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

Answer: 5.00

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 $D>>\lambda \Rightarrow 2D>>2\lambda$
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 $0,1,2$
$\Rightarrow \frac{D}{\sqrt{x^2+D^2}}=\frac{m}{2}$
$\Rightarrow \frac{D^2}{x^2+D^2}=\frac{m^2}{4}$
$4D^2=m^2{x}^2+m^2{D}^2$
$\Rightarrow x=\sqrt{\frac{4D^2-m^2{D}^2}{m^2}}$
$x=\frac{D}{m}\sqrt{4-m^2}$
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 $P$.