12. We will make use of arctangents and sines in our solution, even though they can be “shortcut” somewhat since the angles are small enough to justify the use of the small angle approximation. (a) Given y/D= 15/300 (both expressed here in centimeters), then θ= tan−1(y/D) = 2.86°. Use of Eq. 36-6 (with a= 6000 nm and λ = 500 nm) leads to ( )6000nm sin 2.86sin1.883rad .500nmaαπ°π===λThus, 2sin0.256 .pmII⎛⎞⎜⎟⎝⎠(b) Consider Eq. 36-3 with “continuously variable”
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