P34_047 - 47 Consider a ray that grazes the top of the pole as shown in the diagram below Here θ1 = 35◦ 1 = 0.50 m and 2 = 1.50 m The length of

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Unformatted text preview: 47. Consider a ray that grazes the top of the pole, as shown in the diagram below. Here θ1 = 35◦ , 1 = 0.50 m, and 2 = 1.50 m. The length of the shadow is x + L. x is given by x = 1 tan θ1 = (0.50 m) tan 35◦ = 0.35 m. According to the law of refraction, n2 sin θ2 = n1 sin θ1 . We take n1 = 1 and n2 = 1.33 (from Table 34–1). Then, sin θ1 sin 35.0◦ θ2 = sin−1 = sin−1 = 25.55◦ . n2 1.33 L is given by L= 2 tan θ2 = (1.50 m) tan 25.55◦ = 0.72 m . The length of the shadow is 0.35 m + 0.72 m = 1.07 m. . .. .. .. .. .. .. . .. .. .. .. .. .. . .. 1 ..... .. ... .... . . . . . .. . . ... . .... . . .. . .... .. . .... . . .. .. . . . .. . .. . .. .. . ..... . ... . .. .. .. . . .. .. .. . .. . ... .. ... .. . . .. . . . . .. . . . . .. . . . .. . .. . . . . .. . . . . .. . . . .. . . . . .. . . . . 2 .. . . . . .. . .. . ..... . . ... . . .. . ... . . .. . . . . . .. . . . . .. . . . . .. . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . . .. . . . .. . .. . . . . .. . . . .. . ..... ... . . .. .. . . . .. θ 1 air water θ shadow .. .. ................................. .................. ................................................... ... . .. ... .. .... .. . .. .... ... ... L x 2 ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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