A Young's double slit interference arrangement with slits ${\mathrm{S}}_1$ and ${\mathrm{S}}_2$ is immersed in water (refractive index $=\frac{4}{3}$ ) as shown in the figure. The positions of maxima on the surface of water are given by ${\mathrm{x}}^2={\mathrm{p}}^2{\mathrm{m}}^2{\mathrm{\lambda }}^2-{\mathrm{d}}^2$ , where $\mathrm{\lambda }$ is the wavelength of light in air (refractive index = 1), $2\mathrm{d}$ is the separation between the slits and m is an integer. The value of p is

Solution

${\mathrm{x}}_1=\sqrt{{\mathrm{x}}^2+{\mathrm{d}}^2}$
${\mathrm{x}}_2=\frac{4}{3}\sqrt{{\mathrm{x}}^2+{\mathrm{d}}^2}$
$∆\mathrm{x}=\left(\frac{4}{3}-1\right)\sqrt{{\mathrm{x}}^2+{\mathrm{d}}^2}=\mathrm{n}\mathrm{\lambda }$
$\frac{1}{3}\sqrt{{\mathrm{x}}^2+{\mathrm{d}}^2}=\mathrm{n}\mathrm{\lambda }$
$\left({\mathrm{x}}^2+{\mathrm{d}}^2\right)=9{\mathrm{n}}^2{\mathrm{\lambda }}^2$
$\therefore \left(\mathrm{P}=3\right)$

Hence, the correct answer is 3.