ps07soln

# ps07soln - PHYSICS 218 SOLUTION TO HW 7 Created 8:15am Last...

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PHYSICS 218 SOLUTION TO HW 7 Created: October 23, 2004 8:15am Last updated: November 1, 2004 8:44pm 1. (a) Let E 0 = | E 0 | e E and B 0 = | B 0 | e M . By comparing to E 0 = E r + iE i and B 0 = B r + iB i , it is easy to verify that | E 0 | = q E 2 r + E 2 i , B 0 = q B 2 r + B 2 i , ϕ E = arctan( E i /E r ) and ϕ M = arctan( B i /B r ). Therefore, we have E x = | E 0 | e i ( kz ωt + ϕ E ) ,B y = | B 0 | e i ( kz + ϕ M ) . Taking the real part of E x and B y , we obtain < E x = | E 0 | cos( kz + ϕ E ) , < B y = | B 0 | cos( + ϕ M ) , from which we see that a E = | E 0 | and a M = | B 0 | . (b) Using the expressions u E = 1 2 ² 0 E 2 x and u M = 1 2 µ 0 B 2 y for the energy densities of electric and magnetic Felds respectively, we have u E = 1 2 ²a 2 E cos 2 ( + ϕ E ) ,u M = 1 2 µ a 2 M cos 2 ( + ϕ M ) . (c) Averaging over one cycle T =2 π/ω , we Fnd that h cos 2 ( + ϕ E ) i = 1 T Z T 0 cos 2 ( + ϕ E )d t = 1 2 . Thus, we have h u E i = 1 4 2 E and h u M i = 1 4 µ a 2 M . (d) We note that for a complex number | z | 2 can always be written as | z | 2 = zz = z z . Sowehave h u E i = 1 4 ²E 0 E 0 and h u M i = 1 4 µ B 0 B 0 . (e) The Poynting vector is given by s = 1 µ ( < E x ) × ( < B y )= 1 µ ( < E x )( < B y ) ˆ k . The z -component of the Poynting vector is thus s ( z, t 1 µ ( < E x < B y 1 µ a E a M cos( + ϕ E ) cos( + ϕ M ). (f) The average of s ( )is h s i = 1 µ a E a M h cos( + ϕ E ) cos( + ϕ M ) i . Using the trigono- metric identity 2 cos A cos B = cos( A + B ) + cos( A B ), we Fnd h cos( + ϕ E ) cos( + ϕ M ) i = 1 2 cos( ϕ E ϕ M ) . (1) The average of s ( ) over one cycle is thus h s i = 1 2 µ a E a M cos( ϕ E ϕ M ). (g) We can rewrite h s i = 1 2 µ ( a E cos ϕ E a M cos ϕ M + a E sin ϕ E a M sin ϕ M 1 2 µ ( E r B r + E i B i ). On the other hand, we observe that < ( E 0 B 0 < ( E 0 B 0 E r B r + E i B i , we then have h s i = 1 2 µ < ( E 0 B 0 1 2 µ < ( E 0 B 0 ). 2. (a) The reﬂection and transmission coeﬃcients with the electric Feld perpendicular to the plane of incidence and κ M 1 = κ M 2 = 1 given by Elmore and Heald (page 274) are R = sin( ϑ 1 ϑ 2 ) sin( ϑ 1 + ϑ 2 ) ,T = 2cos ϑ 1 sin ϑ 2 sin( ϑ 1 + ϑ 2 ) . Using Snell’s law n 1 sin ϑ 1 = n 2 sin ϑ 2 ,wehave R = n 2 cos ϑ 2 n 1 cos ϑ 1 n 2 cos ϑ 2 + n 1 cos ϑ 1 = 2 n 1 cos ϑ 1 n 2 cos ϑ 2 + n 1 cos ϑ 1 . At normal incidence, ϑ 1 = ϑ 2 = 0, so we recover R = n 1 n 2 n 1 + n 2 = 2 n 1 n 1 + n 2 . 1

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PHYSICS 218 SOLUTION TO HW 7 (b) When the electric feld is parallel to the plane oF incidence, Elmore and Heald (page 275) gives R k = tan( ϑ 1 ϑ 2 ) tan( ϑ 1 + ϑ 2 ) ,T k = 2 cos ϑ 1 sin ϑ 2 sin( ϑ 1 + ϑ 2 ) cos( ϑ 1 + ϑ 2 ) .
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ps07soln - PHYSICS 218 SOLUTION TO HW 7 Created 8:15am Last...

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