Question

When light rays approach the slits as shown in the figure below, making an angle $\theta$ with the central axis, the fringe width will be given by $\frac{x\lambda D}{yd}$, the value of $x+y$ is

Answer 1

Path difference, $\Delta x=d\sin \theta +\frac{yd}{D}$

For maxima,
$\Delta x=n\lambda$

$d\sin \theta +\frac{yd}{D}=n\lambda$

$\Rightarrow \frac{yd}{D}=(n\lambda -d\sin \theta )$

$\Rightarrow y=(n\lambda -d\sin \theta )\frac{D}{d}$

Hence, maxima will occur at

$y=-D\sin \theta ,\frac{\lambda D}{d}-D\sin \theta ,\frac{2\lambda D}{d}-D\sin \theta$

Hence, fringe width $=y_1-y_0$

$=\frac{\lambda D}{d}-D\sin \theta +D\sin \theta$

$=\frac{\lambda D}{d}$

Comparing with $\frac{x\lambda D}{yd}$

Thus, $x+y=1+1=2$