Verified Solution • 2025–26 Session moderate

QJEEM21Ph9ae9: JEE Main 2021 Physics - Class 12 JEEM21Ph9ae9 Solution: In a Young's double slit experiment two slits are separated by…

In a Young's double slit experiment two slits are separated by 2 mm and the screen is placed one meter away. When a light of wavelength 500 nm is used, the fringe separation will be :

Multiple Choice Options

  • A 0.50 mm
  • B 0.25 mm
  • C 1 mm
  • D 0.75 mm

Step-by-Step Verified Solution

Fringe width ( β\beta ) = λDd{{\lambda D} \over d}

d = 2 ×\times 10 −- 3 m

λ\lambda = 500 ×\times 10 −- 9 m

D = 1 m

Now

β\beta = 500×10−9×12×10−3{{500 \times {{10}^{ - 9}} \times 1} \over {2 \times {{10}^{ - 3}}}}

β\beta = 52×10−4{5 \over 2} \times {10^{ - 4}}

β\beta = 2.5 ×\times 10 −- 4

β\beta = 0.25 mm

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