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MIT2_71S09_gsol8 - 2.710 Optics Problem Set 8 Solutions 1(a...

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2.710 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 1 + cos 2 π x Λ ˆ I in = 1 2 δ ( u ) + 1 4 δ u - 1 Λ + 1 4 δ u + 1 Λ ˆ I out = ˆ I in · OTF = 1 2 δ ( u ) + a 4 δ u - 1 Λ + a 4 δ u + 1 Λ I out ( x ) = 1 2 1 + a cos 2 π x Λ m u = 1 Λ = ( 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 - 0 . 3125), then m u =50mm - 1 = (1 - 0 . 6250) = 0 . 3750 = 37.5 % (b) The cut-off frequency for incoherent imaging is u 0 = 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 π x 40 μ m + 1 2 cos 2 π 3 x 40 μ m (a) The contrast m is given by: m = I max - I min I max + I min At the input, m = 1 - 0 1+0 = 1. 1
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(b) The Fourier transform of I ( x ) is: ˜ I ( u ) = 1 8 δ u - 1 40 + δ u + 1 40 + δ u - 3 40 + δ u + 3 40 + 1 2 δ ( u ) The Fourier transform of the output intensity is: ˜ I 0 ( u ) = (MTF) · ˜ I ( u ) = 1 2 δ ( u ) + (0 . 25) 1 8 δ u - 1 40 + δ u + 1 40 = 1 2 δ ( u ) + 1 16 1 2 δ u - 1 40 + 1 2 δ u + 1 40 I 0 ( x ) = 1 2 + 1 16 cos 2 π x 40 m out = ( 1 2 + 1 16 ) - ( 1 2 - 1 16 ) ( 1 2 + 1 16 ) + ( 1 2 - 1 16 ) = 1 8 = 0 . 125 (c) The incoherent transfer function is an autocorrelation of the coherent transfer function. The coherent transfer function in this case is probably a triangle function with half the cut-off frequency. H ( u ) = F{ h ( x ) } 2
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3. h ( x ) = sinc 2 x b (a) Incoherent iPSF ˜ h ( x ) = | h ( x ) | 2 = sinc 4 x b (b) MTF = ˜ H ( u ) ˜ H ( u ) = F{ ˜ h ( x ) } = F sinc 2 x b · sinc 2 x b = F sinc 2 x b ⊗ F sinc 2 x b = b Λ( bu ) b Λ( bu ) b Λ( bu ) MTF 3
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2.710 Optics Spring °09 Solutions to Problem Set #8 Due Wednesday, May 13, 2009 Problem 4: a) Consider the system shown in Figure 1. At the input we place a sinusoidal amplitude grating of perfect contrast, g t ( x ) = 1 2 h 1 + cos ° 2 ° x ° ±i ; (1) where ± = 0 : 5 ² m, f 1 = f 2 = f = 20cm, ° = 10 ² m, and the two apertures at the pupil plane have a diameter of 1cm and are separated by 1cm from the optical axis.
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