JEE Main 2016PhysicsWave OpticsMediumMCQ

JEE Main 2016 — Wave Optics Question with Solution

JEE Main 2016 (09 Apr Online)

Question

In Young's double-slit experiment, the distance between slits and the screen is 1 m and monochromatic light of wavelength 600 nm 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 between the slits. If the angular resolution of the eye is 160°, then the value of d0 is close to

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Show full solutionCorrect option: C
Correct answer
C2 mm

Step-by-step explanation

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 θ 0 = β D = λ d 0

d 0 = λ θ 0 = 600× 10 −9 ( 1 60 × π 180 )

d0=600×60×180×10−9π m=2 mm

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About this question

This is a previous-year question from JEE Main 2016, covering the Wave Optics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.