Q24E - University Physics with Modern Physics

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?

Step-by-Step Solution

Answer

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

Step 1: Given Data

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

Step 2: 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.

Step 3: 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 $51.6°$.