Question

In Young's double slit experiment, two slits of equal intensities produce interference pattern forming central maxima at point $O$. A transparent film $(\mu =1.5)$ is placed in front of one of the slits. Now the intensity of light at point $O$ is $75\%$ of the intensity of a bright fringe. The wavelength of light used is $6000\mathring{\text{A}}$. The thickness of the film cannot be,

• A. $0.2\mu \text{m}$
• B. $1.0\mu \text{m}$
• C. $1.4\mu \text{m}$
• D. $1.6\mu \text{m}$

Answer 1

The correct option is D $1.6\mu \text{m}$

Let $I$ is the intensity of each slit. So, the intensity at a maxima will be $4I$

$\therefore 75\%$ intensity will correspond to a point where intensity is $3I$.

Now, resultant intensity at a point is given by,

$I_R=I_0+I_0+2\sqrt{I_0}\sqrt{I_0}\cos \left(\varphi \right)$

$\Rightarrow 3I_0=2I_0(1+\cos \varphi )$

$\cos \left(\varphi \right)=\frac{1}{2}$

$\therefore \varphi =\frac{\pi }{3},2\pi -\frac{\pi }{3},2\pi +\frac{\pi }{3},4\pi -\frac{\pi }{3}........$

And, $\Delta x=\frac{\lambda }{6},\lambda -\frac{\lambda }{6},\lambda +\frac{\lambda }{6},2\lambda -\frac{\lambda }{6},........$

Now, path difference due to the film is given by,
$\Delta x=(\mu -1)t\Rightarrow t=\frac{\Delta x}{(\mu -1)}$

For $\Delta x=\frac{\lambda }{6},t=\frac{{10}^{-7}}{1.5-1}=0.2\mu \text{m}$

For $\Delta x=\lambda -\frac{\lambda }{6},t=\frac{5\times {10}^{-7}}{1.5-1}=1\mu \text{m}$

For $\Delta x=\lambda +\frac{\lambda }{6},t=\frac{7\times {10}^{-7}}{1.5-1}=1.4\mu \text{m}$

For $\Delta x=2\lambda -\frac{\lambda }{6},t=\frac{11\times {10}^{-7}}{1.5-1}=2.2\mu \text{m}$

Hence, $\left(D\right)$ is the correct answer.