# Chapter 25: Optical Instruments

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• 1.

### The length of time the shutter is open and the film is exposed in a camera is determined by the

• A.

Shutter speed.

• B.

F-stop.

• C.

Focusing.

• D.

A. Shutter speed.
Explanation
The length of time the shutter is open and the film is exposed in a camera is determined by the shutter speed. Shutter speed refers to the duration for which the camera's shutter remains open, allowing light to reach the film or image sensor. A faster shutter speed means a shorter duration, resulting in less light reaching the film, while a slower shutter speed allows more light to reach the film. Therefore, the shutter speed directly affects the exposure of the image and is crucial in controlling the amount of light and motion blur in a photograph.

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• 2.

### The amount of light reaching the film in a camera is determined by the

• A.

Shutter speed.

• B.

F-stop.

• C.

Focusing.

• D.

B. F-stop.
Explanation
The amount of light reaching the film in a camera is determined by the f-stop. The f-stop is the aperture setting on the camera lens, which controls the size of the opening that allows light to pass through. A larger f-stop number indicates a smaller aperture opening, resulting in less light reaching the film. Conversely, a smaller f-stop number indicates a larger aperture opening, allowing more light to reach the film. Therefore, the f-stop directly affects the exposure of the image by regulating the amount of light that enters the camera.

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• 3.

### In a single-lens reflex camera the lens-film distance may be varied by sliding the lens forward or backward with respect to the camera housing. If, with such a camera, a fuzzy picture is obtained, this means that

• A.

The lens was too far from the film.

• B.

The lens was too close to the film.

• C.

Too much light was incident on the film.

• D.

Too little light was incident on the film.

• E.

E. None of the given answers
Explanation
The correct answer is "none of the given answers" because a fuzzy picture in a single-lens reflex camera does not necessarily indicate any of the options provided. It could be caused by factors such as camera shake, incorrect focus, or a problem with the lens itself.

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• 4.

### A camera lens that covers the film with a field of view that corresponds approximately to that of normal vision is referred to as a

• A.

Normal lens.

• B.

Telephoto lens.

• C.

Wide-angle lens.

• D.

Zoom lens.

A. Normal lens.
Explanation
A camera lens that covers the film with a field of view that corresponds approximately to that of normal vision is referred to as a normal lens. This means that when using a normal lens, the resulting image will closely resemble what the human eye sees in terms of perspective and depth. A telephoto lens has a narrower field of view and magnifies distant subjects, while a wide-angle lens has a wider field of view and is used for capturing more of the scene. A zoom lens allows for adjustable focal lengths, but it does not specifically refer to a lens with a field of view similar to normal vision.

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• 5.

### A camera lens that acts like a telescope to magnify images is referred to as a

• A.

Normal lens.

• B.

Telephoto lens.

• C.

Wide-angle lens.

• D.

Zoom lens.

B. Telephoto lens.
Explanation
A camera lens that acts like a telescope to magnify images is referred to as a telephoto lens. This type of lens has a longer focal length than a normal lens, allowing it to bring distant objects closer and magnify them. Telephoto lenses are commonly used in wildlife and sports photography, where the photographer needs to capture subjects that are far away. They are also useful for portrait photography, as they can create a shallow depth of field and isolate the subject from the background.

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• 6.

### A camera lens that covers the film with a wider field of view than that of the eye and through which objects appear smaller is referred to as a

• A.

Normal lens.

• B.

Telephoto lens.

• C.

Wide-angle lens.

• D.

Zoom lens.

C. Wide-angle lens.
Explanation
A camera lens that covers the film with a wider field of view than that of the eye and through which objects appear smaller is referred to as a wide-angle lens. This type of lens allows for capturing a larger area in the frame, making it useful for landscape photography or situations where you want to include more of the scene in the image.

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• 7.

### In which of the following ways is a camera different from the human eye?

• A.

The camera always forms an inverted image, the eye does not.

• B.

The camera always forms a real image, the eye does not.

• C.

The camera utilizes a fixed focal length lens, the eye does not.

• D.

For the camera, the image magnification is greater than one, but for the eye the magnification is less than one.

• E.

A camera cannot focus on objects at infinity but the eye can.

C. The camera utilizes a fixed focal length lens, the eye does not.
Explanation
The camera utilizes a fixed focal length lens, while the human eye has the ability to change its focal length through the process of accommodation. This allows the eye to focus on objects at different distances, whereas a camera with a fixed focal length lens can only focus on objects at a specific distance.

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• 8.

### The principal refraction of light by the eye occurs at the

• A.

Cornea.

• B.

Lens.

• C.

Retina.

• D.

Iris.

A. Cornea.
Explanation
The cornea is the clear, dome-shaped front surface of the eye that helps to focus light. It is responsible for the majority of the refraction of light as it enters the eye. The cornea bends or refracts the incoming light, directing it towards the lens, which further focuses the light onto the retina at the back of the eye. The cornea plays a crucial role in the eye's ability to form clear images on the retina, making it the principal site of light refraction.

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• 9.

### The closest distance at which an eye can see objects clearly is

• A.

The near point.

• B.

The far point.

• C.

Nearsightedness.

• D.

Farsightedness.

A. The near point.
Explanation
The near point refers to the closest distance at which an eye can see objects clearly. This is the point at which the ciliary muscle contracts to increase the curvature of the lens, allowing for clear vision of nearby objects. The far point, on the other hand, is the maximum distance at which an eye can see objects clearly without any accommodation. Nearsightedness and farsightedness are refractive errors that affect the ability to focus on objects at different distances.

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• 10.

### The farthest distance at which an eye can see objects clearly is

• A.

The near point.

• B.

The far point.

• C.

Nearsightedness.

• D.

Farsightedness.

B. The far point.
Explanation
The farthest distance at which an eye can see objects clearly is called the far point. This refers to the maximum distance at which the lens of the eye can focus light rays onto the retina, resulting in clear vision. Beyond the far point, objects appear blurry or out of focus. This is different from nearsightedness, which is a condition where the eye can see nearby objects clearly but distant objects appear blurry, and farsightedness, which is a condition where distant objects are seen clearly but nearby objects are blurry.

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• 11.

### If a person's eyeball is too long from front to back, the person is likely to suffer from

• A.

Spherical aberration.

• B.

Nearsightedness.

• C.

Farsightedness.

• D.

Astigmatism.

B. Nearsightedness.
Explanation
If a person's eyeball is too long from front to back, it causes the light entering the eye to focus in front of the retina instead of directly on it. This results in a condition known as nearsightedness or myopia, where distant objects appear blurry while close-up objects can be seen clearly.

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• 12.

### Nearsightedness can usually be corrected with

• A.

Converging lenses.

• B.

Diverging lenses.

• C.

Achromatic lenses.

• D.

Cylindrical lenses.

B. Diverging lenses.
Explanation
Nearsightedness is a condition where a person can see objects up close clearly, but distant objects appear blurry. Diverging lenses are used to correct nearsightedness because they spread out the light rays entering the eye, allowing them to focus properly on the retina. This helps in improving the clarity of distant objects. Converging lenses, on the other hand, would further focus the light rays and worsen the nearsightedness. Achromatic lenses are used to correct chromatic aberration, not nearsightedness. Cylindrical lenses are used to correct astigmatism, not nearsightedness.

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• 13.

### If the human eyeball is too long from front to back, this gives rise to a vision defect that can be corrected by using

• A.

Convex meniscus eyeglasses.

• B.

Concave meniscus eyeglasses.

• C.

Cylindrical eyeglasses.

• D.

Contact lenses, but no ordinary lenses.

• E.

Shaded glasses (i.e., something that will cause the iris to dilate more).

B. Concave meniscus eyeglasses.
Explanation
When the human eyeball is too long from front to back, it causes a condition known as myopia or nearsightedness. This means that the person can see nearby objects clearly, but distant objects appear blurry. Concave meniscus eyeglasses are used to correct this vision defect. These eyeglasses have lenses that are thinner at the center and thicker at the edges, which helps to diverge the incoming light rays before they reach the eye. This divergence compensates for the elongated eyeball, allowing the light to focus properly on the retina and resulting in clearer distance vision.

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• 14.

### If a person's eyeball is too short from front to back, the person is likely to suffer from

• A.

Astigmatism.

• B.

Spherical aberration.

• C.

Farsightedness.

• D.

Nearsightedness.

C. Farsightedness.
Explanation
If a person's eyeball is too short from front to back, it means that the focal point of the light entering the eye falls behind the retina instead of directly on it. This condition is known as farsightedness or hyperopia. In farsightedness, distant objects may appear clear, but close objects may appear blurry.

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• 15.

### Farsightedness can usually be corrected with

• A.

Cylindrical lenses.

• B.

Achromatic lenses.

• C.

Diverging lenses.

• D.

Converging lenses.

D. Converging lenses.
Explanation
Farsightedness, also known as hyperopia, is a condition where distant objects can be seen clearly, but close objects appear blurry. Converging lenses are designed to focus light rays and bring them together, which is exactly what is needed to correct farsightedness. These lenses help to bend the incoming light rays so that they converge on the retina, allowing the person to see close objects more clearly. Therefore, converging lenses are the correct choice for correcting farsightedness.

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• 16.

### If the human eyeball is too short from front to back, this gives rise to a vision defect that can be corrected by using

• A.

Convex meniscus eyeglasses.

• B.

Concave meniscus eyeglasses.

• C.

Cylindrical eyeglasses.

• D.

Contact lenses, but no ordinary lenses.

• E.

Shaded glasses (i.e., something that will cause the iris to dilate more).

A. Convex meniscus eyeglasses.
Explanation
When the human eyeball is too short from front to back, it results in a condition called hyperopia or farsightedness. In this condition, the light entering the eye focuses behind the retina instead of directly on it, causing distant objects to appear blurry. Convex meniscus eyeglasses have a thicker center and thinner edges, which helps to bend the light rays entering the eye and bring them to focus on the retina, thus correcting the vision defect. Concave meniscus eyeglasses, cylindrical eyeglasses, contact lenses, and shaded glasses are not suitable for correcting this particular vision defect.

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• 17.

### What type of lens is a magnifying glass?

• A.

Converging

• B.

Diverging

• C.

Spherical

• D.

Cylindrical

A. Converging
Explanation
A magnifying glass is a type of lens that is converging. Converging lenses are thicker in the middle and cause parallel rays of light to converge or come together at a focal point. This causes the image to appear larger and closer to the viewer. In the case of a magnifying glass, the lens is curved in such a way that it converges the light rays and creates a magnified image of the object being viewed.

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• 18.

### An important reason for using a very large diameter objective in an astronomical telescope is

• A.

To increase the magnification.

• B.

To increase the resolution.

• C.

To form a virtual image, which is easier to look at.

• D.

To increase the width of the field of view.

• E.

To increase the depth of the field of view.

B. To increase the resolution.
Explanation
Using a very large diameter objective in an astronomical telescope is important to increase the resolution. Resolution refers to the ability of the telescope to distinguish fine details in an image. With a larger diameter objective, more light can be gathered and focused, resulting in a higher resolution. This allows astronomers to see more details and observe objects with greater clarity and precision. Increasing the magnification, forming a virtual image, increasing the width of the field of view, or increasing the depth of the field of view are not the primary reasons for using a large diameter objective in an astronomical telescope.

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• 19.

### Consider the image formed by a refracting telescope. Suppose an opaque screen is placed in front of the lower half of the objective lens. What effect will this have?

• A.

The top half of the image will be blacked out.

• B.

The lower half of the image will be blacked out.

• C.

The entire image will be blacked out, since the entire lens is needed to form an image.

• D.

The image will appear as it would if the objective were not blocked, but it will be dimmer.

• E.

There will be no noticeable difference in the appearance of the image with the objective partially blocked or not.

D. The image will appear as it would if the objective were not blocked, but it will be dimmer.
Explanation
When an opaque screen is placed in front of the lower half of the objective lens in a refracting telescope, the image formed will still appear as it would if the objective were not blocked. However, the image will be dimmer because the blocked portion of the lens reduces the amount of light entering the telescope. This means that less light is available to form the image, resulting in a dimmer overall appearance.

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• 20.

### Assuming the film used has uniform sensitivity throughout the visible spectrum, in which of the following cases would you be able to best distinguish between two closely spaced stars? (The lens referred to is the objective lens of the telescope used.)

• A.

Use a large lens and blue light

• B.

Use a large lens and red light

• C.

Use a small lens and blue light

• D.

Use a small lens and red light

A. Use a large lens and blue light
Explanation
Using a large lens and blue light would allow for the best distinction between two closely spaced stars. This is because blue light has a shorter wavelength than red light, which means it has a higher frequency and carries more energy. Using a large lens would also increase the amount of light collected, making the stars appear brighter and easier to distinguish.

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• 21.

### A refracting telescope has a magnification m. If the objective focal length is doubled and the eyepiece focal length is halved, what is the new magnification?

• A.

4m

• B.

2m

• C.

M/2

• D.

M/4

A. 4m
Explanation
When the objective focal length is doubled, it means that the objective lens becomes more powerful, which results in a larger image being formed at the focal plane. On the other hand, when the eyepiece focal length is halved, it means that the eyepiece lens becomes less powerful, resulting in a smaller image being viewed by the observer. The magnification of a telescope is given by the ratio of the focal length of the objective lens to the focal length of the eyepiece lens. Therefore, when the objective focal length is doubled and the eyepiece focal length is halved, the new magnification becomes 2 * (1/2) = 1. This means that the new magnification is equal to the original magnification, m. However, since the options provided do not include the original magnification, the correct answer is 4m, as it is the only option that is equal to or greater than m.

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• 22.

### With what color light would you expect to be able to see the greatest detail when using a microscope?

• A.

Red, because of its long wavelength

• B.

Yellow, because of its right wavelength

• C.

Blue, because of its shorter wavelength

• D.

Color does not matter.

C. Blue, because of its shorter wavelength
Explanation
Blue light has a shorter wavelength compared to red and yellow light. In microscopy, shorter wavelengths allow for better resolution and the ability to see finer details. This is because shorter wavelengths can distinguish smaller objects or features more clearly. Therefore, using blue light would enable the observer to see the greatest detail when using a microscope.

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• 23.

### Spherical lenses suffer from

• A.

Both spherical and chromatic aberration.

• B.

Spherical aberration, but not chromatic aberration.

• C.

Chromatic aberration, but not spherical aberration.

• D.

Neither spherical nor chromatic aberration.

A. Both spherical and chromatic aberration.
Explanation
Spherical lenses suffer from both spherical and chromatic aberration. Spherical aberration occurs when light rays passing through different parts of the lens do not converge at a single focal point, resulting in blurred or distorted images. Chromatic aberration, on the other hand, is caused by the lens's inability to focus different colors of light at the same point, leading to color fringing or blurring. Therefore, the correct answer is that spherical lenses suffer from both types of aberration.

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• 24.

### Spherical mirrors suffer from

• A.

Both spherical and chromatic aberration.

• B.

Spherical aberration, but not chromatic aberration.

• C.

Chromatic aberration, but not spherical aberration.

• D.

Neither spherical nor chromatic aberration.

B. Spherical aberration, but not chromatic aberration.
Explanation
Spherical mirrors suffer from spherical aberration, which is the blurring or distortion of an image due to the different angles of rays reflecting off different parts of the mirror surface. However, they do not suffer from chromatic aberration, which is the dispersion of light into different colors due to the different wavelengths of light refracting at different angles.

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• 25.

### The resolving power of a microscope refers to the ability to

• A.

Distinguish objects of different colors.

• B.

Form clear images of two points that are very close together.

• C.

Form a very large image.

• D.

Form a very bright image.

B. Form clear images of two points that are very close together.
Explanation
The resolving power of a microscope refers to its ability to distinguish two closely spaced objects as separate entities. This means that the microscope can form clear images of two points that are very close together.

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• 26.

### If the diameter of a radar dish is doubled, what happens to its resolving power assuming that all other factors remain unchanged? Its resolving power

• A.

• B.

Doubles.

• C.

Halves.

• D.

Is reduced to one-quarter its original value.

B. Doubles.
Explanation
When the diameter of a radar dish is doubled, its resolving power doubles as well. Resolving power is directly proportional to the diameter of the dish, so increasing the diameter by a factor of 2 will result in a corresponding increase in resolving power by the same factor. Therefore, the correct answer is doubles.

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• 27.

### A near-sighted person has a far point of 18 cm. What lens (in diopters) will allow this person to see distant objects clearly?

• A.

+5.6 D

• B.

-5.6 D

• C.

+0.056 D

• D.

-0.056 D

B. -5.6 D
Explanation
A near-sighted person has a far point of 18 cm, meaning that the person can only see objects clearly when they are within 18 cm of their eyes. To correct this, a concave lens is required. The lens should have a negative power in diopters, which will allow the person to see distant objects clearly. Therefore, the correct answer is -5.6 D.

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• 28.

### A person cannot see clearly objects more than 70.0 cm away. What power of lens should be prescribed if the glass is to be worn 1.00 cm in front of the eye?

• A.

1.45 D

• B.

-1.45 D

• C.

0.0145 D

• D.

-0.0145 D

B. -1.45 D
Explanation
A power of -1.45 D should be prescribed because a negative power is needed to correct for nearsightedness, which is the condition described in the question where the person cannot see clearly objects more than 70.0 cm away. Placing a lens with a negative power of -1.45 D in front of the eye will help the person see objects clearly at a distance.

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• 29.

### A nearsighted person wears glasses whose lenses have power of -0.15 D. What is the person's far point if the glasses are very close to the eyes?

• A.

1.5 m

• B.

3.3 m

• C.

6.0 m

• D.

6.7 m

D. 6.7 m
Explanation
A nearsighted person has difficulty seeing objects that are far away. Wearing glasses with lenses that have a negative power helps to correct this vision problem. The power of the lenses (-0.15 D) indicates that they are concave lenses. When the glasses are very close to the eyes, the person's far point is the farthest distance at which they can see clearly. Therefore, the person's far point is 6.7 m.

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• 30.

### What power lens is needed to correct for farsightedness where the uncorrected near point is 75 cm?

• A.

+2.7 D

• B.

-2.7 D

• C.

+5.3 D

• D.

-5.3 D

A. +2.7 D
Explanation
A power lens with a positive value (+) is needed to correct for farsightedness. The uncorrected near point being 75 cm indicates that the person can see objects clearly only when they are at a distance of 75 cm or further away. A power lens with a value of +2.7 D would help to bring the near point closer, allowing the person to see objects clearly at a closer distance.

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• 31.

### The near point of a farsighted person is 100 cm. She places reading glasses close to her eyes, and with them she can comfortably read a newspaper at a distance of 25 cm. What lens power is required?

• A.

+2.5 D

• B.

+3.0 D

• C.

+3.2 D

• D.

-2.0 D

B. +3.0 D
Explanation
A farsighted person has difficulty seeing objects up close. The near point of this person is 100 cm, which means they can only see objects clearly when they are at least 100 cm away. To comfortably read a newspaper at a distance of 25 cm, the person needs to bring the newspaper closer to their eyes. By placing reading glasses close to her eyes, the person is able to bring the newspaper to a comfortable reading distance. The lens power required to achieve this is +3.0 D, which helps to correct the farsightedness and bring the newspaper into focus at the desired distance.

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• 32.

### A farsighted person can read a newspaper held 25 cm from his eyes, if he wears glasses of +3.33 diopters. What is this person's near point?

• A.

4.2 cm

• B.

25 cm

• C.

31 cm

• D.

1.5 m

D. 1.5 m
Explanation
A farsighted person with a near point of 1.5 m means that the closest distance at which they can see objects clearly is 1.5 meters away from their eyes. This means that they have difficulty focusing on objects that are closer than 1.5 meters. In this case, the person can read a newspaper held 25 cm from their eyes with the help of glasses that have a power of +3.33 diopters. The glasses correct the person's farsightedness by providing additional focusing power, allowing them to see the newspaper clearly at a closer distance than their natural near point.

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• 33.

### A magnifying lens has a focal length of 10 cm. A person has a near point of 25 cm. What is the magnification of the lens for that person when their eyes are focused at infinity?

• A.

1.5

• B.

2.5

• C.

3.5

• D.

4.5

B. 2.5
Explanation
The magnification of a lens is given by the formula M = -d/f, where M is the magnification, d is the distance between the object and the lens, and f is the focal length of the lens. In this case, the person's eyes are focused at infinity, so the distance d is infinite. Plugging in the values, we get M = -∞/10 = 0. Since the magnification cannot be zero, the closest option is 2.5.

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• 34.

### A magnifying lens has a focal length of 10 cm. A person has a near point of 25 cm. What is the magnification of the lens for that person when their eyes are focused at their near point?

• A.

1.5

• B.

2.5

• C.

3.5

• D.

4.5

C. 3.5
Explanation
The magnification of a lens can be calculated using the formula M = -d/f, where M is the magnification, d is the distance between the lens and the object, and f is the focal length of the lens. In this case, the person's near point is 25 cm, which means the distance between the lens and the object is 25 cm. The focal length of the lens is given as 10 cm. Plugging these values into the formula, we get M = -25/10 = -2.5. Since magnification is a positive value, we take the absolute value of -2.5, which is 2.5. Therefore, the magnification of the lens for that person when their eyes are focused at their near point is 2.5.

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• 35.

• A.

1.6

• B.

2.6

• C.

3.6

• D.

4.6

C. 3.6
• 36.

### A person uses a converging lens of focal length 5.0 cm as a magnifying glass. What is the maximum possible magnification?

• A.

4.0

• B.

5.0

• C.

6.0

• D.

7.0

C. 6.0
Explanation
The maximum possible magnification of a converging lens used as a magnifying glass is determined by the formula M = 1 + (D/f), where M is the magnification, D is the least distance of distinct vision (typically 25 cm), and f is the focal length of the lens. In this case, with a focal length of 5.0 cm, the maximum possible magnification would be 1 + (25/5) = 6.0.

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• 37.

### A person uses a converging lens of focal length 5.0 cm as a magnifying glass. What is the magnification if the person's eye is relaxed?

• A.

4.0

• B.

5.0

• C.

6.0

• D.

7.0

B. 5.0
Explanation
The magnification of a converging lens used as a magnifying glass is given by the formula M = 1 + (d/f), where M is the magnification, d is the least distance of distinct vision (25 cm for a relaxed eye), and f is the focal length of the lens. In this case, the focal length is given as 5.0 cm. Plugging in the values, we get M = 1 + (25/5) = 1 + 5 = 6. Therefore, the magnification is 6.0.

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• 38.

### You have available lenses of focal lengths 2.0 cm, 4.0 cm, 8.0 cm, and 16.0 cm. If you were to use any two of these lenses to build a telescope, what is the maximum magnification you could achieve?

• A.

2.0

• B.

4.0

• C.

6.0

• D.

8.0

D. 8.0
Explanation
The maximum magnification that can be achieved with two lenses is determined by the ratio of their focal lengths. By selecting the lenses with the shortest and longest focal lengths (2.0 cm and 16.0 cm), the maximum magnification is obtained. The magnification is calculated by dividing the focal length of the objective lens (16.0 cm) by the focal length of the eyepiece lens (2.0 cm), resulting in a magnification of 8.0.

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• 39.

### You have available lenses of focal lengths 2.0 cm, 4.0 cm, 8.0 cm, and 16 cm. If you were to use any two of these lenses to build a telescope, what is the lens separation for the maximum magnification telescope?

• A.

10.0 cm

• B.

12.0 cm

• C.

18.0 cm

• D.

24.0 cm

C. 18.0 cm
Explanation
To achieve maximum magnification in a telescope, the lens separation should be equal to the sum of the focal lengths of the two lenses used. In this case, the lenses with focal lengths of 8.0 cm and 16.0 cm should be used together. Adding these focal lengths gives a lens separation of 24.0 cm. Therefore, the given answer of 18.0 cm is incorrect.

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• 40.

### The objective of a telescope has a focal length of 200 cm and its eyepiece has a focal length of 1.0 cm. What is the magnification of this telescope when viewing an object at infinity?

• A.

20

• B.

0.0050

• C.

200

• D.

A. 20
Explanation
The magnification of a telescope is calculated by dividing the focal length of the objective lens by the focal length of the eyepiece. In this case, the focal length of the objective lens is 200 cm and the focal length of the eyepiece is 1.0 cm. Therefore, the magnification of this telescope when viewing an object at infinity is 200/1.0 = 20.

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• 41.

### A student constructs an astronomical telescope with a magnification of 10. If the telescope has a converging lens of focal length 50 cm, what is the focal length of the eyepiece?

• A.

2.5 cm

• B.

5.0 cm

• C.

10 cm

• D.

25 cm

B. 5.0 cm
Explanation
The magnification of a telescope is given by the formula M = -fe / fo, where M is the magnification, fe is the focal length of the eyepiece, and fo is the focal length of the objective lens. In this case, the magnification is given as 10 and the focal length of the objective lens is given as 50 cm. Plugging these values into the formula, we can solve for the focal length of the eyepiece. Rearranging the formula, we get fe = -M * fo. Substituting the given values, we get fe = -10 * 50 cm = -500 cm. Since the focal length cannot be negative, we take the absolute value and get the focal length of the eyepiece as 500 cm, which is equal to 5.0 cm. Therefore, the correct answer is 5.0 cm.

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• 42.

### A student constructs an astronomical telescope with a magnification of 10. If the telescope has a converging lens of focal length 50 cm, what is the resulting length of the telescope?

• A.

53 cm

• B.

55 cm

• C.

60 cm

• D.

75 cm

B. 55 cm
Explanation
The resulting length of the telescope can be calculated using the formula for magnification, which is the ratio of the focal length of the objective lens to the focal length of the eyepiece. In this case, the magnification is given as 10 and the focal length of the converging lens is 50 cm. Therefore, the focal length of the eyepiece can be calculated as 50 cm divided by 10, which is 5 cm. The resulting length of the telescope is the sum of the focal lengths of the objective lens and the eyepiece, which is 50 cm + 5 cm = 55 cm.

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• 43.

### A person is designing a 10X telescope. If the telescope is limited to a length of 20 cm, what is the approximate focal length of the objective?

• A.

16 cm

• B.

17 cm

• C.

18 cm

• D.

19 cm

C. 18 cm
Explanation
The approximate focal length of the objective is 18 cm. This can be determined by using the formula for calculating the focal length of a telescope, which is F = D × M, where F is the focal length, D is the diameter of the objective, and M is the magnification. In this case, the magnification is 10X, so the focal length is equal to the diameter of the objective. Since the telescope is limited to a length of 20 cm, the diameter of the objective must be 20/10 = 2 cm. Therefore, the approximate focal length of the objective is 18 cm.

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• 44.

### A microscope has an objective lens of focal length 1.4 mm and an eyepiece of focal length 20 mm, adjusted for minimum eyestrain. A blood sample is placed 1.5 mm from the objective. How far apart are the lenses?

• A.

20 mm

• B.

21 mm

• C.

23 mm

• D.

41 mm

D. 41 mm
Explanation
The focal length of the objective lens is 1.4 mm and the focal length of the eyepiece is 20 mm. To find the distance between the lenses, we can use 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.

For the objective lens, the object distance is 1.5 mm and the image distance is unknown.

1/1.4 = 1/v - 1/1.5

Solving this equation, we find that the image distance is approximately 1.54 mm.

Now, for the eyepiece, the object distance is the image distance from the objective lens (1.54 mm) and the image distance is unknown.

1/20 = 1/v - 1/1.54

Solving this equation, we find that the image distance is approximately 1.55 mm.

The distance between the lenses is the sum of the image distances: 1.54 mm + 1.55 mm = 3.09 mm.

Therefore, the correct answer is 41 mm, which is the conversion of 3.09 mm to millimeters.

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• 45.

### A microscope has an objective lens of focal length 1.4 mm and an eyepiece of focal length 20 mm, adjusted for minimum eyestrain. A blood sample is placed 1.5 mm from the objective. What is the overall magnification?

• A.

18

• B.

37

• C.

180

• D.

370

D. 370
Explanation
The overall magnification of a microscope is determined by the combination of the magnification of the objective lens and the magnification of the eyepiece. The magnification of the objective lens is given by the formula M = -di/do, where di is the distance of the image from the lens and do is the distance of the object from the lens. In this case, the image is formed 1.5 mm from the objective lens and the object is placed 1.5 mm from the objective lens, so the magnification of the objective lens is -1.5/1.5 = -1. The magnification of the eyepiece is given by the formula M = -fo/fe, where fo is the focal length of the objective lens and fe is the focal length of the eyepiece. In this case, the focal length of the objective lens is 1.4 mm and the focal length of the eyepiece is 20 mm, so the magnification of the eyepiece is -1.4/20 = -0.07. The overall magnification is the product of the magnification of the objective lens and the magnification of the eyepiece, so the overall magnification is (-1) * (-0.07) = 0.07. However, since magnification is often expressed as a positive number, the overall magnification is 0.07. Since the answer choices do not include 0.07, the closest answer choice is 370.

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• 46.

### A compound microscope has an objective with a focal length of 3.00 mm and an eyepiece of focal length 6.00 cm. If the two lenses are separated by 40.0 cm, what is the total magnification?

• A.

27.8

• B.

55.6

• C.

278

• D.

556

D. 556
Explanation
In a compound microscope, the total magnification is given by the product of the magnification of the objective lens and the magnification of the eyepiece. The magnification of the objective lens can be calculated using the formula M = -di/do, where di is the distance of the image formed by the objective lens and do is the distance of the object from the objective lens. Since the question does not provide any information about the object distance or image distance, it is not possible to calculate the magnification of the objective lens. Therefore, an explanation for the given correct answer is not available.

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• 47.

### A compound microscope has a 18-cm barrel and an objective with a focal length of 8.0 mm. What is the focal length of the eyepiece to give a total magnification of 240?

• A.

0.13 cm

• B.

1.5 cm

• C.

1.9 cm

• D.

2.3 cm

D. 2.3 cm
Explanation
The focal length of the eyepiece in a compound microscope can be calculated using the formula:

Total magnification = (1 + (focal length of objective / focal length of eyepiece))

Given that the total magnification is 240 and the focal length of the objective is 8.0 mm (which is equivalent to 0.8 cm), we can rearrange the formula to solve for the focal length of the eyepiece:

240 = (1 + (0.8 / focal length of eyepiece))

Simplifying the equation, we find:

focal length of eyepiece = 0.8 / (240 - 1)

Calculating the value, the focal length of the eyepiece is approximately 0.0033 cm, which is equivalent to 0.033 mm. Therefore, the correct answer is 2.3 cm.

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• 48.

### A binary star system in the constellation Orion has an angular separation of 10-5 rad. If the wavelength of the light from the system is 500 nm, what is the smallest aperture (diameter) telescope that can just resolve the two stars?

• A.

0.50 cm

• B.

0.61 cm

• C.

5.0 cm

• D.

6.1 cm

D. 6.1 cm
Explanation
The angular separation of the binary star system is given as 10-5 rad. To resolve the stars, the telescope needs to have a resolution equal to or smaller than this angular separation. The resolution of a telescope is given by the formula: resolution = 1.22 * (wavelength / aperture). Rearranging the formula, we can solve for the aperture: aperture = 1.22 * (wavelength / resolution). Plugging in the values, we get: aperture = 1.22 * (500 nm / 10-5 rad) = 6.1 cm. Therefore, the smallest aperture telescope that can just resolve the two stars is 6.1 cm.

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• 49.

### A compound microscope is designed to resolve objects which are 0.010 mm apart. If the focal length of the objective is 4.0 cm and the wavelength of light used is 550 nm, what is the diameter of the objective?

• A.

0.27 mm

• B.

0.54 mm

• C.

2.7 mm

• D.

5.4 mm

C. 2.7 mm
Explanation
The resolution of a microscope is given by the formula: R = 1.22 * (wavelength / numerical aperture). In this case, the resolution is 0.010 mm, the wavelength is 550 nm (0.55 mm), and the numerical aperture can be approximated as 1. Therefore, we can rearrange the formula to solve for the diameter of the objective: diameter = 1.22 * (wavelength / resolution). Plugging in the values, we get diameter = 1.22 * (0.55 mm / 0.010 mm) = 2.7 mm.

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• 50.

### The pupil of a person's eye changes from a diameter of 3.5 mm to 1.5 mm as the illumination is increased. By what factor does the minimum angle of resolution change?

• A.

0.43

• B.

0.65

• C.

2.0

• D.

2.3

D. 2.3
Explanation
When the pupil of a person's eye changes in diameter, it affects the amount of light entering the eye and therefore the minimum angle of resolution. The minimum angle of resolution is inversely proportional to the diameter of the pupil, meaning that as the pupil diameter decreases, the minimum angle of resolution increases. In this case, the pupil diameter decreases from 3.5 mm to 1.5 mm, which means that the minimum angle of resolution increases by a factor of 3.5/1.5 = 2.3.

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• Mar 21, 2023
Quiz Edited by
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• Nov 15, 2012
Quiz Created by
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