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JEE Main 20169 AprilOnline
Exam
JEE Main
Year
2016
Date
9 April
Mode
Online
Subject
Physics
Chapter
Wave Optics

JEE Main 2016 · 9 April · Online · Physics · Wave Optics

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

In Young's double-slit experiment, the distance between slits and the screen is 1m1m and monochromatic light of wavelength 600nm600nm is being used. A person standing near the slits is looking at the fringe pattern. When the separation between the slits is varied, the interference pattern disappears for a particular distance d0{d}_{0} between the slits. If the angular resolution of the eye is 160°\dfrac{1}{60} °, then the value of d0{d}_{0} is close to

Options

  1. A1mm1mm
  2. B3mm3mm
  3. C2mm2mm
  4. D4mm4mm

Correct answer

C. 2mm2mm

Step-by-step solution

When the angular fringe width in the interference pattern becomes less than the angular resolving power of the person's eye, the fringes will disappear.

The angular fringe width θ0=βD=λd0{\theta}_{0} = \dfrac{\beta}{D} = \dfrac{\lambda}{{d}_{0}}

d0=λθ0=600×10−9(160×π180){d}_{0} = \dfrac{\lambda}{{\theta}_{0}} = \dfrac{600 \times {10}^{- 9}}{( \dfrac{1}{60} \times \dfrac{\pi}{180} )}

d0=600×60×180×10−9πm=2mm{d}_{0} = \dfrac{600 \times 60 \times 180 \times {10}^{- 9}}{\pi}m = 2\text{mm}

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