Physics · Wave Optics

JEE Main 2024 — 9 April, Shift 2 — Question 53

Monochromatic light of wavelength 500 nm is used in Young's double slit experiment. An interference pattern is obtained on a screen When one of the slits is covered with a very thin glass plate (refractive index $=1.5$ ), the central maximum is shifted to a position previously occupied by the $4^{\text {th }}$ bright fringe. The thickness of the glass-plate is $\qquad$ . $\mu \mathrm{m}$ .

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

$(\mu-1) \mathrm{t}=\mathrm{n} \lambda$

$(1.5-1) \mathrm{t}=4 \times 500 \times 10^{-9} \mathrm{~m}$

$\mathrm{t}=4000 \times 10^{-9} \mathrm{~m}$

$\mathrm{t}=4 \mu \mathrm{~m}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Physics
Chapter: Wave Optics
Topic: Young's Double Slit Experiment and Its Modifications