Replace the cosine grating in the previous problem with a square bar grating, that is, a series

Replace the cosine grating in the previous problem with a €œsquare€ bar grating, that is, a series of many fine alternating opaque and transparent bands of equal width. We now filter out all terms in the transform plane but the zeroth and the two first-order diffraction spots. These we determine to have relative irradiances of 1.00, 0.36, and 0.36: compare them with Figs. 7.40a and 7.42. Derive an expression for the general shape of the irradiance distribution on the image plane€”make a sketch of it. What will the resulting fringe system look like?

Figs. 7.40a

Figs. 7.42

Transcribed Image Text:

f(x) a = 4 Л3 1 сm -1/4 0 A/4 -1/4 0 1/4 λ 1 (cm) |Ao 14, k%3D 2т || Аз Аз mk 0 к 2k 3k 4k 5k 8k 10k 0 2т 4т бт 8т 10т 16т 20т Fourier coefficients Am 0.50 0.25 -7k -3k 3k 7k T -6k –5k -4k -2k -k k 2k 4k 5k 6k т mk