hw2_soln - HW 2 Solution 1 a The light circle must have the...

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HW 2 Solution c θ n=1.33 R 1m 1) a) The light circle must have the cone angle within the critical angle between water and air, i.e. and radius of circle R = . D 59 . 48 ) 333 . 1 / 1 ( sin 1 = = c θ ) ( 134 . 1 59 . 48 tan 1 m = × o b) This a Brewster angle problem which requires p-polarization. The Brewster angle o 87 . 36 ) 333 . 1 / 1 ( tan 1 = = B θ c) 49 48 1 333 . 1 1 333 . 1 1 2 = + = T 2) w ) ( 04987 . 0 ) 10 630 /( 10 / 630 , 100 9 8 2 m w z nm m o o o o o = × = = = = π λ π λ μ 36.87 o water air a) At z=1m, ) ( 008 . 2 ) 04987 . 0 / 1 ( 1 10 ) / ( 1 2 4 2 mm z z w w o o = + = + = . b) At z=1cm, w=102( μ m) c) ) ( 011 . 4 ) /( 2 mrad w o o = = π λ θ which is not valid for -0.04987<z<0.04987(m). 3) m z nm mm w mm w o oy ox 5 , 638 , 125 . 0 , 05 . 0 = = = = λ 40.62mm 16.25mm ) ( 07694 . 0 / ) ( 01231 . 0 ) 10 638 /( ) 10 05 . 0 ( / 2 9 2 3 2 m w z m w z o oy oy o ox ox = = = × × = = λ π π λ π ) ( 31 . 20 ) / ( 1 2 mm z z w w ox ox x = + = , ) ( 124 . 8 ) / ( 1 2 mm z z w w oy oy y = + = 4) Mach-Zehnder: Optical path for beam 2 = 1 1 2 2 d n d + , optical path for beam 1 = 1 d Constructive interference condition: o o m d d n d m d d n d k λ π = + = + 1 1 1 2 1 1 1 2 2 2 ) 2 ( Michelson: Optical path for beam 1 = 1 2 L commom +
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