p235w11_practice_exam_3_2_sol

# What is the width of the slit a b c d e 35 m 42 m 93

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Unformatted text preview: 3.1 m 4 cm This is the distance to the second minimum in a single slit diffraction pattern. The angles to these minima are given by sinmin = n/dslit. So this width of the slit is given by dslit = n/sinmin. For this, the angle is tanmin = 0.04/1.5 = 1.53. This implies a slit width of 3.4x10-5 m, or about the value in answer a. 10: (5 points) In the picture below, a man holds a converging lens 10 cm in front of his eye. Which of the following is a possible value for the focal length of this lens? Explain your answer. a. b. c. d. e. 50 cm 30 cm 3 cm 12 cm 8 cm This image is upright and substantially magnified. To be upright, it must be within the focal length of the lens. To be substantially magnified, it must be relatively close to the focal length (the magnification goes to 1 when d0 goes to zero). This makes 12 or 30 cm the best choices, while 50 cm is also at least possible… 11: (5 points) Consider the table of X-ray absorption lengths given below. Imagine the Xrays pass through 2 cm of fat, 4 cm of bone, and 3 more cm of fat. What fraction of the incident rays get through? I I 0 e ( 0.02/0.0520.04/0.017 0.03/0.052) 0.036 12: (5 points) Light from a very distant source passes through a converging lens and an image of the object is formed very near the focal point of the lens. a: (3 points) As I move the object closer to the lens, in which direction does...
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## This test prep was uploaded on 02/15/2014 for the course PHYSICS 235 taught by Professor Mckay during the Spring '08 term at University of Michigan.

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