243S2008-05-soln

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

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Unformatted text preview: EE243, Sp 2008 HW#5 Solutions Problem 1 (a) L= 0.8+0.8 =1.6 µ m. Therefore spatial frequency ν =1/L = 0.62 / µ m For incoherent illumination, spatial frequency cutoff = 2 NA/ λ = (2 × 0.35)/ 0.436 µ m= 1.6/ µ m The third harmonic of the square mask pattern = 0.62 × 3 = 1.87 / µ m and the higher harmonics will not be collected by the projection lens. I image = I o [ 1 + 4 π sin ( 2 π 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 π 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 α with the optical axis, x-dependence of the E-field can be described by : E(x) ∝ e -i2 π ν α x where ν α 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 β satisfying : P sin β = n λ . By considering only the n=0, +1, amd –1 orders , we have three beams each can be represented by : E = e -i2 π ν α x E +1 = 1 2 e -i2 π [ ν α + ν m ] x x Optical Axis α Plane wave e -i2 π ν α x x Optical Axis α e -i2 π ν α x e -i2 π ( ν α- ν m ) x e -i2 π ( ν α + ν m ) x E E +1 E-1 E...
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243S2008-05-soln - EE243 Sp 2008 HW#5 Solutions Problem 1(a...

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