Question

A Young's double slit interference arrangement with slits $S_1$ and $S_2$ is immersed in water (refractive index = 4/3) as shown in 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), 2d is the separation between the slits and m is an integer. The value of p is

Answer 1

Answer: 3

For maxima at point p
Path difference between waves $\Delta x=m\lambda$
$S_{2}p-S_{1}p=m\lambda$
$\mu \sqrt{x^2+d^2}-\sqrt{x^2+d^2}=m\lambda$
$\sqrt{x^2+d^2}\left(\mu -1\right)=m\lambda$
$\sqrt{x^2+d^2}=m\lambda \left(\because \mu =\frac{4}{3}\right)$
$x^2=3^2{m}^2{\lambda }^2-d^2$
On comparing we get $p=3$