A plane light wave with wavelength `lambda` falls normally on a phase diffraction grating whose side view is shown in Fig. The grating is cut on a glass plate with refractive index `n`. Find the depth `h` of the lines at which the intensity of the central Fraunhofer maximum is equal to zero. what is in this case the diffraction angle corresponding to the first maximum?

Explanation

The intensity of the central Fraunhofer maximum will be zero if the waves from successive grooves (not in the same plane) differ in phase by an rod multiple of `pi`. Then since the phase difference is
`delta = (2pi)/(lambda)(n-1)h`
fro the central ray we have
`(2pi)/(lambda)(n-1)h = (k-(1)/(2))2pi, k=1, 2, 3,........`
or `h=(lambda)/(n-1) (k -(1)/(2))`.
The path difference between the rays `1` & `2` is approximately (neglecuing terms of order `theta^(2))`
`a sin theta+a - na`
`=a sin theta -(n-1)a`
Thus for a maximum
`a sin theta-(k' -(1)/(2))lambda = mlambda`
or `a sin theta = (m+k'+(1)/(2))lambda`
`k' = 0,1,2,.........`
`m = 0, +-1, +-2,......`
The first maximum after the central minimum is obtained when `m +k' = 0`
We get `a sin theta_(1) = (1)/(2)lambda`