Question

A screen is at a distance $D=80\, cm$ from a diaphragm having two narrow slits $S_{1}$ and $S_{2}$ which are $d=2\, mm$ 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

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 $C$.
$\Delta \phi=2 \pi\left(\frac{ S }{\beta}\right)$ ...(4)
$\{\beta=$ fringe width $\}$
$\beta=\frac{\lambda' D }{ d }=\frac{\left(\lambda / \mu_{ w }\right) D }{ d }$
$=\frac{\left[\frac{5000\,\mathring{A}}{4 / 3}\right] \times 0.8}{2 \times 10^{-3}}=150\, \mu m$
From (4), $\Delta \phi=2 \pi \times \frac{(25\, \mu m )}{(150\, \mu m )}=\frac{\pi}{3}$
Intensity at $C, I=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}} \cos \phi$
$= I _{0}+ I _{0}+2 \sqrt{ I _{0} I _{0}} \cos \pi / 3$
$=2 I _{0}+2 I _{0} \cdot \frac{1}{2}=3 I _{0}$
$I _{\max }=\left(\sqrt{ I _{1}}+\sqrt{ I _{2}}\right)^{2}$
$=\left(\sqrt{ I _{0}}+\sqrt{ I _{0}}\right)^{2}=4 I _{0}$
So, $\frac{ I }{ I _{\max }}=\frac{3 I _{0}}{4 I _{0}}=\frac{3}{4}$