91Ó°ÊÓ

Angle for first-order maximum $\left({\mathrm{θ}}_1\right)$is $11.3°$.

A set of a large number of parallel slits of equal width and which are equidistant between the centers is known as grating.The angular position of the fringe is given by the following relation.

$\mathit{s}\mathit{i}\mathit{n}{\mathit{θ}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{m}\mathbf{λ}}{\mathbf{d}}$

Here, $\mathit{λ}$ is the wavelength of light used and is the slit width.

Given:${\mathrm{θ}}_1=11.3^{∘}$

Now, the angular position of the bright fringes is given by

$sin{\mathrm{θ}}_m=\frac{m\mathrm{λ}}{d}$

Whereas $m=0,Â\pm 1,Â\pm 2,...$

The first-order maximum is for$m=Â\pm 1$

Hence,

$sin{\mathrm{θ}}_1=\frac{\mathrm{λ}}{d}â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(1\right)$

And the fourth-maximum is for

$\sin {\mathrm{θ}}_4=\frac{4\mathrm{λ}}{d}â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(1\right)$

Divide (2) by (1),

$$\begin{gathered} \sin {\mathrm{θ}}_4=4sin{\mathrm{θ}}_1 \\ {\mathrm{θ}}_4={\sin }^{-1}\left(4\sin {\mathrm{θ}}_1\right) \\ ={\sin }^{-1}\left(4\sin 11.3°\right) \\ =51.6° \end{gathered}$$

Thus, the angular position of the fourth-order maximum is $51.6°$.