Question

In the figure shown, a parallel beam of light is incident on the plane of the slits of Young's double slit experiment. Light incident on the slit $S_1$ passes through a medium of variable refractive index $\mu =1+ax$ (where ${}^{\prime }{x}^{\prime }$ is the distance from the plane of slits as shown), up to a distance ‘𝑙’ before falling on $S_1$. Rest of the space is filled with air. If at ${}^{\prime }{O}^{\prime }$ a minima is formed , then the minimum value of the positive constant a (in terms of $l$ and wave length ${}^{\prime }{\lambda }^{\prime }$ in air) is

• A. $\frac{l^2}{\lambda }$
• B. $\frac{\lambda }{l}$
• C. None of these
• D. $\frac{\lambda }{l^2}$

Answer 1

The correct option is D $\frac{\lambda }{l^2}$

We know that: $\Delta x=(2n+1)\frac{\lambda }{2}$

Optical path shift due to introduction of film
$=(\mu -1)t$

Given: $\mu =1+ax$

Total length of slab $=l$

For a small element ${}^{\prime }d{x}^{\prime }$ change in path difference can be calculated as,

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

$\Delta x=\left[\right(1+ax)-1]dx=axdx$

For the whole length,
$\Delta x={\int }_0^{1}axdx=\frac{\left(a{l}^2\right)}{2}$

For a minima to be at ${}^{\prime }{O}^{\prime }$

$\Delta x=\frac{(2n+1)\lambda }{2}$

i.e,$\frac{a{l}^2}{2}=\frac{(2n+1)\lambda }{2}$

For minimum ${}^{\prime }{a}^{\prime },n=0$

$\Rightarrow \frac{\left(a{l}^2\right)}{2}=\frac{\lambda }{2}$

$\Rightarrow a=\frac{\lambda }{l^2}$