Light with wavelength `530nm` falls on a transparent diffraction grating with period `1.50 mu m`. Find the angle, relative to the grating normal. At which the Fraunhofer maximum of highest order is observed provided the light falls on the grating
(a) at right angles,
(b) at the angle `60^(@)` to the normal.

Explanation

Here the simple formula
`d sin theta = m_(1)lambda` holds.
Thus `1.5 sin theta = mxx 0.530 sin theta = (m xx 0.530)/(1.5)`
Highest permissible `m` is `m = 2` because `sin theta gt 1` if `m = 3`. Thus
`sin theta = (1.06)/(1.50)` for `m =2`, This gives `theta = 45^(@)` nearby.
(b)Here `d(sin theta_(0) - sin theta) = n lambda`
Thus `sin theta = sin theta_(0) - (n lambda)/(d)`
`= sin 60^(@) - n xx (0.53)/(1.5)`
`= 0.86602 - n xx 0.353333`
For `n = 5, sin theta =- 0.900645`
for `n = 6, sin theta lt -1`.
Thus the highest order is `n = 5` and we get
`theta sin^(-1) (-0.900645) ~= - 64^(@)`