Question

A screen is at a distance $D=80\text{cm}$ from a diaphragm having two narrow slits, $S_{1}and{S}_2$ which are $d=2\text{mm}$ apart. Slit $S_1$ is covered by a transparent sheet of thickness $t_1=2.5\mu \text{m}and{S}_2$ by another transparent sheet of thickness $t_2=1.25\mu \text{m}$ . sheets are made of same material having refractive index $\mu =1.40$. Water is filled in space between diaphragm and screen. A monochromatic light beam of wavelength $\lambda =5000A^{\circ }$ is incident normally on the diaphragm. Assuming intensity of beam to be uniform, calculate ratio of intensity of center of screen to intensity of individual slit, $({\mu }_w=\frac{4}{3})$

Answer 1

Let $C$ be the center of screen.

Path difference at $C$,

$\Delta x=t_1({\mu }_{rel}-1)-t_2({\mu }_{rel}-1)$

$\Delta x={\mu }_{rel}(t_1-t_2)-(t_1-t_2)=(t_1-t_2)({\mu }_{rel}-1)$

$\Delta x=(2.5-1.25)(\frac{14\times 3}{4\times 10}-1)$

$\Delta x=1.25\times \frac{1}{20}=\frac{1}{16}\mu \text{m}$

$\varphi =\frac{2\pi }{\lambda }\times \Delta x=\frac{2\pi \times 4}{5000\times 3\times {10}^{-10}}\times \frac{1}{16}\times {10}^{-6}$

$\Rightarrow I_{max}=4I_0$

$IatC,I_c=2I_0(1+\cos \frac{\pi }{3})=3I_0$

Required ratio = $\frac{I_c}{I_0}=3$