Question

A coherent beam light of intercity and absolute wavelength $\lambda =5000A^o$ is being incident on the slits making an angle ${30}^o$ with horizontal. If screen is placed at a distance $D=1m$ from the slits is $3\times {10}^{-4}$.
A thin film of glass ${\mu }_g=3/2$ and thickness $0.41mm$ is placed near one slit as shown in the figure, find
(a) the position of central maxima.
(b) the intensity at point $O$.

Answer 1

(a)Let central maxima lies at point $P$ at a distance $x$ from the central line. Then optical path difference $\Delta p$ at point $P$ is
$(S_{2}P-t)+t{\mu }_g-{\mu }_{W}d\sin \theta -S_{1}p$
$\Rightarrow S_{2}P-S_{1}P+({\mu }_g-1)t-{\mu }_{W}d\sin \theta$
$\Rightarrow \frac{xd}{D}+({\mu }_g-1)t-{\mu }_{W}d\sin \theta$
For central maxima $\Delta p=0$
$x=\frac{D}{d}[\left({\mu }_{W}d\sin \theta \right)-({\mu }_g-1)t]$
$=\frac{1}{3\times {10}^{-4}}[\frac{4}{3}\times 3\times {10}^{-4}\times \frac{1}{2}-0.5\times 0.41\times {10}^{-3}]$
$=\frac{1}{3\times {10}^{-4}}[2\times {10}^{-4}-2.05\times {10}^{-4}]$
$=-\frac{5\times {10}^{-6}}{3\times {10}^{-4}}=-1.66\times {10}^{-2}=-1.66cm$
So central maxima lies at a distance $1.66cm$ below the central line
(b)At point $O$, optical path difference is
$({\mu }_g-1)t-{\mu }_{W}d\sin \theta =-5\times {10}^{-6}m$
So intensity at at $0$
$I_0=I+I+2\sqrt{II}\cos \frac{2\pi }{5\times {10}^{-7}}(-5\times {10}^{-6})=4I$