Question

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

Short answer

It is proved that a grating made up of alternately transparent andopaque strips of equal width eliminates all the even orders of maximaexcept$m=0$.

Step-by-step solution

Given data:

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

Diffraction from a grating and single slit:

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)

Step 3:Proof that the even order grating diffraction maximas disappear

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.