Plane mirrors produce images which

are always the same size as the actual object.

are always smaller than the actual object.

are always larger than the actual object.

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

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-half of the focal length of lens A.

one-fourth of the focal length of lens A.

twice the focal length of lens A.

four times the focal length of lens A.

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

through the center of curvature.

Chapter 23: Light: Geometric Optics

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1/124 Questions

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.

About This Quiz

Explore the fundamentals of geometric optics in 'Chapter 23: Light: Geometric Optics'. This quiz assesses understanding of reflection, refraction, and image formation using mirrors and lenses. Key concepts include the ray model of light, laws of reflection, and characteristics of images formed by plane mirrors.

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2.

If the magnification is a negative value, the image is
- Upright.
- Inverted.
Correct Answer

A. Inverted.

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.

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1
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.
Correct Answer

A. 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.

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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.
Correct Answer

A. Convex mirror.

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.

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5.

A spherical mirror on which reflection takes place on the inner surface of the sphere is referred to as a
- Convex mirror.
- Concave mirror.
Correct Answer

A. Concave mirror.

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.

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6.

If the image distance is positive, the image formed is a
- Real image.
- Virtual image.
Correct Answer

A. Real image.

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.

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7.

If the image distance is negative, the image formed is a
- Real image.
- Virtual image.
Correct Answer

A. Virtual image.

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.

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8.

Light travels fastest
- In a vacuum.
- Through water.
- Through glass.
- Through diamond.
Correct Answer

A. In a vacuum.

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.

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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.
Correct Answer

A. Virtual image.

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.

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10.

If the magnification is a positive value, the image is
- Upright.
- Inverted.
Correct Answer

A. Upright.

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.

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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.
Correct Answer

A. Smaller than 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.

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12.

Lenses that are thinner at the center than the edges are called
- Converging lenses.
- Diverging lenses.
Correct Answer

A. Diverging lenses.

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.

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13.

Lenses that are thickest at the center called
- Converging lenses.
- Diverging lenses.
Correct Answer

A. Converging lenses.

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.

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14.

The principle on which lenses work is
- Refraction.
- Polarization.
- Dispersion.
- Total internal reflection.
Correct Answer

A. Refraction.

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.

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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
Correct Answer

A. 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.

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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.
Correct Answer

A. 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.

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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.
Correct Answer

A. Are always the same size as the actual object.

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.

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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
Correct Answer

A. 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.

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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.
Correct Answer

A. One-half of 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.

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20.

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.
Correct Answer

A. 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.

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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.
Correct Answer

A. Virtual and erect.

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.

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22.

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
Correct Answer

A. 10 cm

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.

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23.

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
Correct Answer

A. 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.

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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
Correct Answer

A. 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.

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2
25.

The principle on which mirrors work is
- Refraction.
- Polarization.
- Dispersion.
- Reflection.
Correct Answer

A. Reflection.

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.

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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.
Correct Answer

A. 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°.

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27.

The principle on which fiber optics is based is
- Refraction.
- Polarization.
- Dispersion.
- Total internal reflection.
Correct Answer

A. Total internal reflection.

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.

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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.
Correct Answer

A. Virtual and erect.

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.

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29.

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
Correct Answer

A. 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.

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30.

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.
Correct Answer

A. Always virtual.

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.

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31.

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.
Correct Answer

A. 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.

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32.

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°
Correct Answer

A. 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."

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33.

Lucite has an index of refraction of 1.50. What is its critical angle of incidence?
- 1.16°
- 15°
- 41.8°
- 87.4°
Correct Answer

A. 41.8°

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.

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34.

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
Correct Answer

A. 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.

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35.

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
Correct Answer

A. 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.

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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
Correct Answer

A. 33 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.

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37.

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.
Correct Answer

A. Must equal 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.

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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
Correct Answer

A. 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.

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39.

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
Correct Answer

A. 18.2 cm

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.

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40.

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.
- Yes, since almost everything we see is virtual because most things do not themselves give off light, but only reflect light coming from some other source.
- Yes, but only indirectly in the sense that if the virtual image is formed on a sheet of photographic film, one could later look at the picture formed.
Correct Answer

A. Yes, the rays that appear to emanate from a virtual image can be focused on the retina just like those from an illuminated object.

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.

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41.

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.
Correct Answer

A. 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.

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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
Correct Answer

A. 20 cm 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.

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43.

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
Correct Answer

A. 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.

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44.

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°
Correct Answer

A. >41.8°

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".

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45.

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°
Correct Answer

A. 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°.

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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
Correct Answer

A. 11 cm

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.

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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
Correct Answer

A. 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.

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48.

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.
Correct Answer

A. 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."

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49.

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
Correct Answer

A. 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.

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Quiz Edited by
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Nov 07, 2012

Quiz Created by
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Chapter 23: Light: Geometric Optics

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