Show that a grating made up of alternately transparent and opaque strips of equal width eliminates all the even orders of maxima (except$\mathit{m}\mathbf{=}\mathbf{0}$).

There is a grating made up of alternately transparent andopaque strips of equal width.

The angular distance $\mathit{\theta }$ of the ${\mathit{m}}_{\mathbf{t}\mathbf{h}}$order diffraction maxima produced from a grating having line separation $\mathit{d}$ is

$\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathit{m}\mathit{\lambda }$ .....(1)

Here, is the wavelength of the incident light.

The angular distance $\mathit{\theta }$ of the${\mathit{k}}_{\mathbf{t}\mathbf{h}}$order single slit diffraction minima for slit width $\mathit{a}$is

$\mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathit{k}\mathit{\lambda }$ .....(2)

Slit separation is the distance between the mid points of two slits. Hence slit separation is equal to twice the slit width, that is

$d=2a$

Thus, equation (1) becomes

$2a\sin \theta =m\lambda$ ….. (3)

Subtract twice of equation (2) from equation (3) to get

$\begin{array}{l}2asin\theta -2asin\theta =m\lambda -2k\lambda \\ \left(m-2k\right)\lambda =0\\ \mathrm{m}=2k\end{array}$$\begin{array}{l}2asin\theta -2asin\theta =m\lambda -2k\lambda \\ \left(m-2k\lambda \right)=0\\ \mathrm{m}=2k\end{array}$

But

$\mathrm{K}=\pm 1\pm 2\pm 3\dots$

Hence

$m=\pm 2,\pm 4,\pm 6$

Thus the even order grating diffraction maxima’s (except$m=0$) overlap with single slit diffraction minima’s and disappear.