sample optics problem

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.

2. If carbon tetrachloride has an index of refraction of 1.461, what is the speed of light through this liquid? (c = 3 x 108m/s)

a. 4.38 x 108 m/s b. 2.05 x 108 m/s c. 4.461 x 108 m/s d. 1.461 x 108 m/s

Answer: B

Use the equation v=c/n where n = 1.461 and c = speed of light in a vacuum (3x108 m/s).

3. A ray of light in air is incident on an air-to-glass 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:

ni * sine(Theta i) = nr * sine(Theta r)

where

ni =1.00 (in air), Theta i=30. degrees, nr =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

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:

nr * sine(48.0 deg) = 1.00 * sine (90 deg)

Solve for nr.

nr = 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 liquid-to-glass 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 glass-to-liquid 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:


where

ni =1.52 (in glass), Theta i=30 degrees (angle in glass), nr =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:


where

ni=1.00 (in air), Theta i=45.0 degrees (angle in air), Theta r=30.0 degrees (in glass)

Substitute and solve for nr.

1.00 * sine(45.0 deg) = nr * sine(30.0 deg)

1.00 * sine(45.0 deg) / sine(30.0 deg) = nr

nr = 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 air-to-glass 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 air-to-glass boundary at an angle of incidence less than the critical angle?



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 glass-to-air boundary at an angle of incidence greater than the critical angle?



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 air-to-glass 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:


where

ni =1.52 (in glass), Theta i=25 degrees (angle in glass), nr =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 glass-air surface, then _____.

a. no light is refracted

b. no light is reflected

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

d. 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.

21. Dispersion occurs when _____.

a. some materials bend light more than other materials

b. a material slows down some colors more than others

c. a material changes some colors more than others

d. light has different speeds in different materials

Answer: B

Dispersion occurs when a single material bends certain colors of light more than other colors of light. The cause is the fact that different colors of light have different speeds within the same material.

22. A 3.0 cm tall object is placed along the principal axis of a thin converging lens of 30.0 cm focal length. If the object distance is 40.0 cm, which of the following best describes the image distance and height, respectively?

a. 17.3 cm and 7.0 cm b. 120. cm and -9.0 cm c. 17.3 cm and 1.3 cm d. 120. cm and -1.0 cm

Answer: B

Use the lens equation:1/di + 1/do = 1/fwhere do = 40.0 cm and f = 30.0 cm. Solve for di.1/di = 1/f - 1/do =1/(30.0 cm) - 1/(40.0 cm) = 0.00833 /cmdi = 1/(0.00833 /cm) = 120. cmThen use the -di/do = hi/ho to find hi

It is now known that ho = 3.0 cm; do = 40.0 cm; di = 120. cm. Substitute and solve.hi = -ho *(di/do) = -(3.0 cm) *(120. cm)/(40.0 cm) = -9.0 cm

23. Which of the following best describes the image for a thin converging lens that forms whenever the object is at a distance less than one focal length from the lens?

a. inverted, enlarged and real b. upright, enlarged and virtual

c. upright, reduced and virtual d. inverted, reduced and real

Answer: B

When an object is located inside of the focal point of a converging lens, the image will be virtual, upright, larger than the object and located on the same side of the lens as the object. In essence, the lens would serve as a magnifying glass.

24. Which of the following best describes the image for a thin diverging lens that forms whenever the magnitude of the object distance is less than that of the lens' focal length?

a. inverted, enlarged and real b. upright, enlarged and virtual

c. upright, reduced and virtual d. inverted, reduced and real

Answer: C

A diverging lens always produces an image with the same characteristics, regardless of the object distance. The image is always virtual, upright and reduced in size.

25. An object is placed at a distance of 30.0 cm from a thin converging lens along its axis. The lens has a focal length of 10.0 cm. What are the values of the image distance and magnification (respectively)?

a. 60.0 cm and 2.00 b. 15.0 cm and 2.00 c. 60.0 cm and -0.500 d. 15.0 cm and -0.500

Answer: D

Use the lens equation:1/di + 1/do = 1/fwhere do = 30.0 cm and f = 10.0 cm.Solve for di :1/di = 1/f - 1/do = 1/(10.0 cm) - 1/(30.0 cm) = 0.0666/cmdi = 1 / (0.0666/cm) = 15.0 cmThen use the M =-di/do to find M(do = 30.0 cm; di = 15.0 cm)Substitute and solve for M:M = -(15.0 cm) / (30.0 cm) = - 0.500

26. An object is placed at a distance of 6.0 cm from a thin converging lens along its axis. The lens has a focal length of 9.0 cm. What are the values, respectively, of the image distance and magnification?

a. -18 cm and 3.0 b. 18 cm and 3.0 c. 3.0 cm and -0.50 d. -18 cm and -3.0

Answer: AUse the lens equation:1/di + 1/do = 1/fwhere do = 6.0 cm and f = 9.0 cm.Solve for di:1/di = 1/f - 1/do = 1/(9.0 cm) - 1/(6.0 cm) = -0.0556/cmdi = 1 / (-0.0556/cm) = -18 cmThen use the M =-di/do to find M (do = 6.0 cm; di = -18 cm)Substitute and solve for M:M = -(-18 cm) / (6.0 cm) = 3.0

27. An object is placed at a distance of 30.0 cm from a thin converging lens along the axis. If a real image forms at a distance of 10.0 cm from the lens, what is the focal length of the lens?

a. 30.0 cm b. 15.0 cm c. 10.0 cm d. 7.50 cm

Answer: D

Use the lens equation:

1/di + 1/do = 1/f

where do = 30.0 cm and di = 10.0 cm.

Solve for f:

1/f = 1/di + 1/do = 1/(10.0 cm) + 1/(30.0 cm) = 0.133/cm

f = 1 / (0.133/cm) = 7.50 cm

28. An object is placed at a distance of 40.0 cm from a thin lens along the axis. If a virtual image forms at a distance of 50.0 cm from the lens, on the same side as the object, what is the focal length of the lens?

a. 22.2 cm b. 45.0 cm c. 90.0 cm d. 200. cm

Answer: D

Use the lens equation:

1/di + 1/do = 1/f

where do = 40.0 cm and di = -50.0 cm (Note that di is a negative number since it is a virtual image - i.e., formed on the same side of the lens as the object.)

Solve for f:

1/f = 1/di + 1/do = 1/(-50.0 cm) + 1/(40.0 cm) = 0.00500/cm

f = 1 / (0.00500/cm) = 200. cm

Part B: Multiple-Multiple Choice

29. Which of the following statements are true of converging lenses? Identify all that apply.

a. Converging lenses are thicker at the center than they are at the edges.

b. If the bottom half of a converging lens is covered, then the top half of the image will not be visible.

c. Converging lenses only produce real images.

d. Converging lenses can produce images which are both magnified and reduced in size.

e. Converging lenses only produce inverted images.

f. Converging lenses have a + focal length.

g. The images formed by a converging lens can be located on either side of the lens relative to the object.

Answer: ADFG

a. This is the basic physical feature that characterizes all converging lenses.

b. Find a pair of eyeglasses and see if you can test the truth of this statement. Covering half the lens will only have the effect of making the image fainter.

c. Converging lenses will produce a virtual image of an object placed in front of F.

d. Converging lenses produce magnified images when the object is in front of 2F and reduced images when the object is behind 2F.

e. Converging lenses can produce upright images of objects placed in front of F.

f. Focal length is + for converging lenses and - for diverging lenses.

g. Converging lenses produce both real images formed on the opposite side of the lens (when the object is placed

beyond F) and virtual images formed on the same side of the lens (when the object is placed in front of F).

30. Which of the following statements are true of diverging lenses? Identify all that apply.

a. Diverging lenses are thicker at the center than they are at the edges.

b. If the bottom half of a diverging lens is covered, then the bottom half of the image will not be visible.

c. Diverging lenses only produce virtual images.

d. Diverging lenses can produce images which are both magnified and reduced in size.

e. Diverging lenses only produce upright images.

f. Diverging lenses have a - focal length.

g. The images formed by a diverging lens can be located on either side of the lens relative to the object.

Answer: CEF

a. Diverging lenses would be thinner at the center and thickest along the top and bottom edges.

b. Just like the case of a converging lens, if half of a diverging lens is covered, the full image is still seen; it is merely fainter.

c. Always. A real image is never produced by a diverging lens.

d. Diverging lenses only produce one type of image - a virtual image which is upright and reduced in size.

e. See above statement.

f. Diverging lenses have a - focal length and converging lenses have a + focal length.

g. Diverging lenses only produce virtual images; these images are located on the object's side of the lens.

31. Which of the following statements are true of real images? Identify all that apply.

a. Real images are inverted.

b. Real images as formed by lenses are located on the opposite side of the lens from the object.

c. Real images are magnified in size.

d. Real images are only formed by converging lenses, never by diverging lenses.

e. An image of a real object is formed; the image distance (s' or di) for real images is a + value.

f. An image of a real object is formed; the image height (h' or hi) for real images is a + value.

g. Real images have a - magnification value.

Answer: ABDEG

a. This is always the case - real images are inverted and virtual images are upright.

b. Real images are always located on the opposite side of the lens; virtual images are located on the object's side of the lens.

c. Real images can be either magnified (converging lens, object between F and 2F), reduced (converging lens, object beyond 2F), or the same size (converging lens, object at 2F) as the object.

d. Diverging lenses can only produce virtual images; converging lenses can produce both virtual and real images.

e. Real images - those formed on the opposite side of the lens have a positive s' value (seep. 789 for more).

f. Real images are inverted; this corresponds to a negative h' value.

g. Real images are inverted; this corresponds to a negative magnification value.

32. Which of the following statements are true of virtual images? Identify all that apply.

a. Virtual images are always upright.

b. Virtual images as formed by lenses are always located on the same side of the lens as the object.

c. Virtual images are only formed by diverging lenses, never by converging lenses.

d. Virtual images are always smaller than the object.

e. An image of a real object is formed; the image distance (s' or di) for virtual images is a - value.

f. An image of a real object is formed; the image height (h' or hi) for virtual images is a - value.

g. Virtual images have a - magnification value.

Answer: ABE

a. Virtual images, whether formed by mirrors (of any type) or lenses, are always upright; real images are always inverted.

b. Virtual images are always located on the object's side of the lens; real images are always located on the opposite side of the lens.

c. Virtual images can be formed by both converging lenses (when the object is inside of F) and diverging lenses

(regardless of the object location).

d. Virtual images can be larger than the object (when formed by converging lenses) or smaller than the object (when formed by diverging lenses).

e. Virtual images are located on the object's side of the lens; this corresponds to a negative s' value.

f. Virtual images are always upright; this corresponds to a positive h' value.

g. Virtual images are always upright; this corresponds to a positive M value.

33. Several characteristics of images are described below. Determine whether these images are real or virtual and whether they are formed by converging, diverging lenses or either type. (In all cases, assume that the object is an upright and real object.)

a. Image is upright and magnified.

b. Image if upright and reduced in size.

c. Image is inverted and magnified.

d. Image has a negative s' (di) value.

e. Image has a negative h' (hi) value.

f. Image has a positive h' (hi) value and a magnification value greater than 1.

Answer: See answers below

a. Converging only; an upright (and virtual) and magnified image can only be formed when the object is between F and the surface of a converging lens. Diverging lenses would only produce upright images which are reduced in size.

b. Diverging only; an upright (and virtual) and reduced image can only be formed by a diverging lens. When converging lenses produce upright images, they are magnified in size.

c. Converging only; diverging lenses can only produce upright images which are reduced in size; an inverted and magnified image can be produced by a converging lens when the object is located between F and 2F.

d. Both lenses; a negative s' value corresponds to a virtual image; both converging and diverging lenses can produce virtual images.

e. Converging only; a negative h' value corresponds to an inverted (and real) image; only a converging lens can produce a real image.

f. Converging only; a positive h' value corresponds to an upright (and virtual) image; the M > 1 statement indicates that the image is magnified. A magnified, upright, virtual image can only be produced by a converging lens when the object is between F and 2F.

34. Which of the following statements are true of total internal reflection (TIR)? Include all that apply.

a. TIR can only occur when light approaches a boundary and is incident within the most dense media.

b. TIR can only occur when the angle of incidence is greater than the critical angle.

c. TIR causes a portion of the light to refract along the boundary and the rest to be reflected.

d. When TIR occurs, the reflected light follows the law of reflection.

e. If TIR occurs at the boundary of water and air, then the light must be within water and heading towards the boundary with air.

f. If TIR occurs at the boundary of glass and air, then it is possible that the light is traveling within air and heading towards the glass.

Answer: ABE

a. There are two conditions which must be met for TIR to occur; this is the first condition.

b. This is the second condition.

c. When TIR occurs, all (total) of the light reflects. There is no (nada, none, zero, zilch) refraction taking place.

d. Yes! Reflection always follows the law of reflection. The angle of incidence equals the angle of reflection.

e. TIR can only take place if the light is incident within the more dense of the two medium - in this case, water.

f. Never; air is the least dense of the two media; this would violate the first of the two conditions for TIR. TIR would only take place at this boundary if the light was in the glass and heading towards the boundary with air.

39. The speed of light in a vacuum is 3.00 x 108 m/s. Determine the speed of light through the following materials.

a. air (n=1.00)

b. ethanol (n=1.36)

c. crown glass (n=1.52)

d. zircon (n=1.91)

e. diamond (n=2.42)

Answers:

For each of these questions, find the speed of light in the material by dividing the 3.00 x108 m/s by the index of refraction of the material. This yields the following results:

a. (3.00 x108 m/s)/1.00 = 3 x 108 m/s

b. (3.00 x108 m/s)/1.36 = 2.21 x 108 m/s

c. (3.00 x108 m/s)/1.52 = 1.97 x 108 m/s

d. (3.00 x108 m/s)/1.91 = 1.57 x 108 m/s

e. (3.00 x108 m/s)/2.42 = 1.24 x 108 m/s

sample optics problem

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