Question

Two coherent narrow slits $S_{1}$ and $S_{2}$ emitting light of wavelength $\lambda$ in the same phase are placed parallel to each other at a small separation of $3 \lambda$. The light is collected on a screen $S$ which is placed at a distance $D(> > \lambda)$ from the slit $S_{1}$ and shown in figure. The distance $x$ such that the intensity at point $P$ is equal to the intensity at $O$ is $\frac{D}{2 \sqrt{5}} \times a$. Find the value of $a$.

Answer 1

For point $P$, path difference
$=\Delta x = S _{1} P - S _{2} P$
$=2 \lambda \cos \theta$
For $O , \Delta x=2 \lambda$
So $O$ is a point of maxima.
Now $D > > \lambda \Rightarrow 2 D > > 2 \lambda$
For $P$ to be a point of maxima,

$S_{1} P-S_{2} P=m \lambda$
$ \Rightarrow 2 \lambda \cos \theta=m \lambda$
$\cos \theta=\frac{m}{2}$, So $m$ can take values $0,1,2$
$\Rightarrow \frac{D}{\sqrt{x^{2}+D^{2}}}=\frac{m}{2}$
$\Rightarrow \frac{D^{2}}{x^{2}+D^{2}}=\frac{m^{2}}{4}$
$4 D ^{2}= m ^{2} x ^{2}+ m ^{2} D ^{2}$
$\Rightarrow x =\sqrt{\frac{4 D ^{2}- m ^{2} D ^{2}}{ m ^{2}}}$
$x=\frac{D}{m} \sqrt{4-m^{2}}$
or $x=D \sqrt{\frac{4}{m^{2}}-1}$
$m=0$ gives $x=\infty$
$m=2$ gives $x=0$ which is for point $O$
$m=1$ gives $x=D \sqrt{3}$ which corresponds to point $P$.