Question

Monochromatic light is at normal incidence on a plane transmission grating. The first-order maximum in the interference pattern is at an angle of $\mathbf{11}\mathbf{.}\mathbf{3}\mathbf{°}$ . What is the angular position of the fourth-order maximum?

Short answer

the fourth-order maximum is situated at an angle of $51.6°$.

Step-by-step solution

Given Data

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

Grating and Angular position of the bright fringes

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.

$\mathit{s}\mathit{i}\mathit{n}{\mathit{\theta }}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{m}\mathbf{\lambda }}{\mathbf{d}}$

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

Determine the angular position of the fourth-order maximum

Given:${\theta }_1=11.3^{\circ }$

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

$sin{\theta }_m=\frac{m\lambda }{d}$

Whereas $m=0,\pm 1,\pm 2,...$

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

Hence,

$sin{\theta }_1=\frac{\lambda }{d}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

And the fourth-maximum is for

$\sin {\theta }_4=\frac{4\lambda }{d}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Divide (2) by (1),

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

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