JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Wave Optics

For constructive interference, the path difference must equal $m\lambda$ (wavelength in air). In water with refractive index $\mu =4/3$, the optical path difference is $\Delta =\mu (\sqrt{d^2+x^2}-\sqrt{d^2+x^2})$. Wait, both slits are in water, so the path difference in water is $\sqrt{d^2+x^2}-\sqrt{d^2+x^2}$ for paths from the two slits. Actually, from the geometry and the condition for maxima: $\frac{4}{3}\sqrt{d^2+x^2}-\sqrt{d^2+x^2}=m\lambda$, giving $\frac{1}{3}\sqrt{d^2+x^2}=m\lambda$. Squaring: $\frac{1}{9}(d^2+x^2)=m^2{\lambda }^2$, so $x^2=9m^2{\lambda }^2-d^2$. Therefore $p=3$.