243S2008-05-soln - EE243, Sp 2008 HW#5 Solutions Problem 1...

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Unformatted text preview: EE243, Sp 2008 HW#5 Solutions Problem 1 (a) L= 0.8+0.8 =1.6 m. Therefore spatial frequency =1/L = 0.62 / m For incoherent illumination, spatial frequency cutoff = 2 NA/ = (2 0.35)/ 0.436 m= 1.6/ m The third harmonic of the square mask pattern = 0.62 3 = 1.87 / m and the higher harmonics will not be collected by the projection lens. I image = I o [ 1 + 4 sin ( 2 x L ) MTF 1 ] with MTF 1 = 1 - 2 0.436 0.62 3.14 0.35 = 0.51 I image = I o [ 1 + 0.65 sin ( 2 x L ) ] I image (max) = 1.65 I o , I image (min) = 0.35 I o (b) CONTRAST = I max- I min I max- I min = 1.65 - 0.35 1.65 + 0.35 = 0.65 Problem 2 Traveling wave approach For a coherent source beam making an angle of with the optical axis, x-dependence of the E-field can be described by : E(x) e -i2 x where can be treated as a photon-source effective spatial frequency. When the inclined beam hits a periodic mask pattern of periodicity P, diffracted beams will result with scattered angles satisfying : P sin = n . By considering only the n=0, +1, amd 1 orders , we have three beams each can be represented by : E = e -i2 x E +1 = 1 2 e -i2 [ + m ] x x Optical Axis Plane wave e -i2 x x Optical Axis e -i2 x e -i2 ( - m ) x e -i2 ( + m ) x E E +1 E-1 E...
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243S2008-05-soln - EE243, Sp 2008 HW#5 Solutions Problem 1...

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