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QJEEAdv15Ph4af4: JEE Advanced 2015 Physics - Class 12 JEEAdv15Ph4af4 Solution: A Young's double slit interference arrangement with slits S 1…

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 = p2m2λ2 − d2, where λ 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
JEE Advanced 2015 Paper 1 Offline Physics - Wave Optics Question 12 English

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JEE Advanced 2015 Paper 1 Offline Physics - Wave Optics Question 12 English Explanation

μ(S2P)−S1P=mλ

⇒μd2+x2−d2+x2=mλ

⇒(μ−1)d2+x2=mλ

⇒(43−1)d2+x2=mλ

or, d2+x2=3mλ

Squaring this equation we get,

x2=9m2λ2−d2

⇒p2=9 or p=3

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