Verified Solution • 2025–26 Session moderate

QJEEM19Ph8633: JEE Main 2019 Physics - Class 12 JEEM19Ph8633 Solution: In a Young’s double slit experiment with slit separation 0.1…

In a Young’s double slit experiment with slit separation 0.1 mm, one observes a bright fringe at angle 140{1 \over {40}} by using light of wavelength λ\lambda 1. When the light of wavelength λ\lambda 2 is used a bright fringe is seen at the same angle in the same set up. Given that λ\lambda 1 and λ\lambda 2 are in visible range (380 nm to 740 nm), their values are -

Multiple Choice Options

  • A 400 nm, 500 nm
  • B 625 nm, 500 nm
  • C 380 nm, 500 nm
  • D 380 nm, 525 nm

Step-by-Step Verified Solution

Path difference = d sin θ\theta ≈\approx d θ\theta

= 0.1 ×\times 140{1 \over {40}} mm = 2500nm

or bright fringe, path difference must be integral multiple of λ\lambda .

∴\therefore    2500 = n λ\lambda 1 = m λ\lambda 2

∴\therefore     λ\lambda 1 = 625 (from n = 4), λ\lambda 2 = 500 (from m = 5)

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