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

For maxima Path defference $= m \lambda$
$\therefore S_{2}A - S_{1}A = m \lambda$

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