Question

In the figure, a parallel beam of light is incident on the plane of the slits of a 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 ${}^{\prime }{l}^{\prime }$ 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 wavelength ${}^{\prime }{\lambda }^{\prime }$ in air) is

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

Answer 1

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

Path difference due to insertion of a glass slab of thickness $t$ and refractive index $n$ is given by $t(n-1)$.

In the above case, consider a section of the variable refractive index medium of infinitesimal thickness at a distance x from the slits' plane.

Path difference due to this section=$\delta \left(\Delta x\right)=dx(\mu -1)=dx(1+ax-1)=dx\left(ax\right)$

Therefore the total path difference due to the variable refractive index=$\Delta x={\int }_0^{l}axdx=\frac{a{l}^2}{2}$

The rays from the two slits now have a path difference of $\frac{a{l}^2}{2}$. To obtain a minima at O, the path difference must be an odd multiple of $\frac{\lambda }{2}$.

$⟹\frac{a{l}^2}{2}=(n+\frac{1}{2})\lambda$

'a' is minimum for n=0,

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