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Part A
1. C, A, F, I, B, K, E, D
2. C
3. B
4. D
5. B
6. A
7a. D
7b. A
7c. B
Part B
1a. theta_2 = 31.7 degrees (Snell's Law)
theta_3 = 90 - theta_2 = 58.3 degrees
theta_4 = 58.3 degrees (Law of Reflection)
theta_5 = 90 - theta_4 = 31.7 degrees
theta_6 = 60.0 degrees (Snell's Law)
1b. critical angle = 37.3 degrees
If the ray is incident at the glass-air interface at an angle
greater than the critical angle, it cannot leave the glass. It
is forced to stay inside the glass: Total Internal Reflection.
2a. Starting at central peak and moving outward, maxima
are m = 0,1,2,3. Starting at first minimum on one side of central
peak, minima are m = 1,2,3.
2b. Use any one of the minima to find its theta:
theta_1 = 0.191 degrees; theta_2 = 0.382 degrees; theta_3 = 0.573
degrees; etc.
Then use any one of the above theta values to find: D = 1.7 x
10^-4 m or 0.17 mm
3a. d_image = 38.9 mm
3b. m = -0.389
3c. h_image = -29 mm, which is smaller than 35 mm, so:
YES.