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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 3.4x105 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 Xray 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.
 Spring '08
 Mckay

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