# hw20 - would be d mD y m ∆ = ∆ In this case m 10 5 m 10...

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PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS CLASS 20. 35.02, 35.08, 35.21; Q36.01, Q36.02 35.02. a) f = c / λ = (3.00 × 10 8 m/s)/(589 × 10 -9 m) f = 5.09 × 10 14 Hz b) n = / n = (589 × 10 -9 m)/1.52 n = 3.88 × 10 -7 m = 388 nm ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ c) v = n f = (3.88 × 10 -7 m)(5.09 × 10 14 Hz) v = 1.97 × 10 8 m/s 35.08. The time needed for a pulse to travel through a layer of thickness L is given by, c L n n c L v L t = = = / , where n is the index of refraction of the layer. For pulse 1, c L c L c L c L c L t 33 . 6 ) 50 . 1 65 . 1 18 . 3 ( 50 . 1 65 . 1 2 59 . 1 1 = + + = + + = For pulse 2, c L c L c L c L c L t 30 . 6 45 . 1 60 . 1 70 . 1 55 . 1 2 = + + + = a) From the numbers, it is clear that, pulse 2 takes less time . b) The difference is, c L c L t 30 . 6 33 . 6 - = c L t 03 . 0 = 35.21. According to equation 35.17, on page 967, the position of a maximum on the screen is given by, d D m y m = . Then for two different wavelengths, the separation of the maxima

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Unformatted text preview: would be, d mD y m ∆ = ∆ . In this case, m 10 . 5 m) 10 480 m 10 600 m)( . 1 ( 3 3 9 9 3---× ×-× = ∆ y ∆ y 3 = 7.2 ×10-5 m = 0.072 mm Q36.01. a) a sin θ = m λ . This formula gives the location of the minima of the diffraction pattern, where m is the path difference between the top and bottom rays, and m is an integer. Therefore if m = 5, a minimum will appear on the screen. b) If m = 4.5, it is midway between two minima, so it will result in a bright area on the screen. Q36.02. a) a sin = m . Increasing f will decrease , so the pattern will contract toward the center . b) The corn syrup will decrease , so the pattern will contract toward the center ....
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hw20 - would be d mD y m ∆ = ∆ In this case m 10 5 m 10...

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