MIT2_71S09_gsol5

MIT2_71S09_gsol5 - | | | | 2.710 Optics Problem Set 5...

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Unformatted text preview: | | | | 2.710 Optics Problem Set 5 Solutions 1. 2 Ļ€ | k 1 | = | k 2 | = Ī» 2 Ļ€ 2 Ļ€ 1 āˆš 3 k 1 = (sin 30 ā—¦ x Ė† + cos 30 ā—¦ z Ė†) = x Ė† + z Ė† Ī» Ī» 2 2 2 Ļ€ k 2 = (cos 45 ā—¦ x Ė† + sin 45 ā—¦ sin 30 ā—¦ y Ė† + sin 45 ā—¦ cos 30 ā—¦ z Ė†) 2 Ī» Ļ€ āˆš 2 āˆš 2 āˆš 6 = x Ė† + y Ė† + z Ė† Ī» 2 4 4 Assuming E 1 = E 2 = 1, āˆš 3 i 2 ik 1 Ā· Ļ€ x ( z ) r iĻ† 1 + E 1 ( x, y, z ) = e ā‰” āˆš e 6 i 2 Ī» Ļ€ Ī» 2 2 = e interference pattern I āˆš 2 āˆš 2 ik 2 Ā· ( z ) r iĻ† 2 x + y + E 2 ( x, y, z ) = e ā‰” e = e 2 4 4 I = | E 1 + E 2 | 2 = | E 1 | 2 + | E 2 | 2 + 2 | E 1 || E 2 | cos( Ļ† 1 āˆ’ Ļ† 2 ) = 2[1 + cos( Ļ† 1 āˆ’ Ļ† 2 )] (a) In the xy-plane, z = 0 2 Ļ€ Ļ† 1 = Ī» ( x/ 2) 2 Ļ€ 1 āˆš 2 āˆš 2 Ļ† 2 = 2 Ī» Ļ€ ( āˆš 2 2 x + āˆš 4 2 y ) Ļ† 1 āˆ’ Ļ† 2 = Ī» 2 āˆ’ 2 x āˆ’ 4 y = Ī” Ļ† I = 2[1 + cos Ī” Ļ† ] so the profile is a sinusoidal profile. The maxima are along the lines whose equation is: 2 Ļ€ 1 āˆš 2 āˆš 2 Ī» 2 āˆ’ 2 x āˆ’ 4 y = 2 mĻ€, where m āˆˆ Z 1 āˆ’ āˆš 2 āˆš 2 x āˆ’ y = mĪ» 2 4 (b) For the plane z = Ī» āŽ« = x + Ī» Ļ† 1 2 Ī» Ļ€ 1 2 āˆš 2 3 āŽ¬ 2 Ļ€ 1 āˆ’ āˆš 2 āˆš 2 2 āˆš 3 āˆ’ āˆš 6 2 Ļ€ āˆš 2 āˆš 2 āˆš 6 āŽ­ Ī” Ļ† = Ī» 2 x + 4 y + 4 Ī» Ļ† 2 = Ī» 2 x + 4 y + 4 Ī» I = 2[1 + cos Ī” Ļ† ] , so the interference pattern is still a sinusoid (i.e. a set of linear fringes). The maxima occur when Ī” Ļ† = 2 Ļ€m, m āˆˆ Z . The equation of the fringe lines are: 2 Ļ€ 1 āˆ’ āˆš...
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MIT2_71S09_gsol5 - | | | | 2.710 Optics Problem Set 5...

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