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 ā 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 ā 2 ā 2 ā 6 = x Ė + y Ė + z Ė Ī» 2 4 4 Assuming E 1 = E 2 = 1, ā 3 i 2 ik 1 Ā· Ļ x ( z ) r iĻ 1 + E 1 ( x, y, z ) = e ā” ā e 6 i 2 Ī» Ļ Ī» 2 2 = e interference pattern I ā 2 ā 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 ā Ļ 2 ) = 2[1 + cos( Ļ 1 ā Ļ 2 )] (a) In the xy-plane, z = 0 2 Ļ Ļ 1 = Ī» ( x/ 2) 2 Ļ 1 ā 2 ā 2 Ļ 2 = 2 Ī» Ļ ( ā 2 2 x + ā 4 2 y ) Ļ 1 ā Ļ 2 = Ī» 2 ā 2 x ā 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 ā 2 ā 2 Ī» 2 ā 2 x ā 4 y = 2 mĻ, where m ā Z 1 ā ā 2 ā 2 x ā y = mĪ» 2 4 (b) For the plane z = Ī» ā« = x + Ī» Ļ 1 2 Ī» Ļ 1 2 ā 2 3 ā¬ 2 Ļ 1 ā ā 2 ā 2 2 ā 3 ā ā 6 2 Ļ ā 2 ā 2 ā 6 ā­ 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 ā Z . The equation of the fringe lines are: 2 Ļ 1 ā ā...
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MIT2_71S09_gsol5 - | | | | 2.710 Optics Problem Set 5...

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