MIT2_71S09_gquiz2_sol

# MIT2_71S09_gquiz2_sol - 2.710 Optics 1 Solution Quiz 2...

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Unformatted text preview: 2.710 Optics 1. Solution: Quiz 2 Solutions Spring `09 z g1 (x, z) = a1 exp i2 sin + cos , 6 6 x z g2 (x, z) = a2 exp i2 - sin + cos , 3 3 x At z = 0, a1 = 1 a2 = 1 2 x 1 x 3 2 2 I(x) = | 1 (x, 0) + g2 (x, 0)| = a1 ei2 2 + a2 e-i2 2 g 2a1 a2 x 1 3 = (a2 + a2 ) 1 + 2 cos 2 + 1 2 2 a1 + a2 2 2 21 1 1 x 1+ 3 2 = 1+ 1+ cos 2 1 4 2 1+ 4 x 5 4 2 = 1 + cos 2 where = 4 5 1+ 3 m= 4 2 period contrast = 5 1+ 3 5 4 5 9 9 1 I(0) = 1 + = = =2 4 4 5 5 4 4 After phase-shifting g1 by /2, we get x 5 4 5 1 I(0) = = 1 I(x) = 1 + cos 2 + 4 5 2 4 4 2. Solution: = 1.0m, f = 5cm, = 10m 1 According to Lecture 19, slide 12, with duty cycle = 1 , 2 q x 1 gt (x) = sinc ei2q 2 q=- 2 q 1 q Gt (u) = sinc u- 2 q=- 2 q-1 2(-1) 2 if q = odd q sin ( q ) q 2 sinc = = 1 if q = 0 q 2 2 0 if q = 0, even 1 3 1 1 Gt (u) = - u+ + u+ 10m 10m 3 -3rd order -1st order 1 1 1 1 3 + (u) + u - - u- + 2 10m 10m 3 0th order (DC) +1st order +3rd order We see that the even orders are missing. x 1 3f 1 f 1 gpp- (x ) Gt = - x + + x + + (x ) f 2 3 1 f 1 3f + x - - x - + 3 f 10-6 5 10-2 = = 5mm, 10 10-6 1 1 1 1 1 gpp- (x ) - (x +15mm)+ (x +5mm)+ (x )+ (x -5mm)- (x -15mm)+ 2 3 3 Since 2 1 1 gpp+ (x ) = gpp- (x )gpm (x ) = (x + 5mm) - (x - 15mm) 3 x 3x x 1 1 +i2 10m gout (x ) Gpp+ = e-i2 10m - e f 3 2 2 1 1 1 1 2x 2 Iout (x ) = |gout (x )| = + +2 - cos 2 3 3 10m 10 2 2x = 2 - 2 cos 2 9 3 10m Reminder: |a1 ei1 + a2 ei2 |2 = (a1 ei1 + a2 ei2 )(a1 e-i1 + a2 e-i2 ) 2 = a2 + a2 + a1 a2 ei(1 -2 ) + a1 a2 ei(2 -1 ) 1 2 = a2 + a2 + 2a1 a2 cos(2 - 1 ) 1 10 ( 92 + Imax - Imin V = = 10 Imax + Imin ( 92 + 2 ) 3 2 2 ) 3 2 10 - ( 92 - 10 + ( 92 - 2 ) 3 2 2 ) 3 2 2 2 32 = 3 = 5 2 9102 Now modify the pupil mask so that its complex transmissivity is: This can be done by inserting a piece of glass of optical thickness equal to in the upper hole. The physical thickness, if n = 1.5, is: 2 (2k + 1) (2k + 1) 1m (n-1)t = (2k+1) t = = = (2k+1)m, where k is any integer 2(n - 1) 2 (1.5 - 1) With this modification, gout (x ) = 3 1 -i2 10xm 1 +i2 10xm e + 3 e result of -phase shift in the +3rd order Iout 10 2 2x = 2 + 2 cos 2 9 3 10m Now the intensity is maximum at x = 0. 3 MIT OpenCourseWare http://ocw.mit.edu Spring 2009 2.71 / 2.710 Optics For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ...
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