961_PartUniversity Physics Solution

# 961_PartUniversity Physics Solution - The Nature and...

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The Nature and Propagation of Light 33-19 (b) Multiplying Eq. (1) by cos β and Eq. (2) by cos α yields: (1) : cos sin cos cos cos sin cos x tt a ωα =− and (2): cos sin cos cos cos sin cos y a ω αω . Subtracting yields: cos cos cos (sin cos sin cos ). xy t a βα ββ (c) Squaring and adding the results of parts (a) and (b) yields: 22 2 2 ( sin sin ) ( cos cos ) (sin cos sin cos ) x y a αβ α α −+ = (d) Expanding the left-hand side, we have: 2 2 (sin cos ) (sin cos ) 2 (sin sin cos cos ) 2 (sin sin cos cos ) 2 cos( ). x y x y xy x y xy αα α β ++ +− + =+− + The right-hand side can be rewritten: 2 2 (sin cos sin cos ) sin ( ). aa −= Therefore, 2 2 2c o s ( ) s i n ( ) . xy x y a Or, 2 2 o s s i n, w h e r e . y a δ δδ + = EVALUATE: (e) 2 0: 2 ( ) 0 , x yx y x y x y =+ = = = which is a straight diagonal line 2 : 2 , which is an ellipse 42 a x y π 222 : ,which is a circle. 2 xya = This pattern repeats for the remaining phase differences. 33.66. IDENTIFY: Apply Snell’s law to each refraction. SET UP: Refer to the figure that accompanies the problem. EXECUTE: (a) By the symmetry of the triangles, ,and . AB CBBA ba arab θ θθθθ = === Therefore, sin sin sin sin . CCA A C A bab a b a nn θθ = = (b) The total angular deflection of the ray is 2 2 4 . AA BCC A A ab aba a b θπθθθ θ θπ Δ =−+− +−= − + (c) From Snell&s Law, sin 1 sin arcsin sin A A b a n n ⎛⎞ =⇒ = ⎜⎟ ⎝⎠ . 1 24 a r c s i n s i n . A A a a n θπθ θ π Δ= + = + (d) 1 2 1 2 14 c o s 024 a r c s i n s i n 02 sin 1 A a dd n n n Δ⎛ ⎞ ==− ⇒=− . 11 sin 16cos 41 . 2 4cos 1 cos n + . 1 3cos 1 n . 1 1 cos ( 1). 3 n = (e) For violet: 1 arccos ( 1) arccos (1.342 58.89 33 n = = ° . violet violet 139.2 40.8 . ° = ° For red: 1 arccos ( arccos (1.330 59.58 n = = ° . red red 137.5 42.5 . Δ = ° EVALUATE: The angles we have calculated agree with the values given in Figure 37.20d in the textbook. 1 is larger for red than for violet, so red in the rainbow is higher above the horizon. 33.67. IDENTIFY: Follow similar steps to Challenge Problem 33.66. SET UP: Refer to Figure 33.20e in the textbook. EXECUTE: The total angular deflection of the ray is 2 6 2 , A AAA A A a bbb a b a b θπθπθθθ θ θ π + − + = + where we have used the fact from the previous problem that all the internal angles are equal and the two external equals are equal. Also using the Snell&s Law

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961_PartUniversity Physics Solution - The Nature and...

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