## Chapter Review

### Concept Items

#### 16.1Reflection

1.

Part A. Can you see a virtual image? Part B. Can you photograph one? Explain your answers.

1. A. yes; B. No, an image from a flat mirror cannot be photographed.
2. A. no; B. Yes, an image from a flat mirror can be photographed.
3. A. yes; B. Yes, an image from a flat mirror can be photographed.
4. A. no; B. No, an image from a flat mirror cannot be photographed.
2.
State the law of reflection.
1. $ϑr=ϑi$, where $ϑr$ is the angle of reflection and $ϑi$ is the angle of incidence.
2. $ϑr>ϑi$, where $ϑr$ is the angle of reflection and $ϑi$ is the angle of incidence.
3. $ϑr<ϑi$, where $ϑr$ is the angle of reflection and $ϑi$ is the angle of incidence.
4. $ϑr=0$, where $ϑr$ is the angle of reflection.

#### 16.2Refraction

3.
Does light change direction toward or away from the normal when it goes from air to water? Explain.
1. The light bends away from the normal because the index of refraction of water is greater than that of air.
2. The light bends away from the normal because the index of refraction of air is greater than that of water.
3. The light bends toward the normal because the index of refraction of water is greater than that of air.
4. The light bends toward the normal because the index of refraction of air is greater than that of water.

#### 16.3Lenses

4.

An object is positioned in front of a lens with its base resting on the principal axis. Describe two rays that could be traced from the top of the object and through the lens that would locate the top of an image.

1. A ray perpendicular to the axis and a ray through the center of the lens
2. A ray parallel to the axis and a ray that does not pass through the center of the lens
3. A ray parallel to the axis and a ray through the center of the lens
4. A ray parallel to the axis and a ray that does not pass through the focal point
5.
A person timing the moonrise looks at her watch and then at the rising moon. Describe what happened inside her eyes that allowed her to see her watch clearly one second and then see the moon clearly.
1. The shape of the lens was changed by the sclera, and thus its focal length was also changed, so that each of the images focused on the retina.
2. The shape of the lens was changed by the choroid, and thus its focal length was also changed, so that each of the images focused on the retina.
3. The shape of the lens was changed by the iris, and thus its focal length was also changed, so that each of the images focused on the retina.
4. The shape of the lens was changed by the muscles, and thus its focal length was also changed, so that each of the images focused on the retina.
6.

For a concave lens, if the image distance, di, is negative, where does the image appear to be with respect to the object?

1. The image always appears on the same side of the lens.
2. The image appears on the opposite side of the lens.
3. The image appears on the opposite side of the lens only if the object distance is greater than the focal length.
4. The image appears on the same side of the lens only if the object distance is less than the focal length.

### Critical Thinking Items

#### 16.1Reflection

7.
Why are diverging mirrors often used for rear-view mirrors in vehicles? What is the main disadvantage of using such a mirror compared with a flat one?
1. It gives a wide range of view. The image is formed closer than the actual object.
2. It gives a narrow range of view. The image is formed farther than the actual object.
3. It gives a narrow range of view. The image formed is closer than the actual object.
4. It gives a wide range of view. The image is formed farther than the actual object.

#### 16.2Refraction

8.
A high-quality diamond may be quite clear and colorless, transmitting all visible wavelengths with little absorption. Explain how it can sparkle with flashes of brilliant color when illuminated by white light.
1. Diamond and air have a small difference in their refractive indices that results in a very small critical angle. The light that enters a diamond may exit at only a few points, and these points sparkle because many rays have been directed toward them.
2. Diamond and air have a small difference in their refractive indices that results in a very large critical angle. The light that enters a diamond may exit at only a few points, and these points sparkle because many rays have been directed toward them.
3. Diamond has a high index of refraction with respect to air, which results in a very small critical angle. The light that enters a diamond may exit at only a few points, and these points sparkle because many rays have been directed toward them.
4. Diamond has a high index of refraction with respect to air, which results in a very large critical angle. The light that enters a diamond may exit at only a few points, and these points sparkle because many rays have been directed toward them.
9.
The most common type of mirage is an illusion in which light from far-away objects is reflected by a pool of water that is not really there. Mirages are generally observed in deserts, where there is a hot layer of air near the ground. Given that the refractive index of air is less for air at higher temperatures, explain how mirages can be formed.
1. The hot layer of air near the ground is lighter than the cooler air above it, but the difference in refractive index is small, which results in a large critical angle. The light rays coming from the horizon strike the hot air at large angles, so they are reflected as they would be from water.
2. The hot layer of air near the ground is lighter than the cooler air above it, and the difference in refractive index is large, which results in a large critical angle. The light rays coming from the horizon strike the hot air at large angles, so they are reflected as they would be from water.
3. The hot layer of air near the ground is lighter than the cooler air above it, but the difference in refractive index is small, which results in a small critical angle. The light rays coming from the horizon strike the hot air at large angles, so they are reflected as they would be from water.
4. The hot layer of air near the ground is lighter than the cooler air above it, and the difference in the refractive index is large, which results in a small critical angle. The light rays coming from the horizontal strike the hot air at large angles, so they are reflected as they would be from water.

#### 16.3Lenses

10.
When you focus a camera, you adjust the distance of the lens from the film. If the camera lens acts like a thin lens, why can it not be kept at a fixed distance from the film for both near and distant objects?
1. To focus on a distant object, you need to increase the image distance.
2. To focus on a distant object, you need to increase the focal length of the lens.
3. To focus on a distant object, you need to decrease the focal length of the lens.
4. To focus on a distant object, you may need to increase or decrease the focal length of the lens.
11.

Part A—How do the refractive indices of the cornea, aqueous humor, and the lens of the eye compare with the refractive index of air?

Part B—How do the comparisons in part A explain how images are focused on the retina?

1. (A) The cornea, aqueous humor, and lens of the eye have smaller refractive indices than air.

(B) Rays entering the eye are refracted away from the central axis, which causes them to meet at the focal point on the retina.
2. (A) The cornea, aqueous humor, and lens of the eye have greater refractive indices than air.

(B) Rays entering the eye are refracted away from the central axis, which causes them to meet at the focal point on the retina.
3. (A) The cornea, aqueous humor, and lens of the eye have smaller refractive indices than air.

(B) Rays entering the eye are refracted toward the central axis, which causes them to meet at the focal point on the retina.
4. (A) The cornea, aqueous humor, and lens of the eye have greater refractive indices than air.

(B) Rays entering the eye are refracted toward the central axis, which causes them to meet at the focal point on the retina.

### Problems

#### 16.1Reflection

12.

Some telephoto cameras use a mirror rather than a lens. What radius of curvature is needed for a concave mirror to replace a 0.800 -m focal-length telephoto lens?

1. 0.400 m
2. 1.60 m
3. 4.00 m
4. 16.0 m
13.

What is the focal length of a makeup mirror that produces a magnification of 2.00 when a person’s face is 8.00 cm away?

1. –16 cm
2. –5.3 cm
3. 5.3 cm
4. 16 cm

#### 16.2Refraction

14.

An optical fiber uses flint glass (n = 1.66) clad with crown glass (n = 1.52) . What is the critical angle?

1. 33.2°
2. 23.7°
3. 0.92 rad
4. 1.16 rad
15. Suppose this figure represents a ray of light going from air (n = 1.0003) through crown glass (n = 1.52) into water, similar to a beam of light going into a fish tank.
Figure 16.39

Calculate the amount the ray is displaced by the glass (Δx) , given that the incident angle is 40.0° and the glass is 1.00 cm thick.

1. 0.839 cm
2. 0.619 cm
3. 0.466 cm
4. 0.373 cm

#### 16.3Lenses

16.

A camera’s zoom lens has an adjustable focal length ranging from 80.0 to 200 mm . What is its range of powers?

1. The lowest power is 0.05 D and the highest power is 0.125 D.
2. The lowest power is 0.08 D and the highest power is 0.20 D.
3. The lowest power is 5.00 D and the highest power is 12.5 D.
4. The lowest power is 80 D and the highest power is 200 D.
17.

Suppose a telephoto lens with a focal length of 200 mm is being used to photograph mountains 10.0 km away. (a) Where is the image? (b) What is the height of the image of a 1,000-m-high cliff on one of the mountains?

1. (a) The image is 0.200 m on the same side of the lens. (b) The height of the image is – 2.00 cm.
2. (a) The image is 0.200 m on the opposite side of the lens. (b) The height of the image is – 2.00 cm.
3. (a) The image is 0.200 m on the opposite side of the lens. (b) The height of the image is +2.00 cm.
4. (a) The image is 0.100 m on the same side of the lens. (b) The height of the image is +2.00 cm.

### Performance Task

#### 16.3Lenses

18.

In this performance task, you will investigate the lens-like properties of a clear bottle.

Materials
• a water bottle or glass with a round cross-section and smooth, vertical sides
• enough water to fill the bottle
• a meter stick or tape measure
• a bright light source with a small bulb, such as a pen light
• a small bright object, such as a silver spoon.

Instructions

Procedure
1. Look through a clear glass or plastic bottle and describe what you see.
2. Next, fill the bottle with water and describe what you see.
3. Use the water bottle as a lens to produce the image of a bright object.
4. Estimate the focal length of the water bottle lens.
1. How can you find the focal length of the lens using the light and a blank wall?
2. How can you find the focal length of the lens using the bright object?
3. Why did the water change the lens properties of the bottle?
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