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JEE Main 201419 AprilOnline
Exam
JEE Main
Year
2014
Date
19 April
Mode
Online
Subject
Physics
Chapter
Wave Optics

JEE Main 2014 · 19 April · Online · Physics · Wave Optics

In Young’s double-slit experiment, the distance between the two...

In Young’s double-slit experiment, the distance between the two identical slits is6.16 . 1 times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single-slit diffraction pattern is :

Options

  1. A33
  2. B66
  3. C2424
  4. D1212

Correct answer

D. 1212

Step-by-step solution

d=d = separation between 2 slits

a=a = slit width

βy={\beta}_{y} = fringe width in YDSE & βD={\beta}_{D} = width of central maxima

βy=\lambdaDd=\lambdaD6.1aβD=2\lambdaDa{\beta}_{y} = \dfrac{\lambdaD}{d} = \dfrac{\lambdaD}{6.1a}{\beta}_{D} = \dfrac{2\lambdaD}{a}

βD=2×6.1βy{\beta}_{D} = 2 \times 6.1{\beta}_{y}

=12.2βy= 12.2{\beta}_{y}
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