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

Correct answer: 0 x _ 1 = x ^ 2 + d ^ 2 x _ 2 = (4)/(3) x ^ 2 + d ^ 2 ∆ x = ((4)/(3) - 1) x ^ 2 + d ^ 2 = nλ (1)/(3) x ^ 2 + d ^ 2 = nλ ( x ^ 2 + d ^ 2 ) = 9 n ^ 2 λ ^ 2 ∴ (P = 3)

JEE Advanced 2015Paper 1

Exam: JEE Advanced
Year: 2015
Paper: Paper 1
Subject: Physics
Chapter: Wave Optics

JEE Advanced 2015 · Paper 1 · Physics · Wave Optics

A Young's double slit interference arrangement with slits S _ 1 and...

A Young's double slit interference arrangement with slits ${S}_{1}$ and ${S}_{2}$ is immersed in water (refractive index $= \dfrac{4}{3}$ ) as shown in the figure. The positions of maxima 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

$A Young's double slit interference arrangement with slits S _ 1 and S _ 2 is immersed in water (refractive index =$