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 Efield 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.
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 Spring '03
 ee243
 point sources, peak intensity, Nils, RU defocus

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