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 position 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

If light travels a distance x in medium, then the effective distance travelled in vacuum is$\mu x$Consider wavelets undergoing interference at a point P on the water surface.


Optical path length $S_{1}P=\sqrt{x^2+d^2}$

Optical path length $S_{2}P=\mu \sqrt{x^2+d^2}$

Path difference $=(\mu -1)=\sqrt{x^2+d^2}$

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

$\mu =\frac{4}{3}\Rightarrow \frac{1}{3}\sqrt{x^2+d^2}=m\lambda$

$\therefore x^2=9m^2{\lambda }^2-d^2=p^2{m}^2{\lambda }^2-d^2$

$\Rightarrow p^2=9\Rightarrow p=3.$