Physics · Wave Optics

JEE Main 2026 — 4 April, Evening Shift — Question 24

In a double slit experiment, when one of the slits is covered by a transparent mica sheet of refractive index 1.56, the central fringe shifts to the position of 7th7^{\mathrm{th}} bright fringe, obtained with both slits uncovered. If the light source wavelength is 450 nm450\ \mathrm{nm}, the thickness of mica sheet is α×10−9 m\alpha \times 10^{-9}\ \mathrm{m}. The value of α\alpha is

Answer: 5625

Numerical answer — enter this value.

Step-by-step solution

Shift Δy=(μ−1)tDd=7λDd\Delta y = (\mu-1)t \frac{D}{d} = 7\lambda \frac{D}{d}. Thus (μ−1)t=7λ(\mu-1)t = 7\lambda. t=7×4501.56−1=31500.56=5625 nmt = \frac{7\times450}{1.56-1} = \frac{3150}{0.56} = 5625\ \mathrm{nm}. So α=5625\alpha = 5625.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Physics
Chapter
Wave Optics
Topic
Young's Double Slit Experiment and Its Modifications
In a double slit experiment, when one of the slits is covered by a… | JEE Main 2026 PYQ with Solution · DhiX AI