Verified Solution • 2025–26 Session moderate

QJEEM20Ph2017: JEE Main 2020 Physics - Class 12 JEEM20Ph2017 Solution: In a Young's double slit experiment, the separation between the …

In a Young's double slit experiment, the separation between the slits is 0.15 mm. in the experiment, a source of light of wavelengh 589 nm is used and the interference pattern is observed on a screen kept 1.5 m away. The separation between the successive bright fringes on the screen is :

Multiple Choice Options

  • A 4.9 mm
  • B 5.9 mm
  • C 6.9 mm
  • D 3.9 mm

Step-by-Step Verified Solution

β=λDd=589×10−9×1.50.15×10−3\beta = {{\lambda D} \over d} = {{589 \times {{10}^{ - 9}} \times 1.5} \over {0.15 \times {{10}^{ - 3}}}} = 5.9 mm

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