WHomework set 13_solns_2011

WHomework set 13_solns_2011 - Homework set # 13 Problems:...

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Homework set # 13 Problems: 22.6, 9, 10, 11, 19, 17 Solutions: 22.6 (a) From geometry, 1.25 m sin40.0 d , so 1.94 m d (b) 50.0 above horizontal , or parallel to the incident ray 22.9 (a) From Snell’s law,   11 2 2 1.00 sin30.0 sin 1.52 sin sin19.24 n n  (b) 0 2 2 632.8 nm 417 nm 1.52 n (c) 8 14 9 0 3.00 10 m s 4.74 10 Hz 632.8 10 m c f in air and in syrup (d) 8 8 2 2 1.98 10 m s 1.52 c v n
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22.10 (a) When light refracts from air   1 1.00 n into the Crown glass, Snell’s law gives the angle of refraction as 1 2 sin sin 25.0 Crown glass n     For first quadrant angles, the sine of the angle increases as the angle increases. Thus, from the above equation, note that 2 will increase when the index of refraction of the Crown glass decreases. From Figure 22.12, this means that the longer wavelengths have the largest angles of refraction, and deviate the least from the original path. (b) From Figure 22.12, observe that the index of refraction of Crown glass for the given wavelengths is: 400 nm: 1.53
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WHomework set 13_solns_2011 - Homework set # 13 Problems:...

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