Question

A screen is at a distance $D=80cm$ from a diaphragm having two narrow slits $S_1$ and $S_2$ which are $d=2mm$ apart.. Slit. $S_1$ is covered by a transparent. sheet of thickness $t_1=2.5\mu m$ and $S_2$ by another sheet of thickness $t_2=1.25\mu m$ as shown in figure. Both sheets are made of same material having refractive index $\mu =1.40$. Water is filled in space between diaphragm and screen. $A$ monodichromatic light beam of wavelength $\lambda =5000\mathring{A}$ is incident normally on the diaphragm. Assuming intensity of beam to be uniform and slits of equal width, calculate ratio of intensity at $C$ to maximum intensity of interference pattern obtained on the screen, where $C$ is foot of perpendicular bisector of $S_1{S}_2$. (Refractive index of water, ${\mu }_w=4/3$ ). If it is $\frac{a}{4}$. Find $a$.

Answer 1

Answer: 3.00

When slit $S_1$ is covered by a sheet of thickness $t_1$, fringe pattern shifts upwards.
The shift $S_1$ is given by
$S_1=\frac{\left(\frac{{\mu }_s}{{\mu }_W}-1\right)t_{1}D}{d}$
$d=$ separation between slits
$\therefore S_1=\frac{\left(\frac{1.4}{4/3}-1\right)(2.5\mu m)\times 0.8}{2\times 10^{-3}}=50\mu m$ ...(1)
When slit $S_2$ is covered by a sheet of thickness $t_2$, the fringe pattern shifts downwards.
The shift $S_2$ is given by
$S_2=\frac{\left(\frac{1.4}{4/3}-1\right)(1.25\mu m)\times 0.8}{2\times 10^{-3}}=25\mu m$ ...(2)
Total shift $=S_1-S_2$
$=S=25\mu m$ (upwards) .....(3)
Phase difference between rays reaching at $C$.
$\Delta \varphi =2\pi \left(\frac{S}{\beta }\right)$ ...(4)
$\{\beta =$ fringe width $\}$
$\beta =\frac{{\lambda }^{\prime }D}{d}=\frac{\left(\lambda /{\mu }_w\right)D}{d}$
$=\frac{\left[\frac{5000\mathring{A}}{4/3}\right]\times 0.8}{2\times 10^{-3}}=150\mu m$
From (4), $\Delta \varphi =2\pi \times \frac{(25\mu m)}{(150\mu m)}=\frac{\pi }{3}$
Intensity at $C,I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$
$=I_0+I_0+2\sqrt{I_0{I}_0}\cos \pi /3$
$=2I_0+2I_0\cdot \frac{1}{2}=3I_0$
$I_{\max }={\left(\sqrt{I_1}+\sqrt{I_2}\right)}^2$
$={\left(\sqrt{I_0}+\sqrt{I_0}\right)}^2=4I_0$
So, $\frac{I}{I_{\max }}=\frac{3I_0}{4I_0}=\frac{3}{4}$