Light with wavelength `535nm` falls normally on a diffraction grating. Find its period if the diffraction angle `35^(@)` corresponds to one of the Fraunhofer maxima and the highest order of spectrum is equal to five.

Light with wavelength `535nm` falls normally on a diffraction grating.

1.	Light with wavelength `535nm` falls normally on a diffraction grating. Find its period if the diffraction angle `35^(@)` corresponds to one of the Fraunhofer maxima and the highest order of spectrum is equal to five.
Answer» The diffraction fromula is `d sin theta_(0) = n_(0)lambda` where `theta_(0) = 35^(@)` is the angle of diffraction corresponding to order `n_(0)` (which is not yet known). Thus `d = (n_(0)lambda)/(sin theta_(0)) = n_(0) xx 0.9327mu m` on using `lambda = 0.535mum` For the `n^(th)` order we get `sin theta = (n)/(n_(0)) sin theta_(0) = (n)/(n_(0)) (0.573576)` If `n_(0) -1`, then `n gt n_(0)` is at least `2` and `sin theta gt 1` so `n = 1` is the highest order of diffraction. If `n_(0) = 2` then `n = 3, 4,` but `sin theta gt 1` for `n = 4` thus the highest order of diffraction is `3`. If `n_(0) = 3`, then `n = 4, 5, 6`. For `n = 6, sin theta = 2 xx 0.57 gt 1`, so not allowed while for `n = 5, sin theta = (5)/(3) xx 0.573576 lt 1` is allowed. Thus in this case the highest order of diffraction is five as given. Hence `n_(0) = 3` and `d = 3 xx 0.9327 = 2.7981 ~~ 2.8mu m`.

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Light with wavelength `535nm` falls normally on a diffraction grating. Find its period if the diffraction angle `35^(@)` corresponds to one of the Fraunhofer maxima and the highest order of spectrum is equal to five.

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