MIT2_71S09_usol8

MIT2_71S09_usol8 - 2.71 Optics Problem Set 8 Solutions 1....

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Unformatted text preview: 2.71 Optics Problem Set 8 Solutions 1. (a) For a diffraction limited system the slopes of the OTF are constant. m u =25mm- 1 = 68 . 75% = 0 . 6875 I in = 1 2 h 1 + cos 2 x i I in = 1 2 ( u ) + 1 4 u- 1 + 1 4 u + 1 I out = I in OTF = 1 2 ( u ) + a 4 u- 1 + a 4 u + 1 I out ( x ) = 1 2 1 + a cos 2 x m u = 1 = ( 1 2 + a 2 )- ( 1 2- a 2 ) ( 1 2 + a 2 ) + ( 1 2- a 2 ) = a the contrast is the normalized value of the OTF at that frequency. Using similar triangles, if m u =25mm- 1 = 0 . 6875 = (1- . 3125), then m u =50mm- 1 = (1- . 6250) = 0 . 3750 = 37.5 % (b) The cut-off frequency for incoherent imaging is u = 80mm- 1 . The cut-off fre- quency of the coherently illuminated system is 40mm- 1 . Hence 50mm- 1 frequen- cies do NOT go through if it is coherently illuminated. 2. I ( x ) = 1 2 1 + 1 2 cos 2 x 40 m + 1 2 cos 2 3 x 40 m (a) The contrast m is given by: m = I max- I min I max + I min At the input, m = 1- 1+0 = 1. 1 (b) The Fourier transform of I ( x ) is: I ( u ) = 1 8 u- 1 40 + u + 1 40 + u- 3 40 + u + 3 40 + 1 2 ( u ) The Fourier transform of the output intensity is: I ( u ) = (MTF) I ( u ) = 1 2 ( u ) + (0 . 25) 1 8 u- 1 40 + u + 1 40 = 1 2 ( u ) + 1 16 1 2 u- 1 40 + 1 2 u + 1 40 I ( x ) = 1 2 + 1 16 cos 2 x 40 m out = ( 1 2 + 1 16 )- ( 1 2- 1 16...
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MIT2_71S09_usol8 - 2.71 Optics Problem Set 8 Solutions 1....

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