Question

Two plane mirrors, a source $S$ of light, emitting monochromatic rays of wavelength $\lambda$ and a screen are arranged as shown in figure. If angle $\theta$ is very small, find the fringe width of the interference pattern formed by reflected rays.

• A. $\frac{\lambda (b+2a)}{4a\theta }$
• B. $\frac{\lambda (b+a)}{2a\theta }$
• C. $\frac{\lambda (b+2a)}{2b\theta }$
• D. $\frac{\lambda (b+a)}{4b\theta }$

Answer 1

The correct option is A $\frac{\lambda (b+2a)}{4a\theta }$

Distance between two virtual sources $S_1$ and $S_2$ is

$d=2\times 2a\cos \theta \sin \theta =2a\sin 2\theta$


Screen distance $D=b+2a{\cos }^2\theta$

$\therefore$ Fringe width,

$\beta =\frac{\lambda D}{d}=\frac{\lambda (b+2a{\cos }^2\theta )}{2a\sin 2\theta }$

If $\theta$ is much small $\cos \theta \approx 1$ and $\sin \theta \approx \theta$

$\therefore \beta =\frac{\lambda (b+2a)}{4a\theta }$

Hence, $\left(\text{A}\right)$ is the correct answer.