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.

Explanation

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`.