WQp dp d wqp dp d

J Ln Jo Jo

We will apply this, as an example, to the case of the first three terms in (3.471) and deduce the aberration coefficients kd, ks 1 and ks2 giving optimum tolerances according to the criterion of (3.468). Then in (3.472) for this case

with p normalized to 1 at the edge of the pupil. Eq. (3.472) gives

! AW \2 kd , 4 i 2 . 9 ,2 , kdkS1 , 3 , , . kS1kS2 fo AVA\

(^WQ) =-2 + 45kS1 +--2kS2 + — + 20kdks2 + ~T" (3.474)

If ks2 is considered as a constant and kd and ks1 are varied to minimize the function, giving maximum tolerances, then

kd 8 kS2 IT + 45ks1 + T" = 0 The solution gives the simple result

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