Chap33 - 33.1 a s m 10 04 2 47 1 s m 10 00 3 8 8 = = = n c v b m 10 42 4 47 1 m 10 50 6 7 7 = = = n 33.2 a m 10 17 5 Hz 10 5.80 s m 10 00 3 7 14 8

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Unformatted text preview: 33.1: a) . s m 10 04 . 2 47 . 1 s m 10 00 . 3 8 8 = = = n c v b) . m 10 42 . 4 47 . 1 ) m 10 50 . 6 ( 7 7-- = = = n 33.2: a) . m 10 17 . 5 Hz 10 5.80 s m 10 00 . 3 7 14 8 vacuum- = = = f c b) . m 10 40 . 3 Hz)(1.52) 10 80 . 5 ( s m 10 00 . 3 7 14 8 glass- = = = fn c 33.3: a) . 54 . 1 m/s 10 1.94 m/s 10 00 . 3 8 8 = = = v c n b) . m 10 47 . 5 ) m 10 55 . 3 ( ) 54 . 1 ( 7 7-- = = = n 33.4: 1.501 m)(1.333) 10 38 . 4 ( 7 Benzene water water Benzene Benzene water water 2- = = = n n n n CS . 5 . 47 equal always are angles reflected and Incident a) = = a r : 33.5 b) . . 66 5 . 42 sin 66 . 1 00 . 1 arcsin 2 sin arcsin 2 2 = - = - =- = a b a b b n n 33.6: s m 10 17 . 2 s 10 5 . 11 m 50 . 2 8 9 = = =- t d v 38 . 1 m/s 10 2.17 m/s 10 00 . 3 8 8 = = = v c n 33.7: b b a a n n sin sin = s m 10 51 . 2 194 . 1 / ) m/s 10 00 . 3 ( so 194 . 1 1 . 48 sin 7 . 62 sin 00 . 1 sin sin 8 8 = = = = = = = n c v v c n n n b a a b 33.8 (a) Apply Snells law at both interfaces. 33.9: a) Let the light initially be in the material with refractive index n a and let the third and final slab have refractive index n b Let the middle slab have refractive index n 1 1 1 sin sin : interface 1st n n a a = b b n n sin sin : interface 2nd 1 1 = . sin sin gives equations two the Combining b b a a n n = b) For N slabs, where the first slab has refractive index n a and the final slab has . sin sin , , sin sin , sin sin , index refractive 2 2 2 2 1 1 1 1 b b N N a a b n n n n n n n = = =-- of angle on the depends travel of direction final The . sin sin gives This b b a a n n = incidence in the first slab and the indicies of the first and last slabs. 33.10: a) . 5 . 25 . 35 sin 33 . 1 00 . 1 arcsin sin arcsin air water air water = = = n n b) This calculation has no dependence on the glass because we can omit that step in the . sin sin sin : chain water wate glass glass air air r n n n = = 33.11: As shown below, the angle between the beams and the prism is A/2 and the angle between the beams and the vertical is A, so the total angle between the two beams is 2A. 33.12: Rotating a mirror by an angle while keeping the incoming beam constant leads rotation. mirror the from arose 2 of deflection additional an where 2 2 becomes beams outgoing and incoming between angle the Therefore . by angle incident in the increase an to + 33.13: . 71.8 62.0 sin 1.58 1.70 arcsin sin arcsin = = = a b a b n n 33.14: 38.2. 45.0 sin 1.52 1.33 arcsin sin arcsin = = = a b a b n n...
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This homework help was uploaded on 04/08/2008 for the course PHY 205 taught by Professor Nearing during the Fall '07 term at University of Miami.

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Chap33 - 33.1 a s m 10 04 2 47 1 s m 10 00 3 8 8 = = = n c v b m 10 42 4 47 1 m 10 50 6 7 7 = = = n 33.2 a m 10 17 5 Hz 10 5.80 s m 10 00 3 7 14 8

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