Question

Minimum thickness of a mica sheet having $\mu =\frac{3}{2}$ which should be placed in front of one of the slits in $YDSE$, such that the intensity at the centre of the screen reduces to half of maximum intensity is (Consider an equal intensity of light from each slit)

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

Answer 1

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

Maximum intensity correspond to the point of maxima where, $\Delta \varphi =0$

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

$\Rightarrow I_{max}=I+I+2I=4I...\left(1\right)$

After placing a glass slab,

$\Delta x=(\mu -1)t$

Thus, the corresponding phase difference at $O$ is given by,

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

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

$\Rightarrow \Delta \varphi =\frac{2\pi }{\lambda }(1.5-1)t$

$\Rightarrow \Delta \varphi =\frac{\pi t}{\lambda }$

According to problem, at $O$, the intensity is $\frac{I_{max}}{2}$

$\Rightarrow \frac{I_{max}}{2}=I+I+2\sqrt{II}\cos \left(\frac{\pi t}{\lambda }\right)$

From equation $\left(1\right)$, $\frac{I_{max}}{2}=2I$

$\Rightarrow 2I=I+I+2\sqrt{II}\cos \left(\frac{\pi t}{\lambda }\right)$

$\Rightarrow 2I\cos \left(\frac{\pi t}{\lambda }\right)=0$

$\Rightarrow \cos \left(\frac{\pi t}{\lambda }\right)=0$

$\cos \theta =0or\theta =(2n-1)\frac{\pi }{2}$

For minimum value of thickness of micasheet, the value of $n=1$
.
$\Rightarrow \frac{\pi t}{\lambda }=\frac{\pi }{2}$

$\therefore t_{min}=\frac{\lambda }{2}$

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