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ps07soln - 1 Physics 316 Cornell University Solution for...

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1 Physics 316 Solution for homework 7 Spring 2005 Cornell University I. EXERCISE 1 How can a device be built that outputs horizontally polarized light of an intensity that is independent of the polar- ization direction of an incoming linearly polarized beam? One way to get a linearly polarized wave independent of the initial polarization is to let light go trough a λ/ 4-plate, and then use a polarizer making an angle of 45 with the fast and the slow axis of the λ/ 4-plate. Let’s show what happens mathematically. The initial beam can be written as: ~ E = E cos( kz - ωt )( ~e x cos θ + ~e y sin θ ) , (1.1) where θ is the angle of polarization (unknown). The λ/ 4-plate introduces a phase difference of π/ 2 between the x -component and the y -component. Thus, the beam state at the output is: ~ E 1 = E~e x cos θ cos( kz - ωt ) + E~e y sin θ sin( kz - ωt ) . (1.2) Now, let this wave go through the polarizer. The amplitude of the x component going through the polarizer will be reduced by cos π/ 4 = 1 / 2. The amplitude of the y component going trough the polarizer will be reduced by the same amount since sin π/ 4 = 1 / 2. Thus, the final output, in the direction of polarization, is: E 2 = E 2 cos θ cos( kz - ωt ) + E 2 sin θ sin( kz - ωt ) = E 2 cos( kz - ωt - θ ) . (1.3) II. EXERCISE 2 Cross two linear polarizers with an angle of 90 so that no light is transmitted. Insert a third polarizer in between these two with an angle θ to the first polarizer. What is the transmission as a function of θ ? θ FIG. 1: A polarizer ( B ) is inserted between two orthogonal polarizers ( A and C ). Define the polarizers as in figure 1. Assume I A is the intensity of the light at the output of polarizer A . Then, I B = I A cos 2 θ . The angle between polarizers B and C is π/ 2 - θ . Therefore, I C = I B cos 2 ( π/ 2 - θ ) = I B sin 2 θ . Hence, I C = I A cos 2 θ sin 2 θ = I A 4 sin 2 2 θ. (2.1) In general, I C 6 = 0. 2 III. EXERCISE 3 A. Given a horizontally polarized beam with ~ E = ~e x E x cos( kz - ωt + φ x ) and ω = ck . Verify that | ~ E ( z, t ) | 2 depends on z and t . Compute the time average < | ~ E ( z, t ) | 2 > t . Does it depend on z ? Let’s first calculate | ~ E ( z, t ) | 2 : | ~ E ( z, t ) | 2 = ~ E · ~ E = E 2 x cos 2 ( kz - ωt + φ x ) . (3.1) We then compute the time average, D | ~ E ( z, t ) | 2 E t . To do so, we need to integrate over one period of oscillation of | ~ E ( z, t ) | 2 . At any given z , the period of oscillation is the smallest T > 0 such that: | ~ E ( z, t ) | 2 = | ~ E ( z, t + T ) | 2 . (3.2) Thus cos 2 ( kz - ωt + φ x ) = cos 2 ( kz - ωt + φ x - ωT ) ωT = π, (3.3) that is, T = π/ω . Let’s calculate the time average: D | ~ E ( z, t ) | 2 E t = 1 T Z t + T t | ~ E ( z, t ) | 2 dt (3.4) = ω π Z kz - ωt + φ x - π kz - ωt + φ x - E 2 x ω cos 2 θdθ (with θ = kz - ωt + φ x ) = E 2 x π Z kz - ωt + φ x kz - ωt + φ x - π 1 2 (1 + cos 2 θ ) = E 2 x π θ 2 fl fl fl fl kz - ωt + φ x kz - ωt + φ x - π + E 2 x 4 π Z 2( kz - ωt + φ x ) 2( kz - ωt + φ x ) - 2 π cos θ 0 0 = E 2 x 2 . (3.5) This result is independent of the position, z . The calculation of the average value of an oscillating quantity occurs very often. You have seen in class that cos 2 θ fi = sin 2 θ fi
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