Question

In the figure, 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 $x$ is the distance from the plane of slits as shown), up to a distance $l$ before falling on $S_1$. The rest of the space is filled with air. If at $O$ minima is formed, then the minimum value of the positive constant $a$ ( in terms of $l$ and wavelength $\lambda$ in air) is

• A. $\frac{\lambda }{l}$
• B. $\frac{\lambda }{l^2}$
• C. $\frac{l^2}{\lambda }$
• D. $\frac{l}{\lambda }$

Answer 1

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

When light passes through a medium of refractive index $\mu$,

$\therefore$ Path difference $=(\mu -1)l$

For a small element $dx$ of medium, the path difference,

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

For the whole length,

$\Delta x={\int }_0^{l}axdx=\frac{a{l}^2}{2}$

For minima at $O$,

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

$\Rightarrow \frac{a{l}^2}{2}=(2n-1)\frac{\lambda }{2}$

For minimum value of $a$, $n=1$.

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

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

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