Question

In Young's double slit experiment, a monchromatic light beam of wavelength $\lambda$ from a distant point source is incident upon the two slits and the interference pattern is viewed on a distant screen. Intensity at a point $P$ is equal to the intensity due to individual slits (equal to $I_0$). A thin glass plate of thickness $t$ and refractive index $\mu$ is placed in front of the slit which is at larger distance from $P$, perpendicular to the light path. Assume no absorption of light energy by the glass and choose the correct statements.

• A. The phase difference between the waves arriving at $P$ before placing plate is $\frac{2\pi }{3}$ radian.
• B. After placing at glass piece, the phase difference between waves arriving at $P$ is $\frac{2\pi }{3}+\frac{2\pi (\mu -1)t}{\lambda }$
• C. Intensity at $P$ after placing glass plate is $4I_0{\cos }^2[\frac{\pi }{3}+\frac{\pi }{\lambda }(\mu -1\left)t\right]$
• D. For the intensity to be minimum at $P$, the thickness of glass plate should be,
  $t=\frac{[2n-1]\lambda }{2[\mu -1]}-\frac{\lambda }{3[\mu -1]}$
  where $n=1,2,3,...$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A The phase difference between the waves arriving at $P$ before placing plate is $\frac{2\pi }{3}$ radian.
B After placing at glass piece, the phase difference between waves arriving at $P$ is $\frac{2\pi }{3}+\frac{2\pi (\mu -1)t}{\lambda }$
C Intensity at $P$ after placing glass plate is $4I_0{\cos }^2[\frac{\pi }{3}+\frac{\pi }{\lambda }(\mu -1\left)t\right]$
D For the intensity to be minimum at $P$, the thickness of glass plate should be,

$t=\frac{[2n-1]\lambda }{2[\mu -1]}-\frac{\lambda }{3[\mu -1]}$
where $n=1,2,3,...$
If $\delta$ is phase difference between the two waves arriving at $P$, then,
$I_0=4I_0{\cos }^2\left[\frac{\delta }{2}\right]$
$\Rightarrow \delta =\frac{2\pi }{3}$
After placing the glass piece, the new phase difference will be:
${\delta }^{\prime }=\frac{2\pi }{3}+\frac{2\pi (\mu -1)t}{\lambda }$
Now, intensity at $P$ will be $4I_0{\cos }^2[\frac{\pi }{3}+\frac{\pi }{\lambda }(\mu -1\left)t\right]$
Intensity at $P$ will be minimum $(=0)$ if,
$\frac{\pi }{3}+\frac{\pi }{\lambda }(\mu -1)t=(2n-1)\frac{\pi }{2}$
where $n=1,2,3,...$

$t=\frac{[2n-1]\lambda }{2[\mu -1]}-\frac{\lambda }{3[\mu -1]}$
where $n=1,2,3,...$