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 $x^2 = p^2m^2λ^2 - d^2$, where $λ$ 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

For maxima at point p
Path difference between waves $Δx = mλ$
$S_2 p - S_1p = mλ$
$\mu\sqrt{x^{2} + d^{2}}- \sqrt{x^{2}+d^{2}} = mλ$
$\sqrt{x^{2}+d^{2}}\left(\mu-1\right) = mλ$
$\sqrt{x^{2}+d^{2}} = mλ\quad\left(\because\,\mu = \frac{4}{3}\right)$
$x^{2} = 3^{2}m^{2}λ^{2} - d^{2}$
On comparing we get $p = 3$