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06 February 2015 Name:
Physics 1112
Exam #1A-MWF
Instructions:
This is a closed-book, closed-notes exam. You will be given a separate sheet of formulas and numerical data that you may consult. Nothing written on the formula sheet will be graded.
There is space after each question to show your work; if you need more space, you may use the back of the page, or request more paper. Please clearly indicate where your work for each problem is. Underline or draw a box around your final answer.
The exam consists of four sections. I recommend that you read all the questions at the start so that you can allocate your time wisely.
You may use a scientific calculator for arithmetic only; your calculator must be non-graphing, non-programmable, and non-algebraic. You are not allowed to share your calculator.
The use of cell phones, PDAs, or any other electronic devices (besides calculators) is forbid- den. All such gadgets must be turned off and put away throughout the exam.
- Do not open the exam until told to begin.
- You have one MWF class period (50 minutes) to finish the exam.
- Make sure every page of the exam has your last name.
- You must provide explanations and/or show work legibly to receive full credit for Parts III–IV.
- Make sure that your answers include appropriate units and significant digits.
- Assume that the index of refraction of air is the same as for a vacuum, unless otherwise stated.
By signing below, you indicate that you understand the instructions for this exam and agree to abide by them. You also certify that you will personally uphold the university’s standards of academic honesty for this exam, and will not tolerate any violations of these standards by others. Unsigned exams will not be graded.
Signature:
UGACard #: Row/Seat:
Copyright ©c 2015 University of Georgia. Unauthorized duplication or distribution prohibited.
06 February 2015 Last Name:
Part I II III IV Total Score Grade
Score /24 /16 /32 /28 /100 /
I: Order/Choice Questions (24 points)
For each question below, write your responses in capital letters in the space provided.
- Which of the following ray(s) is/are definitely not drawn correctly? If none are incor- rectly drawn, write “none”. Assume a thin glass lens in air.
C,E
B B
A
A
C
D E
F C
F
D E
(12 pts, 2 for each correct ray listed and not listed) A concave glass lens in air will be a diverging lens, which means that it will bend rays away from the central axis if they do not pass through the center of the lens. Because it is a thin lens, rays that pass through the center should not be deflected.
- A ray of light passes from glass into air, striking the surface at normal incidence (0◦). Of the following quantities, choose all that change as the light enters the air. If none change, write “none”. A. Wavelength B. Frequency C. Propagation speed D. Propagation direction
A,C
(12 pts, 3 for each correct quantity listed and not listed) Because the index of refraction is defined based on the speed of propagation in a material, the propagation speed will, by definition, change as the ray enters a material with a different index. As discussed in class, the wavelength will also change, but the frequency will not. Because the ray is striking the surface at zero angle of incidence (normal incidence), the direction of propagation will not change.
06 February 2015 Last Name:
II: Multiple-Choice Questions (Explanations)
- A laser beam coming from air into a block of higher-index material will bend towards the normal, and a smaller index of refraction in the material will cause the light to bend less, yielding a larger angle of refraction.
- The length of a telescope is given by the sum of the focal lengths of the lenses:
L = fobj + feye.
The magnification is determined by the ratio of the focal lengths:
M = −fobj /feye.
In order to increase L while maintaining a constant ratio, both fobj and feye must be increased.
- Because the star is moving away from Earth, we know the observed frequency must be smaller than the source frequency. We can use the frequency version of the Doppler shift formula to determine the ratio:
fobs = fsrc
( 1 −
u c
) =⇒
fobs fsrc
u c
- 20 c c
- An accurate ray diagram (rather than a hasty sketch) could be used to demonstrate this. Alternatively, you could reason from the lens equation. We know that diverging lenses form virtual images of real objects, which means that di will be negative, and we also know that the focal length is negative. The lens equation tells us that
1 f
do
di
di
f
do
Since inverting that equation is a little messy, instead let’s consider the inverse of the magnification. 1 M
do di
= −do
( 1 f
do
) = 1 −
do f
do |f | Decreasing do will decrease 1/M , which means that M itself must increase.
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III: Mirror (32 points)
An object is placed in front of a mirror, and the image formed on the other side of the mirror is four times as large as the object. The object and image are separated by a distance of D = 100. cm.
(a) Is the image real or virtual? Is it upright or inverted? Briefly explain your answers. (8 pts) Because the image is formed on the other side of the mirror from the real object, it must be virtual. This also implies that the image distance di is a negative number. Thus the magnification, M = −
di do is positive, which means the image is upright.
(b) Calculate the distance between the object and the mirror. (8 pts) Let’s call the unknown object distance x. Because the total distance between object and image is D, the image distance can be written as di = −(D − x) = x − D. (The image distance is supposed to be positive based on part (a).) Now, let’s use the magnification to solve for the object distance:
M = −
di do
−(D − x) x
D
x
D
x
= 1 + M =⇒ x =
D
1 + M
Putting in the numbers given,
x =
100 cm 1 + 4
= 20. 0 cm.
(c) Calculate the focal length of the mirror. (8 pts) Knowing the object distance from part (b), the image distance is
di = x − D = 20. 0 cm) − 100 cm = − 80. 0 cm.
The focal length can then be found from the mirror equation:
1 f
do
di
- 0 cm
- 0 cm
- 0 cm
f =
- 0 cm 3
= 26. 7 cm.
06 February 2015 Last Name:
IV: Prism (28 points)
A beam of red light in air is incident on a water-filled prism in the shape of a 45◦-90◦-45◦ triangle. The prism’s index of refraction for the red light is nR = 1.320. Upon entering the prism, the red light is directed perfectly horizontally across the prism, as shown in the figure.
(a) Calculate the angle of incidence for the incidence for the red light.
(10 pts) Because the ray is horizontal inside the prism, it forms a smaller 45 ◦- 90 ◦- 45 ◦^ triangle with the sides of the prism. From this, we can see that the angle of refraction is θR = 45◦. Knowing the angle of refraction, we can use Snell’s Law to calculate the angle of incidence for the red light,
ni sin θi = nR sin θR =⇒ sin θi =
nR sin θR ni
Putting in numbers, we obtain
sin θi =
1 .320 sin 45. 0 ◦
- 000
= 0. 9334 =⇒ θi = 68. 97 ◦.
45°
θ 45°^ 45° 45° 45°
θ (^) e
(b) For violet light, the index of refraction of the prism is nV = 1.350. If a beam of violet light strikes the prism at the same angle of incidence, find the dispersion angle, ∆θ, between the refracted red and violet beams. Is the refracted violet beam above or below the red beam? (10 pts) The violet light ought to bend more toward the normal than the red light, because of the higher index of refraction. We can find the angle of refraction by applying Snell’s Law with the angle of incidence from part (a).
ni sin θi = nV sin θV =⇒ sin θV =
ni sin θi nV
=⇒ θV = 43. 74 ◦.
Now we just take the difference between the two angles to find the dispersion angle:
∆θ = θR − θV = 1. 26 ◦
Since θV is smaller than θR, the violet ray is below the red ray, as we thought.
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(c) Calculate the angle of refraction for the red light when it exits the right-hand side of the prism. (8 pts) This can be done without further calculation. From the picture, we can see that the angle of incidence is 45 ◦, the same as the angle of refraction at the first interface. Snell’s Law is not dependent on direction, so the angle of refraction for the exiting ray is the same as the initial angle of incidence at the first interface. That is,
nR sin θR = ne sin θe =⇒ θe = θi = 68. 97 ◦
