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JEE Main 202528 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Morning Shift
Subject
Physics
Chapter
Wave Optics

JEE Main 2025 · 28 January · Morning Shift · Physics · Wave Optics

A double slit interference experiment performed with a light of...

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

Correct answer

0

Step-by-step solution

Position of the nth n^{\text{th }} bright fringe w.r.t. central maxima in a YDSE is yn=nλDdy_n=n \frac{\lambda D}{d}. ⇒y10′y10=λ′λ\Rightarrow \quad \frac{y_{10}^{\prime}}{y_{10}}=\frac{\lambda^{\prime}}{\lambda} or y10′=λ′λy10\quad y_{10}^{\prime}=\frac{\lambda^{\prime}}{\lambda} y_{10}=(660nm600nm)10mm=11mm\begin{aligned} & =\left(\frac{660 \mathrm{ nm}}{600 \mathrm{ nm}}\right) 10 \mathrm{ mm} \\ & =11 \mathrm{ mm}\end{aligned}$$
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