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JEE Main 202125 JulyMorning Shift
Exam
JEE Main
Year
2021
Date
25 July
Shift
Morning Shift
Subject
Physics
Chapter
Wave Optics

JEE Main 2021 · 25 July · Morning Shift · Physics · Wave Optics

In the Young's double slit experiment, the distance between the...

In the Young's double slit experiment, the distance between the slits varies in time as d(t)=d0+a0sin\omegatd\left(t\right) = {d}_{0} + {a}_{0}sin\omegat; where d0,ω{d}_{0}, \omega and a0{a}_{0} are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as:

Options

  1. A\dfrac{2\lambdaD\left({d}_{0}\right)}{\left({d}_{0}^{2} - {a}_{0}^{2}\right)}
  2. B\dfrac{2\lambdaD{a}_{0}}{\left({d}_{0}^{2} - {a}_{0}^{2}\right)}
  3. C\lambdaDd02a0\lambdaD{d}_{0}^{2}{a}_{0}
  4. D\lambdaDd0+a0\dfrac{\lambdaD}{{d}_{0} + {a}_{0}}

Correct answer

B. \dfrac{2\lambdaD{a}_{0}}{\left({d}_{0}^{2} - {a}_{0}^{2}\right)}

Step-by-step solution

Fringe width,

β=\lambdaDd\beta = \dfrac{\lambdaD}{d}

{\beta}_\max ⇒ {d}_\min and {\beta}_\min ⇒ {d}_\max

d=d0+a0sin\omegatd = {d}_{0} + {a}_{0}sin\omegat

{d}_\max = {d}_{0} + {a}_{0} and {d}_\min = {d}_{0} - {a}_{0}

∴ {\beta}_\min = \dfrac{\lambdaD}{{d}_{0} + {a}_{0}} and ∴ {\beta}_\max = \dfrac{\lambdaD}{{d}_{0} - {a}_{0}}

{\beta}_\max - {\beta}_\min = \dfrac{\lambdaD}{{d}_{0} - {a}_{0}} - \dfrac{\lambdaD}{{d}_{0} + {a}_{0}} = \dfrac{2\lambdaD{a}_{0}}{{d}_{0}^{2} - {a}_{0}^{2}}

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