Physics · Wave Optics

JEE Main 2025 — 28 January, Morning Shift — Question 71

A double slit interference experiment performed with a light of wavelength 600 nm forms an interference fringe pattern on a screen with $10^{\text {th }}$ bright fringe having its centre at a distance of 10 mm from the central maximum. Distance of the centre of the same $10^{\text {th }}$ bright fringe from the central maximum when the source of light is replaced by another source of wavelength 660 nm would be $\qquad$ mm .

Answer: 11

Numerical answer — enter this value.

Step-by-step solution

In case of YDSE the distance of $\mathrm{n}^{\text {th }}$ maxima from

central maxima is given by

$\mathrm{Y}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}$

Here $\mathrm{n}, \mathrm{D}$ & d are same

So, $\mathrm{y} \times \lambda$

$\Rightarrow \frac{\mathrm{y}_{2}}{\mathrm{y}_{1}}=\frac{\lambda_{2}}{\lambda_{1}} \Rightarrow \frac{\mathrm{y}_{2}}{10 \mathrm{~mm}}=\frac{660 \mathrm{~nm}}{600 \mathrm{~nm}}$

$\Rightarrow y_{2}=11 \mathrm{~mm}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Morning Shift
Subject: Physics
Chapter: Wave Optics
Topic: Young's Double Slit Experiment and Its Modifications