JEE Advanced 2015 · Paper 1 · Physics Q7

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PhysicsInteger+4Question 7 of 60

A Young's double slit interference arrangement with slits $S_1$ and $S_2$ 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 $x^2=p^2m^2\lambda^2-d^2$ , 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

A vertical screen with slits S₁ (above the water surface by d) and S₂ (below it by d), separation 2d. To the right, water fills the lower region and air the upper; a point on the water surface at horizontal distance x from the screen is marked.

Your answer

Answer and worked solution

Official answer: 3

Paths. A point on the water surface at distance $x$ is $\sqrt{x^2+d^2}$ from each slit. $S_1$ is in air and $S_2$ is under water, so the optical paths are $\sqrt{x^2+d^2}$ and $\frac43\sqrt{x^2+d^2}$ .

Maxima.
$\frac13\sqrt{x^2+d^2}=m\lambda\implies x^2=9m^2\lambda^2-d^2.$
So $p=3$ .

Answer: 3

Topics:Interference and Young's double slit

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JEE Advanced 2015 · Paper 1, Physics question 7