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Physics · Wave Optics

JEE Main 2025 — 24 January, Evening Shift — Question 47

Young's double slit interference apparatus is immersed in a liquid of refractive index 1.44. It has slit separation of 1.5 mm . The slits are illuminated by a parallel beam of light whose wavelength in air is 690 nm . The fringe-width on a screen placed behind the plane of slits at a distance of 0.72 m , will be :

  1. Option A:

    0.23 mm

    Correct
  2. Option B:

    0.33 mm

  3. Option C:

    0.63 mm

  4. Option D:

    0.46 mm

Answer: A

Step-by-step solution

β=(λ0μ)×Dd=690×10−9×0.721.44×1.5×10−3=0.23 mm\quad \beta=\left(\frac{\lambda_{0}}{\mu}\right) \times \frac{D}{d}=\frac{690 \times 10^{-9} \times 0.72}{1.44 \times 1.5 \times 10^{-3}}=0.23 \mathrm{~mm}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
24 January, Evening Shift
Subject
Physics
Chapter
Wave Optics
Topic
Young's Double Slit Experiment and Its Modifications

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