Question

Minimum thickness of a mica sheet (which should be placed in front of one of the slits in YDSE) required to reduce the intensity at the centre of screen to half of maximum intensity is (the refractive index of the sheet is $3/2$) Assume that the sheet does not absorb any light

• A. $\frac{\lambda }{3}$
• B. $\frac{\lambda }{4}$
• C. $\frac{\lambda }{2}$
• D. $\frac{\lambda }{8}$

Answer 1

The correct option is C $\frac{\lambda }{2}$

Given
$I_{\text{net}}=\frac{I_{\text{max}}}{2}$
$\mu =\frac{3}{2}$

As we know that in interference $I_{\text{max}}=4I_0$
So according to question
$I_{\text{net}}=\frac{4I_0}{2}=2I_0$

Phase difference $=\frac{2\pi }{\lambda }(\mu -1)t$

Now the resultant intensity is given by

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

$I=2I_0=I_0+I_0+2\sqrt{I_0^2}\cos \left(\frac{2\pi }{\lambda }\right(\mu -1\left)t\right)$

$\frac{2\pi }{\lambda }(\mu -1)t=\frac{\pi }{2}$

$t=\frac{\lambda }{2}$

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