Refraction and Lenses Review
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Questions #1#20
Questions #21#34
Questions #35#46
Part A: Multiple Choice
1. The best definition of refraction is ____.
a. passing through a boundary

b. bouncing off a boundary

c. changing speed at a boundary

d. changing direction when crossing a boundary

Answer: D
Bouncing off a boundary (choice b) is reflection. Refraction involves passing through a boundary (choice a) and changing speed (choice c); however, a light ray can exhibit both of these behaviors without undergoing refraction (for instance, if it approaches the boundary along the normal). Refraction of light must involve a change in direction; the path must be altered at the boundary.

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2. If carbon tetrachloride has an index of refraction of 1.461, what is the speed of light through this liquid? (c = 3 x 10^{8} m/s)
a. 4.38 x 10^{8} m/s

b. 2.05 x 10^{8} m/s

c. 4.461 x 10^{8} m/s

d. 1.461 x 10^{8} m/s

Answer: B
Use the equation v=c/n where n = 1.461 and c = speed of light in a vacuum (3x10^{8} m/s).

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3. A ray of light in air is incident on an airtoglass boundary at an angle of 30. degrees with the normal. If the index of refraction of the glass is 1.65, what is the angle of the refracted ray within the glass with respect to the normal?
a. 56 degrees

b. 46 degrees

c. 30. degrees

d. 18 degrees

Answer: D
Use Snell's law:
n_{i} * sine(Theta i) = n_{r} * sine(Theta r)
where
n_{i }=1.00 (in air), Theta i=30. degrees, n_{r }=1.65
Substitute and solve for Theta r.
sine(Theta r) = 1.00 * sine(30. degrees) / 1.65 = 0.3030
Theta r = invsin(0.3030) = 17.6 degrees

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4. If the critical angle for internal reflection inside a certain transparent material is found to be 48.0 degrees, what is the index of refraction of the material? (Air is outside the material).
a. 1.35

b. 1.48

c. 1.49

d. 0.743

Answer: A
The critical angle is the angle of incidence (which is always in the more dense material) for which the angle of refraction is 90 degrees.
Apply this to Snell's law equation:
n_{r} * sine(48.0 deg) = 1.00 * sine (90 deg)
Solve for n_{r}.
n_{r }= 1.00 / sin(48.0 deg) = 1.35

5. Carbon disulfide (n = 1.63) is poured into a container made of crown glass (n = 1.52). What is the critical angle for internal reflection of a ray in the liquid when it is incident on the liquidtoglass surface?
a. 89.0 degrees

b. 68.8 degrees

c. 21.2 degrees

d. 4.0 degrees

Answer: B
The critical angle is the angle of incidence (which is always in the more dense material) for which the angle of refraction is 90 degrees.
Apply this to Snell's law equation:
1.63* sine(Theta critical.) = 1.52 * sine (90)
Solve for Theta critical.
sine(Theta critical.) = 1.52/1.63 = 0.9325
Theta critical. = invsine(0.9325) = 68.8 degrees

6. Carbon tetrachloride (n = 1.46) is poured into a container made of crown glass (n = 1.52). If the light ray in glass incident on the glasstoliquid boundary makes an angle of 30 degrees with the normal, what is the angle of the corresponding refracted ray with respect to the normal?
a. 55.5 degrees

b. 29.4 degrees

c. 31.4 degrees

d. 19.2 degrees

Answer: C
Use Snell's law:
n_{i }* sine(Theta i) = n_{r} * sine(Theta r)
where
n_{i }=1.52 (in glass), Theta i=30 degrees (angle in glass), n_{r }=1.46 (in carbon tetrachloride)
Substitute and solve for Theta r
1.52 * sin(30 deg) = 1.46 * sin(Theta r)
[1.52 * sin(30 deg)]/1.46 = sin(Theta r)
0.5205 = sin(Theta r)
Theta r = invsine(0.5205) = 31.4 degrees

7. A light ray in air is incident on an air to glass boundary at an angle of 45.0 degrees and is refracted in the glass of 30.0 degrees with the normal. What is the index of refraction of the glass?
a. 2.13

b. 1.74

c. 1.23

d. 1.41

Answer: D
Use Snell's law:
n_{i} * sine(Theta i) = n_{r} * sine(Theta r)
where
n_{i}=1.00 (in air), Theta i=45.0 degrees (angle in air), Theta r=30.0 degrees (in glass)
Substitute and solve for n_{r}.
1.00 * sine(45.0 deg) = n_{r} * sine(30.0 deg)
1.00 * sine(45.0 deg) / sine(30.0 deg) = n_{r}
n_{r }= 1.41

8. A beam of light in air is incident at an angle of 35 degrees to the surface of a rectangular block of clear plastic (n = 1.5). The light beam first passes through the block and reemerges from the opposite side into air at what angle to the normal to that surface?
a. 42 degrees

b. 23 degrees

c. 35 degrees

d. 59 degrees

Answer: C
The light ray bends towards the normal upon entering and away from the normal upon exiting. If the opposite sides are parallel to each other and surrounded by the same material, then the angle at which the light enters is equal to the angle at which the light exits.

9. A light ray in air enters and passes through a block of glass. What can be stated with regard to its speed after it emerges from the block?
a. speed is less than when in glass

b. speed is less than before it entered glass

c. speed is same as that in glass

d. speed is same as that before it entered glass

Answer: D
The speed of a light wave (like any wave) is dependent upon the medium through which it moves. In the case of a light wave, the speed is least in the most dense medium. Thus, the light moves slower in glass than in air. However, upon exiting the glass and entering the air, the light returns to the original speed. The speed is the same for the same medium.

10. Which of the following describes what will happen to a light ray incident on an airtoglass boundary?
a. total reflection

b. total transmission

c. partial reflection, partial transmission

d. partial reflection, total transmission

Answer: C
Upon reaching a boundary, a wave undergoes both reflection and transmission. The only exception is for light in the more dense medium and at angles of incidence greater than the critical angle; in such a case, total internal reflection occurs. Since the question does not specify such conditions, one would have to answer c.

11. Which of the following describes what will happen to a light ray incident on an airtoglass boundary at an angle of incidence less than the critical angle?
a. total reflection

b. total transmission

c. partial reflection, partial transmission

d. partial reflection, total transmission

Answer: C
At a boundary between any two materials, there will be both reflection and transmission. The only exception is when the light is incident in the more dense material and at an incident angle greater than the critical angle.

12. Which of the following describes what will happen to a light ray incident on an glasstoair boundary at an angle of incidence greater than the critical angle?
a. total reflection

b. total transmission

c. partial reflection, partial transmission

d. partial reflection, total transmission

Answer: A
At a boundary between any two materials, there will be both reflection and transmission. The only exception is when the light is incident in the more dense material and at an incident angle greater than the critical angle. Since the light in this problem is in the more dense medium (glass) and at an angle greater than the critical angle, total internal reflection will occur. No transmission will occur at this boundary for such angles.

13. What is the angle of incidence on an airtoglass boundary if the angle of refraction in the glass (n = 1.52) is 25 degrees?
a. 16 degrees

b. 25 degrees

c. 40 degrees

d. 43 degrees

Answer: C
Use Snell's law:
n_{i} * sine(Theta i) = n_{r} * sine(Theta r)
where
n_{i }=1.52 (in glass), Theta i=25 degrees (angle in glass), n_{r }=1.00 (in air)
Substitute and solve for Theta r.
1.52 * sine(25 degrees) = 1.00 * sine(Theta r)
1.52 * sine(25 degrees) / 1.00 = sine(Theta r)
0.0.6424 = sine(Theta r)
Theta r = invsine(0.6424) = 40.0 degrees

14. A ray of white light, incident upon a glass prism, is dispersed into its various color components. Which one of the following colors experiences the greatest amount of refraction?
a. orange

b. violet

c. red

d. green

Answer: B
The shorter wavelengths of light undergo the most refraction. Thus, violet is refracted the most and red light is refracted the least. The fact that the various component colors of white light refract different amounts leads to the phenomenon of dispersion.

15. When light from air hits a smooth piece of glass (n = 1.5) with the ray perpendicular to the glass surface, which of the following will occur?
a. reflection and transmission at an angle of 0 degrees

b. dispersion

c. refraction at an angle of 41.8 degrees

d. all of the above will occur

Answer: A
A portion of the light is reflected and a portion of the light is transmitted into the new medium. Since the angle of incidence is 0 degrees, there is no bending of the ray. Noticeable dispersion only occurs when there is refraction of light at two consecutive boundaries which are nonparallel.

16. If total internal reflection occurs at a glassair surface, then _____.

no light is refracted

no light is reflected

light is leaving the air and hitting the glass with an incident angle greater than the critical angle

light is leaving the air and hitting the glass with an incident angle less than the critical angle
Answer: A
Total internal reflection occurs only when light passes from a more dense medium to a less dense medium (this is why c and d can be ruled out) at an angle of incidence greater than the critical angle. When TIR occurs, all the light is reflected and none of the light is refracted.

17. When light from air hits a smooth piece of glass with the ray perpendicular to the glass surface, the part of the light passing into the glass _____.
a. will not change its speed

b. will not change its direction

c. will not change its wavelength

d. will not change its intensity

Answer: B
When the angle of incidence is 0 degrees (as in this case), there is no bending. The ray still slows down and changes its wavelength. A portion of the wave reflects and so there is a change in intensity within the new medium. Yet, there is no refraction or bending; the direction does not change.

For Questions #18  #20, consider the diagram below.
18. This person suffers from the problem of ____.
a. nearsightedness

b. farsightedness

c. cataracts

d. delusions

Answer: B
A farsighted person has difficulty seeing objects which are placed nearby. The images of nearby objects forms behind the retina as shown in this case. Nearsighted people have the opposite problem in that the images from very distant objects form in front of the retina.

19. This problem could most easily be corrected by the use of a(n) ____.
a. converging lens

b. diverging lens

c. achromatic lens

d. good night's sleep

Answer: A
Since the image of this farsighted individual is forming behind the retina, an artificial lens with increased converging ability will cause the image to form closer to the lens and upon the retina. Farsighted individuals correct for their vision defect through the use of converging lenses.

20. If the image was formed in front of the retina rather than behind the retina, then the person would need to correct the vision problem by using a
a. converging lens

b. diverging lens

c. achromatic lens

d. alarm clock

Answer: B
Nearsighted individuals suffer from an image formed in front of the retina. They must correct for the problem by wearing an artificial lens which provides for some diverging of light prior to reaching the lens of the eye. This will move the image further from the lens of the eye and back towards the retina.

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