Question

A Young's double slit interference arrangement with slits $S_1$ and $S_2$ is immersed in water (refractive index $=4\sqrt{3}$) as shown in the figure. The positions of maxima on the surface of water are given by $x^2=p^2{m}^2{\lambda }^2-d^2$, where $\lambda$ is the wavelength of light in air(refractive index $=1$), $2$d is the separation between the slits and m is an integer. The value of p is?

Answer 1

Path length,

$S_{2}O=\mu \sqrt{x^2+d^2}$

where O is the point of intersection on the water surface.

Path length,

$S_{1}O=\sqrt{x^2+d^2}$

path difference,

$\Delta x=S_{2}O-S_{1}O$

For constructive interference,

$\Delta x=m\lambda$

$\Rightarrow \sqrt{x^2+d^2}(\mu -1)=m\lambda$

Given,

$\mu =\frac{4}{3}$

that gives,

$x^2+d^2={\left(\frac{m\lambda }{\frac{4}{3}-1}\right)}^2$

$or,x^2=9m^2{\lambda }^2-d^2$

But given, $x^2=p^2{m}^2{\lambda }^2-d^2$

On comparing we get $p=3$