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# ch22-sol - Chapter 22 Solution 22.9(a From Snells law n2 =...

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Chapter 22 Solution 22.9 (a) From Snell’s law, ( ) 11 2 2 1.00 sin30.0 sin 1.52 si ns i n19.24 n n θ ° == = ° (b) 0 2 2 632.8 nm 417 nm n λ = (c) 8 14 9 0 3.00 10 m s 4.74 10 Hz 632. 81 0 m c f × = × × in air and in syrup (d) 8 8 2 2 1.98 10 m s c v n × = × 22.11 (a) 0 water n = , so () ( ) 0 1.333 438 nm 584 nm water w ater n λλ = (b) 0 ben zen e ben zene nn and 438 nm 1.12 390 nm benzene w ater n n = 22.13 From Snell’s law, ( ) 2 2 1.00 sin40.0 i n2 9 . ° 4 1.309 n n −− ⎡⎤ ° = ⎢⎥ ⎣⎦ ± ±²³ ²´µ ² ·¸¹¸ ³ ´ and from the law of reflection, 1 40.0 φ = Hence, the angle between the reflected and refracted rays is 2 180 29. 44 0 . ° 01 1 1 αθ ° = ° 22.15 The index of refraction of zircon is 1.923 n = (a) 8 8 1.56 10 m s c v n × = × (b) The wavelength in the zircon is 0 329.1 nm n n =

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(c) The frequency is 8 14 -9 0 3.00 10 m s 4.74 10 m s 632. 81 0 m n vc f λλ × === = × × 22.21 From Snell’s law, the angle of incidence at the air-oil interface is () 1 1 sin 1.48 sin20.0 si n3 4 1.00 oi lo i l air n n θ ⎡⎤ = ⎢⎥ ⎣⎦ ° == 0 . ° and the angle of refraction as the light enters the water is ( ) 11 ns i n 22.3 1.333 i l water n n
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ch22-sol - Chapter 22 Solution 22.9(a From Snells law n2 =...

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