p36_053

# p36_053 - . . . . . . . . . . . . L 2 a L 2 b θ θ • T R...

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53. The wave that goes directly to the receiver travels a distance L 1 and the reﬂected wave travels a distance L 2 . Since the index of refraction of water is greater than that of air this last wave suFers a phase change on reﬂection of half a wavelength. To obtain constructive interference at the receiver, the diFerence L 2 L 1 must be an odd multiple of a half wavelength. Consider the diagram below. The right triangle on the left, formed by the vertical line from the water to the transmitter T, the ray incident on the water, and the water line, gives D a = a/ tan θ . The right triangle on the right, formed by the vertical line from the water to the receiver R, the reﬂected ray, and the water line leads to D b = x/ tan θ .S ince D a + D b = D , tan θ = a + x D . ←−−−− D a −−−−→←−− D b −−→ | | | | a | | | | | | x | | ................................................................................................................................ L 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Unformatted text preview: . . . . . . . . . . . . L 2 a L 2 b θ θ • T R • We use the identity sin 2 θ = tan 2 θ/ (1 + tan 2 θ ) to show that sin θ = ( a + x ) / p D 2 + ( a + x ) 2 . This means L 2 a = a sin θ = a p D 2 + ( a + x ) 2 a + x and L 2 b = x sin θ = x p D 2 + ( a + x ) 2 a + x . Therefore, L 2 = L 2 a + L 2 b = ( a + x ) p D 2 + ( a + x ) 2 a + x = p D 2 + ( a + x ) 2 . Using the binomial theorem, with D 2 large and a 2 + x 2 small, we approximate this expression: L 2 ≈ D +( a + x ) 2 / 2 D . The distance traveled by the direct wave is L 1 = p D 2 + ( a − x ) 2 . Using the binomial theorem, we approximate this expression: L 1 ≈ D + ( a − x ) 2 / 2 D . Thus, L 2 − L 1 ≈ D + a 2 + 2 ax + x 2 2 D − D − a 2 − 2 ax + x 2 2 D = 2 ax D . Setting this equal to ( m + 1 2 ) λ , where m is zero or a positive integer, we ±nd x = ( m + 1 2 )( D/ 2 a ) λ ....
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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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