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JEE Main 201511 AprilOnline
Exam
JEE Main
Year
2015
Date
11 April
Mode
Online
Subject
Physics
Chapter
Wave Optics

JEE Main 2015 · 11 April · Online · Physics · Wave Optics

In a Young's double slit experiment with light of wavelength λ ,...

In a Young's double slit experiment with light of wavelength λ,\lambda , the separation of slits is dd and distance of screen is DD such that D≫d≫λD ≫ d ≫ \lambda . If the Fringe width is β\beta , the distance from point of maximum intensity to the point where intensity falls to half of the maximum intensity on either side is:

In a Young's double slit experiment with light of wavelength λ , the separation of slits is d and distance of screen is

Options

  1. Aβ4\dfrac{\beta}{4}
  2. Bβ3\dfrac{\beta}{3}
  3. Cβ6\dfrac{\beta}{6}
  4. Dβ2\dfrac{\beta}{2}

Correct answer

A. β4\dfrac{\beta}{4}

Step-by-step solution

As {I}_\max = 4I

Then {I}_{net} = \dfrac{1}{2}{I}_\max = 2I

⇒2I=2I(1+cosϕ)⇒ 2I = 2I\left(1 + cosϕ\right)

⇒1+cosϕ=1⇒ 1 + cosϕ = 1

⇒cosϕ=0⇒ cosϕ = 0

⇒ϕ=π2,3π2,5π2,…⇒ ϕ = \dfrac{\pi}{2}, \dfrac{3\pi}{2}, \dfrac{5\pi}{2}, \ldots

ϕ=2πλ\Deltaxϕ = \dfrac{2\pi}{\lambda}\Deltax

⇒π2=2πλ\Deltax⇒ \dfrac{\pi}{2} = \dfrac{2\pi}{\lambda}\Deltax

⇒\Deltax=λ4,3λ4,5λ4,…⇒ \Deltax = \dfrac{\lambda}{4}, \dfrac{3\lambda}{4}, \dfrac{5\lambda}{4}, \ldots

In a Young's double slit experiment with light of wavelength λ , the separation of slits is d and distance of screen is

Also path difference \Deltax=dsinθ=λ4\Deltax = dsin \theta = \dfrac{\lambda}{4}

At desired location

⇒dynD=λ4⇒ \dfrac{d{y}_{n}}{D} = \dfrac{\lambda}{4}

⇒yn=\lambdaD4d=β4{∵β=\lambdaDd}⇒ {y}_{n} = \dfrac{\lambdaD}{4d} = \dfrac{\beta}{4}\left\{∵ \beta = \dfrac{\lambdaD}{d}\right\}

⇒⇒ Separation between central maxima and desired y=β4y = \dfrac{\beta}{4}
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