JEE Main 2025 Jan 28 Shift 1, Physics Q25: Young’s Double Slit Experiment

JEE Main2025Jan 28, Shift 1Physics

A double slit interference experiment performed with a light of wavelength 600 nm forms an interference fringe pattern on a screen with 10th 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.

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Solution

In the Young's double-slit experiment, the position of the $n$ -th bright fringe is $y_{n} = \frac{nλ D}{d}$ , where $D$ is the slit-to-screen distance and $d$ is the slit separation.

For fixed $n$ , $D$ and $d$ : $y \propto λ$ .

$\frac{y _{2}}{y _{1}} = \frac{λ _{2}}{λ _{1}} \Rightarrow y_{2} = y_{1} \cdot \frac{λ _{2}}{λ _{1}} = 10 mm \cdot \frac{660}{600} = 11$ mm.

$∴ y_{2} = 11$ mm.

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JEE Main 2025 Jan 28 Shift 1, Physics Q25: Young’s Double Slit Experiment

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