Question

A Young's double slit interference arrangement with slits $S_1$ and $S_2$ is immersed in water (refractive index $=\frac{4}{3}$) as shown in the figure. The positions of maximum 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 Path defference $=m\lambda$
$\therefore S_{2}A-S_{1}A=m\lambda$

$\therefore \left[\left(n-1\right)\sqrt{d^2+x^2}+\sqrt{d^2+x^2}\right]-\sqrt{d^2-x^2}=m\lambda$
$\therefore \left(n-1\right)\sqrt{\left(d^2+x^2\right)}=m\lambda$
$\therefore \left(\frac{4}{3}-1\right)\sqrt{d^2+x^2}m\lambda$
$\therefore \sqrt{d^2{x}^2}=3m\lambda$
$\therefore d^2+x^2=9m^2{\lambda }^2$
$\therefore x^2=9m^2{\lambda }^2-d^2$
$\therefore P^2=9$
$\Rightarrow P=3$