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 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), $2d$ is the separation between the slits and $m$ is an integer. The value of $p$ is

Answer 1

According to the given situation, a maxima obtained at the screen. Where, $S_{1}O$ and $S_{2}O$ are geometrically equal.
In the given condition an optical path difference will creat due to change in medium. lat assume $\Delta X_0$ is the optical path difference and the distance between slits and screen is $X$, so
$\Delta X_0=S_{1}0\mu -S_{2}0\mu$

$X_1=\sqrt{X^2+d^2}$
$X_2=\mu \sqrt{X^2+d^2}=\frac{4}{3}\sqrt{X^2+d^2}$
optical Path difference ($\Delta X_0$) =$X_2-X_1$

$\Rightarrow (\frac{4}{3}-1)\sqrt{X^2+d^2}=m\lambda$
$\Rightarrow \sqrt{X^2+d^2}=3m\lambda$
$X^2=9m^2{\lambda }^2-d^2$ ...(1)
by compairing (1) to the given equation
$x^2=p^2{m}^2{\lambda }^2-d^2$ (given)
$\Rightarrow p^2=9$
$\Rightarrow p=3$