Question

When a thin transparent sheet of refractive index $\mu =\frac{3}{2}$ is placed near one of the slit in YDSE, the intensity at centre of screen reduces to half of maximum intensity. Then, minimum thickness of sheet should be

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

Answer 1

Answer: (C)

$l_R=4l_0{\cos }^2\left(\frac{\Delta \theta }{2}\right)$
As given $l_R$ at centre is $2l_0$, then $2I_0=4l_0{\cos }^2\left(\frac{\Delta \theta }{2}\right)$
$\Rightarrow$ $\left(\frac{\Delta \theta }{2}\right)=\frac{\pi }{4}$ [for minimum result]
$\Delta \theta =\frac{\pi }{2}$
$\Delta x=\Delta \theta \times \frac{\lambda }{2\pi }=\frac{\lambda }{4}$
For slab $\Delta x=\left(\mu -1\right)t$
$\left(\frac{3}{2}-1\right)t=\frac{\lambda }{4}$
$t=\frac{\lambda }{2}$