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
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.