mm112 2λλ=for which the smallest possible choices are m1= 5 and m2= 4. Returning to Eq. 36-25, then, we find ()4110 mm)sinsinsin0.3621 .0.0056 mmmdθ−−−−⎛⎞λ5(4.0×====°⎜⎟⎝⎠(c) There are no refraction angles greater than 90°, so we can solve for “mmax” (realizing it might not be an integer): max422sin900.0056 mm1110 mmddm−°=≈λλ5.0×where we have rounded down. There are no values of m(for light of wavelength
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