Physics · Wave Optics

JEE Main 2026 — 28 January, Evening Shift — Question 49

A beam of light consisting of wavelengths 650 nm and 550 nm illuminates the Young's double slits with separation of 2 mm such that the interference fringes are formed on a screen, placed at a distance of 1.2 m from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is $\_\_\_\_$ $\times 10^{-5} \mathrm{~m}$ .

Answer: 429

Numerical answer — enter this value.

Step-by-step solution

$y=n \frac{\lambda D}{d}$ $\mathrm{y}_{1}=\mathrm{y}_{2}$ $\mathrm{n}_{1} \lambda_{1} \frac{\mathrm{D}}{\mathrm{d}}=\mathrm{n}_{2} \lambda_{2} \frac{\mathrm{D}}{\mathrm{d}}$ $\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{\lambda_{2}}{\lambda_{1}}=\frac{550}{650}=\frac{11}{13}$ $\mathrm{y}=11 \times \frac{\lambda_{1} \mathrm{D}}{\mathrm{d}}=\frac{11 \times 650 \times 10^{-9} \times 1.2}{2 \times 10^{-3}}$ $y=429 \times 10^{-5}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Evening Shift
Subject: Physics
Chapter: Wave Optics
Topic: Young's Double Slit Experiment and Its Modifications