Two wavelengths λ 1 and λ 2 are used in Young's double slit experiment λ 1 =450 nm and λ 2 =650 nm . The minimum order of fringe produced by λ 2 which overlaps with the fringe produced by λ 1 is n . The value of n is

Physics · Wave Optics

JEE Main 2024 — 4 April, Shift 1 — Question 55

Two wavelengths $\lambda_{1}$ and $\lambda_{2}$ are used in Young's double slit experiment $\lambda_{1}=450 \mathrm{~nm}$ and $\lambda_{2}=650 \mathrm{~nm}$ . The minimum order of fringe produced by $\lambda_{2}$ which overlaps with the fringe produced by $\lambda_{1}$ is $n$ . The value of $n$ is $\qquad$

Answer: 9

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{n}_{2} \lambda_{2}=\mathrm{n}_{1} \lambda_{1}$

$\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{450}{650}=\frac{9}{13}$

$\mathrm{n}_{2}=9$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Physics
Chapter: Wave Optics
Topic: Young's Double Slit Experiment and Its Modifications