Question

In a $YDSE$, both slits produce equal intensities on the screen. A $100\%$ transparent thin film is placed in front of one of the slits. Now, the intensity on the centre becomes $75\%$ of the previous intensity. The wavelength of light is $6000\dot{\text{A}}$ and refractive index of glass is $1.5$. The minimum thickness of the glass slab is

• A. $0.2\mu \text{m}$
• B. $0.3\mu \text{m}$
• C. $0.4\mu \text{m}$
• D. $0.5\mu \text{m}$

Answer 1

The correct option is A $0.2\mu \text{m}$

As we know,

$I=I_{max}{\cos }^2\left(\frac{\Delta \varphi }{2}\right)$

According to problem, when glass is placed, the inetensity

$I=\frac{75}{100}{I}_{max}=\frac{3}{4}{I}_{max}$

$\Rightarrow \frac{3}{4}{I}_{max}=I_{max}{\cos }^2\left(\frac{\Delta \varphi }{2}\right)$

$\Rightarrow \cos \left(\frac{\Delta \varphi }{2}\right)=\frac{\sqrt{3}}{2}$

$\Rightarrow \frac{\Delta \varphi }{2}=\frac{\pi }{6}(2n+1)[n=0,1,2,3,\mathrm{4....}]$

$\therefore \Delta \varphi =\frac{\pi }{3}(2n+1)$

The path difference between interfering waves will be,

$\Delta x=\frac{\Delta \varphi }{\frac{2\pi }{\lambda }}$

$\Rightarrow \Delta x=\frac{\pi (2n+1)\times \lambda }{3\left(2\pi \right)}=\frac{\lambda }{6}[\because \text{For minimum}\Delta x,n=0]$

Now, path difference created by glass slab is,

$\Delta x_{slab}=(\mu -1)t$

$\Rightarrow (\mu -1)t=\frac{\lambda }{6}$

$\Rightarrow t=\frac{\lambda }{6(\mu -1)}=\frac{6000\times {10}^{-10}}{6(1.5-1)}$

$=0.2\times {10}^{-6}\text{m}=0.2\mu \text{m}$

Hence, $\left(\text{A}\right)$ is the correct answer.