243S2008-06-soln

# 243S2008-06-soln - EE243 HW#6 Solutions Problem 1 Sp2008...

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EE243 HW#6 Solutions Sp2008 Problem 1 For defects smaller than the Rayleigh limit ( 0.61 λ NA ), the patterns behaves like a collection of coherent points sources. The intensity on the image plane is just the square of the summation of the electric fields contributed by sources at x i : I (x) = | i E i (x - x i ) | 2 . Since all open areas are the same (i.e., same number of point sources), and the peak electric field is at center of symmetry of the patterns, the shape which has the smallest “average” distance from the center of pattern will have the highest E-field and hence peak intensity. To see this, let us just consider the case of two point sources. Since each point sources contributes a peak -like distribution, the closer the two points will give a higher E total . Answer : The circular opening has the highest peak intensity because it has the smallest “average” distance from the center of pattern.

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Problem 2 a) The slopes should be around 2.5 and 1.8 per λ /NA. b)
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## This note was uploaded on 11/02/2010 for the course EECS 243 taught by Professor Ee243 during the Spring '03 term at University of California, Berkeley.

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243S2008-06-soln - EE243 HW#6 Solutions Problem 1 Sp2008...

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