Question

A monochromatic beam of light of $6000\mathring{A}$ is used in YDSE set-up. The two slits are covered with tow films of equal thickness $t$ but of different refractive indices as shown in the figure. Considering the intensity of the incident beam on the slits to be $I_0$, find the point on the screen at which intensity is $I_0$ and is just above the central maxima. (Assume that there is no change in intensity of the light after passing through the films).
Consider $t=6\mu m,d=1mm,D=1m$, where $d$ and $D$ have their usual meaning. Give your answer in mm.

Answer 1

Given, $\lambda =6000\times {10}^{-9}m,t=6\times {10}^{-6}m,d=1\times {10}^{-3}m,D=1m,{\mu }_1=\frac{3}{2}=1.5,{\mu }_2=\frac{4}{3}=1.3$

Path difference at the central bright fringe,$\Delta =0$

After placing the film,

$\Delta =(S_{2}P-t+t{\mu }_2)-(S_{1}P-t+t{\mu }_1)=(S_{2}P-S_{1}P)+t({\mu }_1-{\mu }_2)=\frac{yd}{D}+6\times {10}^{-6}(1.5-1.3)$

or, $0=\frac{yd}{D}+1.2\times {10}^{-6}$

$\Rightarrow \frac{yd}{D}=-1.2\times 1^{-6}$

$y=-\frac{1.2\times {10}^{-6}\times D}{d}y=-\frac{1.2\times {10}^{-6}\times 1}{1\times {10}^{-3}}=-1.2mm$