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Test Prep

# Test Prep

### Multiple Choice

#### 16.1Reflection

1.

In geometric optics, a straight line emerging from a point is called a (an) ________.

1. ray
2. focal point
3. image
4. object distance
2.

An image of a 2.0 -cm object reflected from a mirror is 5.0 cm tall. What is the magnification of the mirror?

1. 0.4
2. 2.5
3. 3
4. 10
3.

Can a virtual image be projected onto a screen with additional lenses or mirrors? Explain your answer.

1. Yes, the rays actually meet behind the lens or mirror.
2. No, the image is formed by rays that converge to a point in front of the mirror or lens.
3. Yes, any image that can be seen can be manipulated so that it can be projected onto a screen.
4. No, the image can only be perceived as being behind the lens or mirror.

#### 16.2Refraction

4.

What does c represent in the equation $n=cv$?

1. the critical angle
2. the refractive index
3. the speed of light in a vacuum
4. the speed of light in a transparent material
5.

What is the term for the minimum angle at which a light ray is reflected back into a material and cannot pass into the surrounding medium?

1. critical angle
2. incident angle
3. angle of refraction
4. angle of reflection
6.

Consider these indices of refraction: glass: 1.52, air: 1.0003, water: 1.333. Put these materials in order from the one in which the speed of light is fastest to the one in which it is slowest.

1. The speed of light in water > the speed of light in air > the speed of light in glass.
2. The speed of light in glass > the speed of light in water > the speed of light in air.
3. The speed of light in air > the speed of light in water > the speed of light in glass.
4. The speed of light in glass > the speed of light in air > the speed of light in water.
7.
Explain why an object in water always appears to be at a depth that is more shallow than it actually is.
1. Because of the refraction of light, the light coming from the object bends toward the normal at the interface of water and air. This causes the object to appear at a location that is above the actual position of the object. Hence, the image appears to be at a depth that is more shallow than the actual depth.
2. Because of the refraction of light, the light coming from the object bends away from the normal at the interface of water and air. This causes the object to appear at a location that is above the actual position of the object. Hence, the image appears to be at a depth that is more shallow than the actual depth.
3. Because of the refraction of light, the light coming from the object bends toward the normal at the interface of water and air. This causes the object to appear at a location that is below the actual position of the object. Hence, the image appears to be at a depth that is more shallow than the actual depth.
4. Because of the refraction of light, the light coming from the object bends away from the normal at the interface of water and air. This causes the object to appear at a location that is below the actual position of the object. Hence, the image appears to be at a depth that is more shallow than the actual depth.

#### 16.3Lenses

8.

For a given lens, what is the height of the image divided by the height of the object ($hiho$) equal to?

1. power
2. focal length
3. magnification
9.

Which part of the eye has the greatest density of light receptors?

1. the lens
2. the fovea
3. the optic nerve
4. the vitreous humor
10.

What is the power of a lens with a focal length of 10 cm?

1. 10 m–1, or 10 D
2. 10 cm–1, or 10 D
3. 10 m, or 10 D
4. 10 cm, or 10 D
11.
Describe the cause of chromatic aberration.
1. Chromatic aberration results from the dependence of the frequency of light on the refractive index, which causes dispersion of different colors of light by a lens so that each color has a different focal point.
2. Chromatic aberration results from the dispersion of different wavelengths of light by a curved mirror so that each color has a different focal point.
3. Chromatic aberration results from the dependence of the reflection angle at a spherical mirror’s surface on the distance of light rays from the principal axis so that different colors have different focal points.
4. Chromatic aberration results from the dependence of the wavelength of light on the refractive index, which causes dispersion of different colors of light by a lens so that each color has a different focal point.

#### 16.1Reflection

12.

Distinguish between reflection and refraction in terms of how a light ray changes when it meets the interface between two media.

1. Reflected light penetrates the surface whereas refracted light is bent as it travels from one medium to the other.
2. Reflected light penetrates the surface whereas refracted light travels along a curved path.
3. Reflected light bounces from the surface whereas refracted light travels along a curved path.
4. Reflected light bounces from the surface whereas refracted light is bent as it travels from one medium to the other.
13.
Sometimes light may be both reflected and refracted as it meets the surface of a different medium. Identify a material with a surface that when light travels through the air it is both reflected and refracted. Explain how this is possible.
1. Light passing through air is partially reflected and refracted when it meets a glass surface. It is reflected because glass has a smooth surface; it is refracted while passing into the transparent glass.
2. Light passing through air is partially reflected and refracted when it meets a glass surface. It is reflected because glass has a rough surface, and it is refracted while passing into the opaque glass.
3. Light passing through air is partially reflected and refracted when it meets a glass surface. It is reflected because glass has a smooth surface; it is refracted while passing into the opaque glass.
4. Light passing through air is partially reflected and refracted when it meets a glass surface. It is reflected because glass has a rough surface; it is refracted while passing into the transparent glass.
14.

A concave mirror has a focal length of 5.00 cm. What is the image distance of an object placed 7.00 cm from the center of the mirror?

1. −17.5 cm
2. −2.92 cm
3. 2.92 cm
4. 17.5 cm
15.

An 8.0 -cm tall object is placed 6.0 cm from a concave mirror with a magnification of –2.0. What are the image height and the image distance?

1. hi = – 16 cm, di = – 12 cm
2. hi = – 16 cm, di = 12 cm
3. hi = 16 cm, di = – 12 cm
4. hi = 16 cm, di = 12 cm

#### 16.2Refraction

16.
At what minimum angle does total internal reflection of light occur if it travels from water $(n=1.33)$ toward ice $(n=1.31)$?
1. $44.6∘$
2. $26.5∘$
3. $13.3∘$
4. $80.1∘$
17.

Water floats on a liquid called carbon tetrachloride. The two liquids do not mix. A light ray passing from water into carbon tetrachloride has an incident angle of 45.0° and an angle of refraction of 40.1°. If the index of refraction of water is 1.33, what is the index of refraction of carbon tetrachloride?

1. 1.60
2. 1.49
3. 1.21
4. 1.46
18.

Describe what happens to a light ray when it is refracted. Include in your explanation comparison of angles, comparison of refractive indices, and the term normal.

1. When a ray of light goes from one medium to another medium with a different refractive index, the ray changes its path as a result of interference. The angle between the ray and the normal (the line perpendicular to the surfaces of the two media) is greater in the medium with the greater refractive index.
2. When a ray of light goes from one medium to another medium with a different refractive index, the ray changes its path as a result of refraction. The angle between the ray and the normal (the line perpendicular to the surfaces of the two media) is less in the medium with the greater refractive index.
3. When a ray of light goes from one medium to another medium with a different refractive index, the ray does not change its path. The angle between the ray and the normal (the line parallel to the surfaces of the two media) is the same in both media.
4. When a ray of light goes from one medium to another medium with a different refractive index, the ray changes its path as a result of refraction. The angle between the ray and the normal (the line perpendicular to the surfaces of the two media) is less in the medium with the lower refractive index.

#### 16.3Lenses

19.

What are two equivalent terms for a lens that always causes light rays to bend away from the principal axis?

1. a diverging lens or a convex lens
2. a diverging lens or a concave lens
3. a converging lens or a concave lens
4. a converging lens or a convex lens
20.

Define the term virtual image.

1. A virtual image is an image that cannot be projected onto a screen.
2. A virtual image is an image that can be projected onto a screen.
3. A virtual image is an image that is formed on the opposite side of the lens from where the object is placed.
4. A virtual image is an image that is always bigger than the object.
21.
Compare nearsightedness (myopia) and farsightedness (hyperopia) in terms of focal point.
1. The eyes of a nearsighted person have focal points beyond the retina. A farsighted person has eyes with focal points between the lens and the retina.
2. A nearsighted person has eyes with focal points between the lens and the retina. A farsighted person has eyes with focal points beyond the retina.
3. A nearsighted person has eyes with focal points between the lens and the choroid. A farsighted person has eyes with focal points beyond the choroid.
4. A nearsighted person has eyes with focal points between the lens and the retina. A farsighted person has eyes with focal points on the retina.
22.
Explain how a converging lens corrects farsightedness.
1. A converging lens disperses the rays so they focus on the retina.
2. A converging lens bends the rays closer together so they do not focus on the retina.
3. A converging lens bends the rays closer together so they focus on the retina.
4. A converging lens disperses the rays so they do not focus on the retina.
23.
Solve the equation $1do+1di=1f$ for $f$ in such a way that it is not expressed as a reciprocal.
1. $f=do+didodi$
2. $f=dodidi+do$
3. $f=(di+do)$
4. $f=dodi$
24.

What is the magnification of a lens if it produces a 12-cm-high image of a 4 -cm -high object? The image is virtual and erect.

1. $–3.00–3.00$
2. $–13.00–13.00$
3. $13.0013.00$
4. $3.003.00$

### Extended Response

#### 16.1Reflection

25. The diagram shows a light bulb between two mirrors. One mirror produces a beam of light with parallel rays; the other keeps light from escaping without being put into the beam.
Figure 16.40
Where is the light source in relation to the focal point or radius of curvature of each mirror? Explain your answer.
1. The bulb is at the center of curvature of the small mirror and at the focal point of the large mirror.
2. The bulb is at the focal point of the small mirror and at the focal point of the large mirror.
3. The bulb is at the center of curvature of the small mirror and at the center of curvature of the large mirror.
4. The bulb is at the focal point of the small mirror and at the center of curvature of the large mirror.
26.
An object is placed $4.00cm$ in front of a mirror that has a magnification of $1.50$. What is the radius of curvature of the mirror?
1. $−24.0cm$
2. $−4.80cm$
3. $4.80cm$
4. $24.0cm$

#### 16.2Refraction

27. A scuba diver training in a pool looks at his instructor, as shown in this figure. The angle between the ray in the water and the normal to the water is 25°.
Figure 16.41
What angle does the ray make from the instructor’s face with the normal to the water $(n=1.33)$ at the point where the ray enters? Assume $n=1.00$ for air.
1. $68∘$
2. $25∘$
3. $19∘$
4. $34∘$
28.

Describe total internal reflection. Include a definition of the critical angle and how it is related to total internal reflection. Also, compare the indices of refraction of the interior material and the surrounding material.

1. When the interior material has a smaller index of refraction than the surrounding material, the incident ray may approach the boundary at an angle (called the critical angle) such that the refraction angle is 90° . The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs.
2. When the interior material has a smaller index of refraction than the surrounding material, the incident ray may approach the boundary at an angle (called the critical angle) such that the refraction angle is less than 90° . The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs.
3. When the interior material has the same index of refraction as the surrounding material, the incident ray approaches the boundary at an angle (called the critical angle) such that the refraction angle is less than 90° . The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs.
4. When the interior material has a greater index of refraction than the surrounding material, the incident ray may approach the boundary at an angle (called the critical angle) such that the refraction angle is 90° . The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs.

#### 16.3Lenses

29.

The muscles that change the shape of the lens in the eyes have become weak, causing vision problems for a person. In particular, the muscles cannot pull hard enough on the edges of the lens to make it less convex.

Part A—What condition does inability cause?

Part B—Where are images focused with respect to the retina?

Part C—Which type of lens corrects this person’s problem? Explain.

1. Part A—This condition causes hyperopia.

Part B—Images are focused between the lens and the retina.

Part C—A converging lens gathers the rays slightly so they focus onto the retina.
2. Part A—This condition causes myopia.

Part B—Images are focused between the lens and the retina.

Part C—A converging lens gathers the rays slightly so they focus onto the retina.
3. Part A—This condition causes hyperopia.

Part B—Images are focused between the lens and the retina.

Part C—A diverging lens spreads the rays slightly so they focus onto the retina.
4. Part A—This condition causes myopia.

Part B—Images are focused between the lens and the retina.

Part C—A diverging lens spreads the rays slightly so they focus onto the retina.
30.
If the lens-to-retina distance is $2.00cm$, what is the power of the eye when viewing an object $50.0cm$ away?
1. $−52.0D$
2. $0.52D$
3. $1.92D$
4. $52.0D$