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 monochromatic light beam of wavelength $\lambda =5000\mathring{A}$ 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 (${\mu }_w=4/3$).

Answer 1

${\mu }_{rslab}=\frac{{\mu }_{slab}}{{\mu }_{water}}=\frac{1.40}{4/3}$

Effective Path Difference:

${\delta }_e=\Delta t\times ({\mu }_{rslab}-1)=6.25\times {10}^{-8}m$

Therefore, Effective Phase Difference:

${\varphi }_e={\delta }_e\times \frac{2\pi }{\lambda }=\frac{\pi }{3}$

Hence $I_o=I_1+I_2+2\times \sqrt{I_1{I}_2}\cos \left({\varphi }_e\right)$

where $I_1=I_2=I$

$\therefore I_o=3I$

Answer is 3