Question

In a Young's double slit experiment, the slits $S_1$ and $S_2$ are illuminated by a parallel beam of light of wavelength $4000\dot{A}$ from the medium of refractive index ${\mu }_1=1.2$. A thin film of thickness $1.2\mu \text{m}$ and refractive index $\mu =1.5$ is placed in front of $S_1$ perpendicular to the path of light. The refractive index of the medium between the plane of slits and screen is ${\mu }_2=1.4$. If the light coming from $S_1$ and $S_2$ have equal intensity $I$, then the intensity at the geometrical centre of screen $O$ is:

• A. $0$
• B. $2I$
• C. $4I$
• D. $I$

Answer 1

The correct option is B $2I$

Path difference between interfering waves at $O$ is,

$\Delta x=[S_{1}O+({\mu }_{rel}-1\left)t\right]-S_{2}O$

$\because S_{1}O=S_{2}O$

$\Rightarrow \Delta x=({\mu }_{rel}-1)t$

$\Rightarrow \Delta x=(\frac{\mu }{{\mu }_2}-1)t$

Therefore, phase difference at $O$ is given by:
$\Delta \varphi =\left(\frac{2\pi }{\lambda }\right)\Delta x$

$\Rightarrow \Delta \varphi =\left(\frac{2\pi }{{\lambda }_2}\right)(\frac{\mu }{{\mu }_2}-1)t$

Where, ${\lambda }_2$ is the wavelength of light in medium between slits and screen.

Using Snell's law,
${\mu }_1{\lambda }_1={\mu }_2{\lambda }_2$

$\Rightarrow {\lambda }_2=\frac{{\mu }_1{\lambda }_1}{{\mu }_2}$

$\Rightarrow \Delta \varphi =\frac{2\pi {\mu }_2}{{\mu }_1{\lambda }_1}(\frac{\mu }{{\mu }_2}-1)t$

$\Rightarrow \Delta \varphi =\frac{2\pi \left(1.4\right)}{1.2(4000\times {10}^{-10})}(\frac{1.5}{1.4}-1)1.2\times {10}^{-6}$

$\Rightarrow \Delta \varphi =\frac{2\pi }{4}=\frac{\pi }{2}$

Therefore, resultant intensity at $O$ is given by,

$I_R=I_1+I_2+2\sqrt{I_1{I}_2}\cos \Delta \varphi$

$\Rightarrow I_R=I+I+2\sqrt{II}\cos \frac{\pi }{2}$

$\Rightarrow I_R=2I$

Hence, option $\left(B\right)$ is correct.