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 slit $S_2$ is covered by another sheet of thickness $t_2=1.25\mu \text{m}$ as shown in figure


Both sheets are made of same material having refractive index $\mu =1.40$. Water is filled in the space between diaphragm and screen. A monochromatic light beam of wavelength $\lambda =5000\dot{\text{A}}$ is incident normally on the diaphragm. Assuming intensity of beam to be uniform, calculate ratio of intensity at $C$ to the maximum intensity of interference pattern obtained on the screen $({\mu }_w=\frac{4}{3})$.

• A. $\frac{4}{3}$
• B. $\frac{2}{3}$
• C. $\frac{3}{4}$
• D. $\frac{1}{3}$

Answer 1

The correct option is C $\frac{3}{4}$

Path difference of light waves reaching at $C$

$\Delta x=t_1({\mu }^{\prime }-1)-t_2({\mu }^{\prime }-1)$

Where, ${\mu }^{\prime }=\frac{\mu }{{\mu }_w}$

$\Rightarrow \Delta x={\mu }^{\prime }(t_1-t_2)-(t_1-t_2)=(t_1-t_2)({\mu }^{\prime }-1)$

$=(2.5-1.25)(\frac{1.4\times 3}{4}-1)$

$=1.25\times \frac{2}{40}=\frac{2.5}{40}$

$\Rightarrow \Delta x=\frac{1}{16}\mu \text{m}$

Phase difference,

$\varphi =\frac{2\pi }{{\lambda }^{\prime }}\Delta x$ where, ${\lambda }^{\prime }=\frac{\lambda }{{\mu }_w}$

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

$\Rightarrow \varphi =\frac{\pi }{3}$

As, $I_{max}=4I_0$

Intensity at $C$,

$I_C=4I_0{\cos }^2\frac{\varphi }{2}=2I_0(1+\cos \frac{\pi }{3})=3I_0$

$\therefore$ Required ratio$=\frac{I_C}{I_{max}}=\frac{3}{4}$

Hence, option $\left(\text{C}\right)$ is correct.