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QJEEM16Ph1595: JEE Main 2016 Physics - Class 12 JEEM16Ph1595 Solution: In Young’s double slit experiment, the distance between slits…

In Young’s double slit experiment, the distance between slits and the screen is 1.0 m and monochromatic light of 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)o, the value of d0 is close to :

Multiple Choice Options

  • A 1 mm
  • B 2 mm
  • C 4 mm
  • D 3 mm

Step-by-Step Verified Solution

We know,

Fringe width, β = λDd0

JEE Main 2016 (Online) 9th April Morning Slot Physics - Wave Optics Question 117 English Explanation

From image,

θ = βD

⇒  θ = λd0

⇒  d0 = λθ

= 600×10−9160×π180

= 2.06 × 10−3

≃ 2 mm

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