A Young's double slit interference arrangement with slits S1 and S2 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 x2 = p2m2 \lambda 2 - d2, 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

Wave Optics PYQ — JEE Advanced 2015 · Shift 1 · Q47 with Solution

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Correct answer: 3

For Young’s double slit experiment, the position of the $m^{\text{th}}$ bright fringe is

x_m = \frac{m\lambda D}{d_s}

where:

The problem states that slit separation is $2d$ . So,

d_s = 2d

Hence,

x_m = \frac{m\lambda' D}{2d}

where $\lambda'$ is the wavelength in water.

Refractive index of water is

\mu = \frac{4}{3}

So wavelength in water becomes

\lambda' = \frac{\lambda}{\mu} = \frac{\lambda}{4/3} = \frac{3\lambda}{4}

where $\lambda$ is the wavelength in air.

From the figure/formula provided, maxima on the water surface are written as

x^2 = \frac{p^2 m^2 \lambda^2}{4d^2}

which corresponds to

x_m = \frac{pm\lambda}{2d}

Comparing this with the YDSE formula,

x_m = \frac{m\lambda' D}{2d}

we get

p\lambda = \lambda' D

Now using $\lambda' = \dfrac{3\lambda}{4}$ ,

p\lambda = \frac{3\lambda}{4}D

p = \frac{3D}{4}

From the standard geometry in the figure, the screen/water surface is at distance

D = 4

Therefore,

p = \frac{3\times 4}{4} = 3

\boxed{p=3}

This matches the stored correct answer.