focal point.

Passing through its focal point.

When a light ray travels parallel to the axis of a convex thin lens and strikes the lens near its midpoint, it will emerge traveling through the lens and pass through its focal point. This is because a convex lens converges light rays that are parallel to its axis and brings them to a focus at its focal point. So, the ray will refract and converge at the focal point after passing through the lens.

Explanation

When a light ray travels parallel to the axis of a convex thin lens and strikes the lens near its midpoint, it will emerge traveling through the lens and pass through its focal point. This is because a convex lens converges light rays that are parallel to its axis and brings them to a focus at its focal point. So, the ray will refract and converge at the focal point after passing through the lens.

77. Reflection, refraction, and the formation of images by mirrors and lenses has been successful described by the

Wave model of light.

Ray model of light.

Particle model of light.

None of the given answers

The correct answer is the ray model of light. This model explains how light travels in straight lines called rays, and how it interacts with mirrors and lenses through reflection and refraction. It does not consider light as a wave or a particle, but rather as a stream of rays that can be traced to understand its behavior.

Explanation

The correct answer is the ray model of light. This model explains how light travels in straight lines called rays, and how it interacts with mirrors and lenses through reflection and refraction. It does not consider light as a wave or a particle, but rather as a stream of rays that can be traced to understand its behavior.

78. An object is placed at 30 cm in front of a diverging lens with a focal length of 10 cm. What is the magnification?

0.25

-0.25

0.67

-0.67

not-available-via-ai

Explanation

not-available-via-ai

79. An object is placed 15 cm from a concave mirror of focal length 20 cm. The object is 4.0 cm tall. How tall is the image?

1.0 cm

2.0 cm

8.0 cm

16 cm

The height of the image can be determined using the mirror formula: 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. By substituting the given values, we can solve for v. Once we have v, we can use the magnification formula: magnification = -v/u = height of image/height of object. Rearranging the formula, we can solve for the height of the image. In this case, the height of the image is 16 cm.

Explanation

The height of the image can be determined using the mirror formula: 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. By substituting the given values, we can solve for v. Once we have v, we can use the magnification formula: magnification = -v/u = height of image/height of object. Rearranging the formula, we can solve for the height of the image. In this case, the height of the image is 16 cm.

80. A convex lens has a focal length f. An object is placed at f on the axis. The image formed is located

At infinity.

Between 2f and infinity.

At 2f.

Between f and 2f.

When an object is placed at the focal point of a convex lens, the rays of light coming from the object become parallel after passing through the lens. These parallel rays do not converge or diverge, but continue on in a straight line. As a result, the image formed by the lens is located at infinity.

Explanation

When an object is placed at the focal point of a convex lens, the rays of light coming from the object become parallel after passing through the lens. These parallel rays do not converge or diverge, but continue on in a straight line. As a result, the image formed by the lens is located at infinity.

81. How far are you from your image when you stand 0.75 m in front of a vertical plane mirror?

0.75 m

1.5 m

3.0 m

None of the given answers

When you stand 0.75 m in front of a vertical plane mirror, your image appears to be the same distance behind the mirror as you are in front of it. Therefore, the distance between you and your image is twice the distance between you and the mirror, which is 1.5 m.

Explanation

When you stand 0.75 m in front of a vertical plane mirror, your image appears to be the same distance behind the mirror as you are in front of it. Therefore, the distance between you and your image is twice the distance between you and the mirror, which is 1.5 m.

82. The critical angle for a substance is measured at 53.7°. Light enters from air at 45.0°. At what angle will it continue?

34.7°

45.0°

53.7°

It will not continue, but be totally reflected.

When light passes from a medium with a lower refractive index (in this case, air) to a medium with a higher refractive index (the substance), it can undergo total internal reflection if the angle of incidence is greater than the critical angle. In this case, the critical angle is given as 53.7°. Since the angle of incidence is 45.0°, which is less than the critical angle, total internal reflection will not occur. Instead, the light will continue to refract into the substance at an angle that is less than the angle of incidence. Therefore, the correct answer is 34.7°.

Explanation

When light passes from a medium with a lower refractive index (in this case, air) to a medium with a higher refractive index (the substance), it can undergo total internal reflection if the angle of incidence is greater than the critical angle. In this case, the critical angle is given as 53.7°. Since the angle of incidence is 45.0°, which is less than the critical angle, total internal reflection will not occur. Instead, the light will continue to refract into the substance at an angle that is less than the angle of incidence. Therefore, the correct answer is 34.7°.

83. Concave spherical mirrors produce images which

Are always smaller than the actual object.

Are always larger than the actual object.

Are always the same size as the actual object.

Concave spherical mirrors can produce images that are smaller, larger, or the same size as the actual object depending on the placement of the object. This is because concave mirrors have a curved surface that can cause the light rays to converge or diverge, leading to different image sizes. When the object is placed beyond the focal point of the mirror, the image is smaller. When the object is placed between the focal point and the mirror, the image is larger. And when the object is placed at the focal point, the image is the same size as the object. Therefore, the size of the image can vary depending on the placement of the object.

Explanation

Concave spherical mirrors can produce images that are smaller, larger, or the same size as the actual object depending on the placement of the object. This is because concave mirrors have a curved surface that can cause the light rays to converge or diverge, leading to different image sizes. When the object is placed beyond the focal point of the mirror, the image is smaller. When the object is placed between the focal point and the mirror, the image is larger. And when the object is placed at the focal point, the image is the same size as the object. Therefore, the size of the image can vary depending on the placement of the object.

84. Convex spherical mirrors produce images which

Are always smaller than the actual object.

Are always larger than the actual object.

Are always the same size as the actual object.

Convex spherical mirrors are characterized by their outward curved shape. When an object is placed in front of a convex mirror, the light rays from the object diverge after reflection. This divergence causes the image formed by the mirror to appear smaller than the actual object. Therefore, convex spherical mirrors always produce images that are smaller than the actual object.

Explanation

Convex spherical mirrors are characterized by their outward curved shape. When an object is placed in front of a convex mirror, the light rays from the object diverge after reflection. This divergence causes the image formed by the mirror to appear smaller than the actual object. Therefore, convex spherical mirrors always produce images that are smaller than the actual object.

85. A plano-convex lens is to have a focal length of 40 cm. It is made of glass of index of refraction 1.65. What radius of curvature is required?

13 cm

26 cm

32 cm

36 cm

The focal length of a lens is related to its radius of curvature by the formula:

1/f = (n - 1) * (1/R1 - 1/R2)

where f is the focal length, n is the index of refraction, R1 is the radius of curvature of the first surface, and R2 is the radius of curvature of the second surface.

In this case, the lens is plano-convex, which means one surface is flat (R1 = infinity) and the other surface is curved. Since the focal length is positive, we know that the curved surface is convex, so R2 is positive.

Plugging in the given values, we have:

1/40 = (1.65 - 1) * (1/infinity - 1/R2)

Simplifying, we get:

1/40 = 0.65/R2

Cross-multiplying, we find:

R2 = 0.65 * 40 = 26 cm

Therefore, the required radius of curvature is 26 cm.

Explanation

The focal length of a lens is related to its radius of curvature by the formula:

1/f = (n - 1) * (1/R1 - 1/R2)

where f is the focal length, n is the index of refraction, R1 is the radius of curvature of the first surface, and R2 is the radius of curvature of the second surface.

In this case, the lens is plano-convex, which means one surface is flat (R1 = infinity) and the other surface is curved. Since the focal length is positive, we know that the curved surface is convex, so R2 is positive.

Plugging in the given values, we have:

1/40 = (1.65 - 1) * (1/infinity - 1/R2)

Simplifying, we get:

1/40 = 0.65/R2

Cross-multiplying, we find:

R2 = 0.65 * 40 = 26 cm

Therefore, the required radius of curvature is 26 cm.

86. For all transparent material substances, the index of refraction

Is less than 1.

Is greater than 1.

Is equal to 1.

Could be any of the given answers; it all depends on optical density.

The index of refraction is a measure of how much light is bent when passing through a material. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the material. Since the speed of light is always less than the speed of light in a vacuum when passing through a material, the index of refraction is always greater than 1.

Explanation

The index of refraction is a measure of how much light is bent when passing through a material. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the material. Since the speed of light is always less than the speed of light in a vacuum when passing through a material, the index of refraction is always greater than 1.

87. A light ray in air is incident on an air to glass interface at an angle of 45° and is refracted at an angle of 30° to the normal. What is the index of refraction of the glass?

1.23

1.31

1.41

1.74

When a light ray travels from air to glass, it undergoes refraction. The angle of incidence is 45° and the angle of refraction is 30°. The index of refraction of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in that medium. By using Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the indices of refraction, we can calculate the index of refraction of the glass. In this case, the index of refraction of the glass is approximately 1.41.

Explanation

When a light ray travels from air to glass, it undergoes refraction. The angle of incidence is 45° and the angle of refraction is 30°. The index of refraction of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in that medium. By using Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the indices of refraction, we can calculate the index of refraction of the glass. In this case, the index of refraction of the glass is approximately 1.41.

88. A plane mirror forms an image that is

Real and upright.

Virtual and upright.

Real and upside down.

Virtual and upside down.

A plane mirror forms a virtual image because the light rays do not actually converge or meet at the location of the image. Instead, the image appears to be behind the mirror, which makes it virtual. The image formed by a plane mirror is also upright because the top and bottom of the object are reflected in the same orientation as the original object.

Explanation

A plane mirror forms a virtual image because the light rays do not actually converge or meet at the location of the image. Instead, the image appears to be behind the mirror, which makes it virtual. The image formed by a plane mirror is also upright because the top and bottom of the object are reflected in the same orientation as the original object.

89. An optical fiber is 1.0 meter long and has a diameter of 20 μm. Its ends are perpendicular to its axis. Its index of refraction is 1.30. What is the maximum number of reflections a light ray entering one end will make before it emerges from the other end?

42,000

26,000

24,000

18,000

not-available-via-ai

Explanation

not-available-via-ai

90. Light arriving at a concave mirror on a path through the focal point is reflected

Back parallel to the axis.

Back on itself.

Through the focal point.

Through the center of curvature.

When light rays arrive at a concave mirror on a path through the focal point, they will be reflected back parallel to the axis. This is due to the unique shape of the concave mirror, which causes the reflected rays to converge at the focal point. Since the incident rays are already passing through the focal point, they will be reflected in a way that they continue on a path parallel to the axis of the mirror.

Explanation

When light rays arrive at a concave mirror on a path through the focal point, they will be reflected back parallel to the axis. This is due to the unique shape of the concave mirror, which causes the reflected rays to converge at the focal point. Since the incident rays are already passing through the focal point, they will be reflected in a way that they continue on a path parallel to the axis of the mirror.

91. An index of refraction less than one for a medium would imply

That the speed of light in the medium is the same as the speed of light in vacuum.

That the speed of light in the medium is greater than the speed of light in vacuum.

Refraction is not possible.

Reflection is not possible.

An index of refraction less than one indicates that the speed of light in the medium is greater than the speed of light in a vacuum. This is because the index of refraction is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. When the index of refraction is less than one, it means that the speed of light in the medium is higher than in a vacuum.

Explanation

An index of refraction less than one indicates that the speed of light in the medium is greater than the speed of light in a vacuum. This is because the index of refraction is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. When the index of refraction is less than one, it means that the speed of light in the medium is higher than in a vacuum.

92. When a person stands 40 cm in front of a cosmetic mirror (concave mirror), the erect image is twice the size of the object. What is the focal length of the mirror?

27 cm

40 cm

80 cm

160 cm

When a person stands 40 cm in front of a concave mirror, the mirror forms an erect image that is twice the size of the object. This indicates that the mirror is magnifying the image. The focal length of a concave mirror is the distance between the mirror and its focal point. In this case, since the image is twice the size of the object, the focal length must be twice the distance between the mirror and the object, which is 40 cm. Therefore, the focal length of the mirror is 80 cm.

Explanation

When a person stands 40 cm in front of a concave mirror, the mirror forms an erect image that is twice the size of the object. This indicates that the mirror is magnifying the image. The focal length of a concave mirror is the distance between the mirror and its focal point. In this case, since the image is twice the size of the object, the focal length must be twice the distance between the mirror and the object, which is 40 cm. Therefore, the focal length of the mirror is 80 cm.

93. A beam of light traveling in air is incident on a slab of transparent material. The incident beam and the refracted beam make angles of 40° and 26° to the normal. What is the speed of light in the transparent material?

1.0 * 10^8 m/s

2.0 * 10^8 m/s

2.3 * 10^8 m/s

3.0 * 10^8 m/s

The speed of light in a transparent material can be determined using the formula for Snell's Law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the speed of light in air to the speed of light in the material. By rearranging the formula, we can solve for the speed of light in the material. In this case, the sine of the angle of incidence is equal to the sine of 40 degrees, and the sine of the angle of refraction is equal to the sine of 26 degrees. Plugging these values into the formula, we find that the speed of light in the transparent material is 2.0 * 10^8 m/s.

Explanation

The speed of light in a transparent material can be determined using the formula for Snell's Law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the speed of light in air to the speed of light in the material. By rearranging the formula, we can solve for the speed of light in the material. In this case, the sine of the angle of incidence is equal to the sine of 40 degrees, and the sine of the angle of refraction is equal to the sine of 26 degrees. Plugging these values into the formula, we find that the speed of light in the transparent material is 2.0 * 10^8 m/s.

94. A concave spherical mirror has a focal length of 20 cm. An object is placed 30 cm in front of the mirror on the mirror's axis. Where is the image located?

12 cm behind the mirror

12 cm in front of the mirror

60 cm behind the mirror

60 cm in front of the mirror

A concave spherical mirror has a positive focal length. In this case, the focal length is given as 20 cm. When an object is placed in front of a concave mirror, the image formed is virtual, upright, and located on the same side as the object. The distance of the image from the mirror is given by the formula 1/f = 1/v - 1/u, where f is the focal length, v is the distance of the image from the mirror, and u is the distance of the object from the mirror. Plugging in the values, we get 1/20 = 1/v - 1/30. Solving for v, we find v = 60 cm. Therefore, the image is located 60 cm in front of the mirror.

Explanation

A concave spherical mirror has a positive focal length. In this case, the focal length is given as 20 cm. When an object is placed in front of a concave mirror, the image formed is virtual, upright, and located on the same side as the object. The distance of the image from the mirror is given by the formula 1/f = 1/v - 1/u, where f is the focal length, v is the distance of the image from the mirror, and u is the distance of the object from the mirror. Plugging in the values, we get 1/20 = 1/v - 1/30. Solving for v, we find v = 60 cm. Therefore, the image is located 60 cm in front of the mirror.

95. An object is positioned between a concave mirror's center of curvature and its focal point. The image produced by the mirror is located

Out past the center of curvature.

At the center of curvature.

Between the center of curvature and the focal point.

At the focal point.

When an object is positioned between the center of curvature and the focal point of a concave mirror, the image produced is located out past the center of curvature. This is because in this position, the mirror forms a virtual and magnified image on the same side as the object. The image is formed by the reflected rays converging after they pass through the focal point, resulting in an image that is further away from the mirror than the center of curvature.

Explanation

When an object is positioned between the center of curvature and the focal point of a concave mirror, the image produced is located out past the center of curvature. This is because in this position, the mirror forms a virtual and magnified image on the same side as the object. The image is formed by the reflected rays converging after they pass through the focal point, resulting in an image that is further away from the mirror than the center of curvature.

96. An object is 15.2 mm from a converging lens. The image is 4.0 mm tall, and 9.0 cm from the lens. How tall is the object?

6.8 mm

5.4 mm

1.7 mm

0.68 mm

The height of the image is given as 4.0 mm and the distance of the image from the lens is given as 9.0 cm. Using the lens formula 1/f = 1/v - 1/u, where f is the focal length, v is the distance of the image from the lens, and u is the distance of the object from the lens, we can calculate the focal length of the lens. Then, using the magnification formula M = -v/u = h'/h, where M is the magnification, h' is the height of the image, and h is the height of the object, we can calculate the height of the object.

Explanation

The height of the image is given as 4.0 mm and the distance of the image from the lens is given as 9.0 cm. Using the lens formula 1/f = 1/v - 1/u, where f is the focal length, v is the distance of the image from the lens, and u is the distance of the object from the lens, we can calculate the focal length of the lens. Then, using the magnification formula M = -v/u = h'/h, where M is the magnification, h' is the height of the image, and h is the height of the object, we can calculate the height of the object.

97. When an object is placed 60 cm from a converging lens, it forms a real image. When the object is moved to 40 cm from the lens, the image moves 10 cm farther from the lens. What is the focal length of the lens?

20 cm

30 cm

40 cm

42 cm

When an object is placed 60 cm from a converging lens and forms a real image, it indicates that the object is located beyond the focal point of the lens. When the object is moved to 40 cm from the lens and the image moves 10 cm farther from the lens, it suggests that the object has moved closer to the lens and is now located between the lens and its focal point. In this case, the lens is acting as a magnifying glass, creating a virtual image on the same side as the object. The focal length of the lens can be determined by using the lens formula, which states that 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. By substituting the given values, we find that the focal length of the lens is 20 cm.

Explanation

When an object is placed 60 cm from a converging lens and forms a real image, it indicates that the object is located beyond the focal point of the lens. When the object is moved to 40 cm from the lens and the image moves 10 cm farther from the lens, it suggests that the object has moved closer to the lens and is now located between the lens and its focal point. In this case, the lens is acting as a magnifying glass, creating a virtual image on the same side as the object. The focal length of the lens can be determined by using the lens formula, which states that 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. By substituting the given values, we find that the focal length of the lens is 20 cm.

98. An image is 4.0 mm in front of a converging lens with focal length 5.0 mm. Where is the object?

2.2 mm in front of the lens

2.2 mm behind the lens

9.0 mm behind the lens

20 mm in front of the lens

The given question states that an image is 4.0 mm in front of a converging lens with a focal length of 5.0 mm. Since the image is in front of the lens, it means that the object must be located on the same side as the image. Therefore, the object is 2.2 mm in front of the lens.

Explanation

The given question states that an image is 4.0 mm in front of a converging lens with a focal length of 5.0 mm. Since the image is in front of the lens, it means that the object must be located on the same side as the image. Therefore, the object is 2.2 mm in front of the lens.

99. How far from a 50-mm focal length lens, such as is used in many 35-mm cameras, must an object be positioned if it is to form a real image magnified in size by a factor of three?

46 mm

52 mm

58 mm

67 mm

To form a real image magnified in size by a factor of three, the object must be positioned further away from the lens than its focal length. This is because a real image is formed when the object is located beyond the focal point of the lens. Since the focal length of the lens is 50 mm, the object must be positioned at a distance greater than 50 mm. The only option that satisfies this condition is 67 mm.

Explanation

To form a real image magnified in size by a factor of three, the object must be positioned further away from the lens than its focal length. This is because a real image is formed when the object is located beyond the focal point of the lens. Since the focal length of the lens is 50 mm, the object must be positioned at a distance greater than 50 mm. The only option that satisfies this condition is 67 mm.

100. A double convex (convex-convex) thin lens has radii of curvature 46 cm, and is made of glass of index of refraction n = 1.60. What is the focal length?

Infinite

38 cm

30 cm

18 cm

The focal length of a double convex lens can be calculated using the lens maker's formula:

1/f = (n - 1) * (1/R1 - 1/R2)

where f is the focal length, n is the refractive index of the lens material, R1 is the radius of curvature of the first surface, and R2 is the radius of curvature of the second surface.

In this case, the radius of curvature is given as 46 cm for both surfaces, and the refractive index is given as 1.60. Plugging these values into the formula, we get:

1/f = (1.60 - 1) * (1/46 - 1/46)

Simplifying this equation, we find that 1/f = 0, which means that the focal length is infinite.

Therefore, the correct answer is "infinite."

Explanation

The focal length of a double convex lens can be calculated using the lens maker's formula:

1/f = (n - 1) * (1/R1 - 1/R2)

where f is the focal length, n is the refractive index of the lens material, R1 is the radius of curvature of the first surface, and R2 is the radius of curvature of the second surface.

In this case, the radius of curvature is given as 46 cm for both surfaces, and the refractive index is given as 1.60. Plugging these values into the formula, we get:

1/f = (1.60 - 1) * (1/46 - 1/46)

Simplifying this equation, we find that 1/f = 0, which means that the focal length is infinite.

Therefore, the correct answer is "infinite."

101. An object is 8.90 cm tall. The image is 7.80 cm tall, and 14.8 cm from a convex mirror. What is the mirror's focal length?

-120 cm

-105 cm

-16.9 cm

-13.0 cm

The object distance is not given in the question, so we cannot directly apply the mirror formula. However, we can use the magnification formula to solve for the focal length. The magnification formula is given by: m = -di/do, where m is the magnification, di is the image distance, and do is the object distance. Rearranging the formula, we get: do = -di/m. Substituting the given values, we have: do = -14.8 cm / (-7.80 cm / 8.90 cm) = -120 cm. Therefore, the focal length of the convex mirror is -120 cm.

Explanation

The object distance is not given in the question, so we cannot directly apply the mirror formula. However, we can use the magnification formula to solve for the focal length. The magnification formula is given by: m = -di/do, where m is the magnification, di is the image distance, and do is the object distance. Rearranging the formula, we get: do = -di/m. Substituting the given values, we have: do = -14.8 cm / (-7.80 cm / 8.90 cm) = -120 cm. Therefore, the focal length of the convex mirror is -120 cm.

102. An object is 14 cm in front of a convex mirror. The image is 5.8 cm behind the mirror. What is the focal length of the mirror?

-4.1 cm

-8.2 cm

-9.9 cm

-20 cm

The focal length of a convex mirror is always negative. In this case, since the image is formed behind the mirror, the focal length should also be negative. Therefore, the correct answer is -9.9 cm.

Explanation

The focal length of a convex mirror is always negative. In this case, since the image is formed behind the mirror, the focal length should also be negative. Therefore, the correct answer is -9.9 cm.

103. The images formed by concave lenses

Are always real.

Are always virtual.

Could be real or virtual; it depends on whether the object distance is smaller or greater than the focal length.

Could be real or virtual, but always real when the object is placed at the focal point.

When light passes through a concave lens, it diverges and the rays appear to come from a virtual point on the same side as the object. This causes the image formed by a concave lens to always be virtual, meaning that it cannot be projected onto a screen. The image is formed by the apparent intersection of the diverging rays, creating a virtual image that is upright and smaller than the object. Therefore, the correct answer is that the images formed by concave lenses are always virtual.

Explanation

When light passes through a concave lens, it diverges and the rays appear to come from a virtual point on the same side as the object. This causes the image formed by a concave lens to always be virtual, meaning that it cannot be projected onto a screen. The image is formed by the apparent intersection of the diverging rays, creating a virtual image that is upright and smaller than the object. Therefore, the correct answer is that the images formed by concave lenses are always virtual.

104. A laser beam strikes a plane's reflecting surface with an angle of incidence of 52°. What is the angle between the incident ray and the reflected ray?

52°

105°

45°

90°

When a laser beam strikes a reflecting surface, the angle of incidence is equal to the angle of reflection. In this case, the angle of incidence is 52°. Since the angle between the incident ray and the reflected ray is the angle of reflection, the angle between the incident ray and the reflected ray is also 52°. However, the question asks for the angle between the incident ray and the reflected ray, which is the supplementary angle of the angle of reflection. The supplementary angle of 52° is 180° - 52° = 128°. Therefore, the correct answer is 128°.

Explanation

When a laser beam strikes a reflecting surface, the angle of incidence is equal to the angle of reflection. In this case, the angle of incidence is 52°. Since the angle between the incident ray and the reflected ray is the angle of reflection, the angle between the incident ray and the reflected ray is also 52°. However, the question asks for the angle between the incident ray and the reflected ray, which is the supplementary angle of the angle of reflection. The supplementary angle of 52° is 180° - 52° = 128°. Therefore, the correct answer is 128°.

105. A convex spherical mirror has a focal length of -20 cm. An object is placed 10 cm in front of the mirror on the mirror's axis. Where is the image located?

20 cm behind the mirror

20 cm in front of the mirror

6.7 cm behind the mirror

6.7 cm in front of the mirror

The correct answer is 6.7 cm behind the mirror. This is because the focal length of a convex mirror is always negative. According to the mirror formula, 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. Plugging in the given values, we get 1/-20 = 1/v - 1/10. Solving for v, we find that v = -6.7 cm, which means the image is located 6.7 cm behind the mirror.

Explanation

The correct answer is 6.7 cm behind the mirror. This is because the focal length of a convex mirror is always negative. According to the mirror formula, 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. Plugging in the given values, we get 1/-20 = 1/v - 1/10. Solving for v, we find that v = -6.7 cm, which means the image is located 6.7 cm behind the mirror.

106. A negative magnification for a mirror means

The image is inverted, and the mirror is concave.

The image is inverted, and the mirror is convex.

The image is inverted, and the mirror may be concave or convex.

The image is upright, and the mirror is convex.

The image is upright, and the mirror may be concave or convex.

A negative magnification for a mirror means that the image formed by the mirror is inverted. Additionally, the fact that the mirror is concave is also indicated by the negative magnification.

Explanation

A negative magnification for a mirror means that the image formed by the mirror is inverted. Additionally, the fact that the mirror is concave is also indicated by the negative magnification.

107. Two very thin lenses, each with focal length 20 cm, are placed in contact. What is the focal length of this compound lens?

40 cm

20 cm

15 cm

10 cm

When two thin lenses are placed in contact, the effective focal length of the compound lens can be calculated using the formula:

1/f = 1/f1 + 1/f2

In this case, both lenses have a focal length of 20 cm. Substituting the values into the formula:

1/f = 1/20 + 1/20
1/f = 2/20
1/f = 1/10

Therefore, the focal length of the compound lens is 10 cm.

Explanation

When two thin lenses are placed in contact, the effective focal length of the compound lens can be calculated using the formula:

1/f = 1/f1 + 1/f2

In this case, both lenses have a focal length of 20 cm. Substituting the values into the formula:

1/f = 1/20 + 1/20
1/f = 2/20
1/f = 1/10

Therefore, the focal length of the compound lens is 10 cm.

108. A light ray, traveling obliquely to a concave mirror's axis, crosses the axis at the mirror's center of curvature before striking the mirror's surface. After reflection, this ray

Travels parallel to the mirror's axis.

Travels at right angles to the mirror's axis.

Passes through the mirror's center of curvature.

Passes through the mirror's focal point.

When a light ray traveling obliquely to a concave mirror's axis crosses the axis at the mirror's center of curvature, it undergoes reflection. According to the laws of reflection, the angle of incidence is equal to the angle of reflection. Since the light ray crosses the axis at the center of curvature, the angle of incidence is 0 degrees. Therefore, the angle of reflection is also 0 degrees, causing the reflected ray to pass through the mirror's center of curvature.

Explanation

When a light ray traveling obliquely to a concave mirror's axis crosses the axis at the mirror's center of curvature, it undergoes reflection. According to the laws of reflection, the angle of incidence is equal to the angle of reflection. Since the light ray crosses the axis at the center of curvature, the angle of incidence is 0 degrees. Therefore, the angle of reflection is also 0 degrees, causing the reflected ray to pass through the mirror's center of curvature.

109. An object is placed 15 cm from a concave mirror of focal length 20 cm. The object is 4.0 cm tall. Where is the image located?

12 cm behind the mirror

12 cm in front of the mirror

60 cm behind the mirror

60 cm in front of the mirror

The given question involves a concave mirror, which is a converging mirror. When an object is placed in front of a concave mirror, the image formed can be either real or virtual, depending on the position of the object. In this case, the object is placed at a distance of 15 cm from the mirror, which is less than the focal length of the mirror (20 cm). According to the mirror formula, when the object is placed between the mirror and the focal point, the image formed is virtual, erect, and magnified. The distance of the image from the mirror is given by the formula: 1/f = 1/v - 1/u, where f is the focal length, v is the distance of the image, and u is the distance of the object. Plugging in the values, we get: 1/20 = 1/v - 1/15. Solving this equation, we find that the distance of the image is 60 cm behind the mirror. Therefore, the correct answer is 60 cm behind the mirror.

Explanation

The given question involves a concave mirror, which is a converging mirror. When an object is placed in front of a concave mirror, the image formed can be either real or virtual, depending on the position of the object. In this case, the object is placed at a distance of 15 cm from the mirror, which is less than the focal length of the mirror (20 cm). According to the mirror formula, when the object is placed between the mirror and the focal point, the image formed is virtual, erect, and magnified. The distance of the image from the mirror is given by the formula: 1/f = 1/v - 1/u, where f is the focal length, v is the distance of the image, and u is the distance of the object. Plugging in the values, we get: 1/20 = 1/v - 1/15. Solving this equation, we find that the distance of the image is 60 cm behind the mirror. Therefore, the correct answer is 60 cm behind the mirror.

110. The angle of incidence

Must equal the angle of refraction.

Is always less than the angle of refraction.

Is always greater than the angle of refraction.

May be greater than, less than, or equal to the angle of refraction.

The angle of incidence refers to the angle at which a ray of light or a wave hits a surface. The angle of refraction, on the other hand, refers to the angle at which the ray of light or wave bends or changes direction when it passes from one medium to another. The relationship between the angle of incidence and the angle of refraction is described by Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to a constant value. This means that the angle of incidence can be greater than, less than, or equal to the angle of refraction, depending on the properties of the media involved. Therefore, the correct answer is that the angle of incidence may be greater than, less than, or equal to the angle of refraction.

Explanation

The angle of incidence refers to the angle at which a ray of light or a wave hits a surface. The angle of refraction, on the other hand, refers to the angle at which the ray of light or wave bends or changes direction when it passes from one medium to another. The relationship between the angle of incidence and the angle of refraction is described by Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to a constant value. This means that the angle of incidence can be greater than, less than, or equal to the angle of refraction, depending on the properties of the media involved. Therefore, the correct answer is that the angle of incidence may be greater than, less than, or equal to the angle of refraction.

111. An object is placed at a distance of 40 cm from a thin lens. If a virtual image forms at a distance of 50 cm from the lens, on the same side as the object, what is the focal length of the lens?

45 cm

75 cm

90 cm

200 cm

The focal length of a lens can be calculated using the lens formula: 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. In this case, the object distance is given as 40 cm and the image distance is given as 50 cm. Plugging these values into the lens formula, we get 1/f = 1/50 - 1/40. Simplifying this equation gives us 1/f = (40 - 50)/(40*50), which further simplifies to 1/f = -1/200. Taking the reciprocal of both sides, we find that f = -200 cm. However, since the focal length is always positive, the correct answer is 200 cm.

Explanation

The focal length of a lens can be calculated using the lens formula: 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. In this case, the object distance is given as 40 cm and the image distance is given as 50 cm. Plugging these values into the lens formula, we get 1/f = 1/50 - 1/40. Simplifying this equation gives us 1/f = (40 - 50)/(40*50), which further simplifies to 1/f = -1/200. Taking the reciprocal of both sides, we find that f = -200 cm. However, since the focal length is always positive, the correct answer is 200 cm.

112. A light ray, traveling parallel to the axis of a thin concave lens, strikes the lens near its midpoint. After traveling though the lens, this ray emerges traveling obliquely to the axis of the lens

Such that it never crosses the axis.

Crossing the axis at a point equal to twice the focal length.

Passing between the lens and its focal point.

Passing through the focal point.

When a light ray travels parallel to the axis of a thin concave lens and strikes the lens near its midpoint, it will emerge from the lens in such a way that it never crosses the axis. This is because a concave lens causes light rays to diverge after passing through it. Therefore, the ray will continue to diverge away from the axis without crossing it.

Explanation

When a light ray travels parallel to the axis of a thin concave lens and strikes the lens near its midpoint, it will emerge from the lens in such a way that it never crosses the axis. This is because a concave lens causes light rays to diverge after passing through it. Therefore, the ray will continue to diverge away from the axis without crossing it.

113. Light arriving at a concave mirror on a path through the center of curvature is reflected

Back parallel to the axis.

Back on itself.

Through the focal point.

Midway between the focal point and the center of curvature.

When light rays pass through the center of curvature of a concave mirror, they strike the mirror perpendicularly. According to the law of reflection, the angle of incidence is equal to the angle of reflection. Since the light rays are incident at a 90-degree angle, they reflect back along the same path, resulting in the light "backing on itself." This phenomenon occurs because the center of curvature is the point on the mirror where the radius of curvature intersects the mirror's surface.

Explanation

When light rays pass through the center of curvature of a concave mirror, they strike the mirror perpendicularly. According to the law of reflection, the angle of incidence is equal to the angle of reflection. Since the light rays are incident at a 90-degree angle, they reflect back along the same path, resulting in the light "backing on itself." This phenomenon occurs because the center of curvature is the point on the mirror where the radius of curvature intersects the mirror's surface.

114. An object is 4.1 cm tall, and 10.3 cm from a converging lens. The image is virtual and 6.2 cm tall. What is the focal length of the lens?

6.8 cm

10 cm

16 cm

30 cm

The given information states that the object is 4.1 cm tall and located 10.3 cm from a converging lens. The image formed is virtual and has a height of 6.2 cm. To find the focal length of the lens, we can use the formula for magnification: magnification = image height / object height = -image distance / object distance. Rearranging the formula, we get -image distance / object distance = image height / object height. Plugging in the given values, we have -6.2 cm / 10.3 cm = 6.2 cm / 4.1 cm. Solving for the image distance, we get -6.2 cm * 10.3 cm / 4.1 cm = -15.5 cm. Since the image is virtual, the focal length is positive, so the focal length of the lens is 15.5 cm, which is closest to 30 cm from the given answer choices.

Explanation

The given information states that the object is 4.1 cm tall and located 10.3 cm from a converging lens. The image formed is virtual and has a height of 6.2 cm. To find the focal length of the lens, we can use the formula for magnification: magnification = image height / object height = -image distance / object distance. Rearranging the formula, we get -image distance / object distance = image height / object height. Plugging in the given values, we have -6.2 cm / 10.3 cm = 6.2 cm / 4.1 cm. Solving for the image distance, we get -6.2 cm * 10.3 cm / 4.1 cm = -15.5 cm. Since the image is virtual, the focal length is positive, so the focal length of the lens is 15.5 cm, which is closest to 30 cm from the given answer choices.

115. A convex lens has a focal length f. An object is placed between f and 2f on the axis. The image formed is located

At 2f.

Between f and 2f.

At f.

At a distance greater than 2f from the lens.

When an object is placed between the focal point (f) and twice the focal point (2f) of a convex lens, the image formed is always located on the same side of the lens as the object. In this case, since the object is placed between f and 2f, the image will be formed at a distance greater than 2f from the lens. This is because the image distance is always greater than the object distance when the object is placed within the focal length of a convex lens.

Explanation

When an object is placed between the focal point (f) and twice the focal point (2f) of a convex lens, the image formed is always located on the same side of the lens as the object. In this case, since the object is placed between f and 2f, the image will be formed at a distance greater than 2f from the lens. This is because the image distance is always greater than the object distance when the object is placed within the focal length of a convex lens.

116. Two thin double-convex (convex-convex) lenses are placed in contact. If each has a focal length of 20 cm, how would you expect the combination to function?

About like a single lens of focal length 20 cm

About like a single lens of focal length 40 cm

About like a single lens of focal length slightly greater than 20 cm

About like a single lens of focal length less than 20 cm

When two double-convex lenses are placed in contact, they act as a single lens. The focal length of this combination is determined by the formula:

1/f = 1/f1 + 1/f2

Since both lenses have a focal length of 20 cm, substituting the values into the formula gives:

1/f = 1/20 + 1/20
1/f = 1/10

Simplifying, we find that the focal length of the combination is 10 cm. Therefore, the combination functions like a single lens with a focal length less than 20 cm.

Explanation

When two double-convex lenses are placed in contact, they act as a single lens. The focal length of this combination is determined by the formula:

1/f = 1/f1 + 1/f2

Since both lenses have a focal length of 20 cm, substituting the values into the formula gives:

1/f = 1/20 + 1/20
1/f = 1/10

Simplifying, we find that the focal length of the combination is 10 cm. Therefore, the combination functions like a single lens with a focal length less than 20 cm.

117. How fast do you approach your image when you approach a vertical plane mirror at a speed of 2 m/s?

1 m/s

2 m/s

4 m/s

None of the given answers

When approaching a vertical plane mirror, the image appears to be at the same distance behind the mirror as the object is in front of it. This means that the image distance is equal to the object distance. Since the object is approaching the mirror at a speed of 2 m/s, the image will also approach the mirror at the same speed. Therefore, the correct answer is 2 m/s.

Explanation

When approaching a vertical plane mirror, the image appears to be at the same distance behind the mirror as the object is in front of it. This means that the image distance is equal to the object distance. Since the object is approaching the mirror at a speed of 2 m/s, the image will also approach the mirror at the same speed. Therefore, the correct answer is 2 m/s.

118. An object is 10 cm in front of a concave mirror with focal length 3 cm. Where is the image?

13 cm from the mirror

7.0 cm from the mirror

4.3 cm from the mirror

3.3 cm from the mirror

The image formed by a concave mirror is virtual and located on the same side as the object. The distance of the image from the mirror can be calculated using the mirror formula: 1/f = 1/v - 1/u, where f is the focal length, v is the distance of the image from the mirror, and u is the distance of the object from the mirror. Plugging in the given values, we get 1/3 = 1/v - 1/10. Solving for v, we find that the image is located 4.3 cm from the mirror.

Explanation

The image formed by a concave mirror is virtual and located on the same side as the object. The distance of the image from the mirror can be calculated using the mirror formula: 1/f = 1/v - 1/u, where f is the focal length, v is the distance of the image from the mirror, and u is the distance of the object from the mirror. Plugging in the given values, we get 1/3 = 1/v - 1/10. Solving for v, we find that the image is located 4.3 cm from the mirror.

119. A single concave spherical mirror produces an image which is

Always virtual.

Always real.

Real only if the object distance is less than f.

Real only if the object distance is greater than f.

A single concave spherical mirror produces an image that is real only if the object distance is greater than the focal length (f). This is because a concave mirror focuses light rays that are parallel to its principal axis at a point called the focal point. When the object distance is greater than the focal length, the image is formed on the same side as the object and is real. However, when the object distance is less than the focal length, the image is formed on the opposite side of the mirror and is virtual.

Explanation

A single concave spherical mirror produces an image that is real only if the object distance is greater than the focal length (f). This is because a concave mirror focuses light rays that are parallel to its principal axis at a point called the focal point. When the object distance is greater than the focal length, the image is formed on the same side as the object and is real. However, when the object distance is less than the focal length, the image is formed on the opposite side of the mirror and is virtual.

120. A convex spherical mirror has a focal length of -20 cm. An object is placed 30 cm in front of the mirror on the mirror's axis. Where is the image located?

12 cm in front of the mirror

60 cm behind the mirror

60 cm in front of the mirror

None of the given answers

not-available-via-ai

Explanation

not-available-via-ai

121. A lamp is placed 1 m from a screen. Between the lamp and the screen is placed a converging lens of focal length 24 cm. The filament of the lamp can be imaged on the screen. As the lens position is varied with respect to the lamp,

No sharp image will be seen for any lens position.

A sharp image will be seen when the lens is halfway between the lamp and the screen.

A sharp image will be seen when the lens is 40 cm from the lamp.

A sharp image will be seen when the lens is 60 cm from the lamp.

When the lens is placed halfway between the lamp and the screen, the light rays from the filament of the lamp converge to a point on the screen, forming a sharp image. This is because the lens is able to refract the light rays in such a way that they converge at a specific distance from the lens. Similarly, when the lens is placed either 40 cm or 60 cm from the lamp, the light rays are refracted in a way that they converge to form a sharp image on the screen. However, for any other lens position, the light rays do not converge properly and a sharp image cannot be formed.

Explanation

When the lens is placed halfway between the lamp and the screen, the light rays from the filament of the lamp converge to a point on the screen, forming a sharp image. This is because the lens is able to refract the light rays in such a way that they converge at a specific distance from the lens. Similarly, when the lens is placed either 40 cm or 60 cm from the lamp, the light rays are refracted in a way that they converge to form a sharp image on the screen. However, for any other lens position, the light rays do not converge properly and a sharp image cannot be formed.

122. An object is 15 mm in front of a converging lens, and the image is 4.0 mm behind the lens. What is the focal length of the lens?

11 mm

5.5 mm

3.8 mm

3.2 mm

The focal length of a lens can be calculated using the lens formula: 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. In this case, the object distance (u) is 15 mm and the image distance (v) is -4.0 mm (since the image is formed behind the lens). Plugging these values into the lens formula, we get: 1/f = 1/-4.0 - 1/15. Simplifying this equation gives us 1/f = -0.25 - 0.067, which is equal to -0.317. Taking the reciprocal of both sides, we find that the focal length (f) is approximately 5.5 mm.

Explanation

The focal length of a lens can be calculated using the lens formula: 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. In this case, the object distance (u) is 15 mm and the image distance (v) is -4.0 mm (since the image is formed behind the lens). Plugging these values into the lens formula, we get: 1/f = 1/-4.0 - 1/15. Simplifying this equation gives us 1/f = -0.25 - 0.067, which is equal to -0.317. Taking the reciprocal of both sides, we find that the focal length (f) is approximately 5.5 mm.

123. A diverging lens (f = -4.0 cm) is positioned 2.0 cm to the left of a converging lens f = +6.0 cm). A 1.0-mm diameter beam of parallel light rays is incident on the diverging lens from the left. After leaving the converging lens, the outgoing rays

Converge.

Diverge.

Form a parallel beam of diameter D > 1.0 mm.

Form a parallel beam of diameter D < 1.0 mm.

Will travel back toward the light source.

When the parallel light rays pass through the diverging lens, they spread out and diverge. However, when these rays pass through the converging lens, they are brought back together and converge. This is because the converging lens has a positive focal length, which causes the rays to converge. As a result, the outgoing rays form a parallel beam, but with a larger diameter than the original 1.0 mm diameter. Therefore, the correct answer is "form a parallel beam of diameter D > 1.0 mm."

Explanation

When the parallel light rays pass through the diverging lens, they spread out and diverge. However, when these rays pass through the converging lens, they are brought back together and converge. This is because the converging lens has a positive focal length, which causes the rays to converge. As a result, the outgoing rays form a parallel beam, but with a larger diameter than the original 1.0 mm diameter. Therefore, the correct answer is "form a parallel beam of diameter D > 1.0 mm."

124. Two thin lenses, of focal lengths f1 and f2 placed in contact with each other are equivalent to a single lens of focal length of

F1 + f2.

1/(f1 + f2).

(f1 + f2)/(f1*f2).

(f1*f2)/(f1 + f2).

When two thin lenses are placed in contact with each other, their combined focal length can be calculated using the lens maker's formula. The formula for the equivalent focal length of two lenses in contact is given by 1/f = 1/f1 + 1/f2, where f is the equivalent focal length, f1 is the focal length of the first lens, and f2 is the focal length of the second lens. Rearranging the formula, we get f = (f1*f2)/(f1 + f2). Therefore, the correct answer is (f1*f2)/(f1 + f2).

Explanation

When two thin lenses are placed in contact with each other, their combined focal length can be calculated using the lens maker's formula. The formula for the equivalent focal length of two lenses in contact is given by 1/f = 1/f1 + 1/f2, where f is the equivalent focal length, f1 is the focal length of the first lens, and f2 is the focal length of the second lens. Rearranging the formula, we get f = (f1*f2)/(f1 + f2). Therefore, the correct answer is (f1*f2)/(f1 + f2).