Verified Solution • 2025–26 Session moderate

QJEEM20Ph4faf: JEE Main 2020 Physics - Class 12 JEEM20Ph4faf Solution: In a Young’s double slit experiment, light of 500 nm is used to …

In a Young’s double slit experiment, light of 500 nm is used to produce an interference pattern. When the distance between the slits is 0.05 mm, the angular width (in degree) of the fringes formed on the distance screen is close to

Multiple Choice Options

  • A 0.17o
  • B 1.7o
  • C 0.57o
  • D 0.07o

Step-by-Step Verified Solution

β\beta = λDd{{\lambda D} \over d}

and θ\theta = βD{\beta \over D}

⇒\Rightarrow θ\theta = λd{\lambda \over d}

= 500×10−90.05×10−3{{500 \times {{10}^{ - 9}}} \over {0.05 \times {{10}^{ - 3}}}}

= 0.01 rad

= 0.57o

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