Question

A Young's double slit experiment is conducted in water $\left({\mu }_1\right)$ as shown in the figure, and a glass plate of thickness $t$ and refractive index ${\mu }_2$ is placed in the path of $S_2$. The magnitude of the phase difference at $O$ is :
(Assume that ${}^{\prime }{\lambda }^{\prime }$ is the wavelength of light in air). $O$ is symmetrical w.r.t. $S_1$ and $S_2$.

• A. $\left|(\frac{{\mu }_2}{{\mu }_1}-1)t\right|\frac{2\pi }{\lambda }$
  
• B. $\left|(\frac{{\mu }_1}{{\mu }_2}-1)t\right|\frac{2\pi }{\lambda }$
  
• C. $|({\mu }_2-{\mu }_1)t|\frac{2\pi }{\lambda }$
  
• D. $|({\mu }_2-1)t|\frac{2\pi }{\lambda }$
  

Answer 1

The correct option is C $|({\mu }_2-{\mu }_1)t|\frac{2\pi }{\lambda }$


Wavelength of light in the slab is given by :-
${\lambda }_s=\frac{\lambda }{{\mu }_1}$
Now, we know that path difference due to a glass slab is given by :-
$\Delta x=(\frac{{\mu }_2}{{\mu }_1}-1)t$
$\because$ Light rays from both the slits have to cover same distance upto point ${}^{\prime }{O}^{\prime }$ if the slab is not their. Therefore, the path difference at point ${}^{\prime }{O}^{\prime }$ will due to the slab only.
$\Rightarrow \Delta \varphi =\frac{2\pi }{{\lambda }_s}\times \Delta x=\frac{2\pi }{\lambda /{\mu }_1}\times (\frac{{\mu }_2}{{\mu }_1}-1)t$
$\Rightarrow \Delta \varphi =\frac{2\pi }{\lambda }{\mu }_1\times \left(\frac{{\mu }_2-{\mu }_1}{{\mu }_1}\right)t$

$\Rightarrow \begin{array}{c}\Delta \varphi =\frac{2\pi |({\mu }_2-{\mu }_1)|t}{\lambda }\end{array}$

Hence, option (C) is the answer.