A double slit interference experiment performed with a light of… | JEE Main 2025 (28 Jan, Shift 1)

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YDSE: Fringe Width & Fringe Positions

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 10 th bright fringe from the central maximum when the source of light is replaced by another source of wavelength 660 nm would be ________ mm.

Numerical answer type. Work it out, then open the answer.

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Answer: 11

$y_n = \frac{n\lambda D}{d}$ , so for the same $n$ , $D$ , $d$ : $y \propto \lambda$ .
$y' = 10 \times \frac{660}{600} = 11$ mm
So the answer is 11 mm.

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YDSE: Fringe Width & Fringe Positions question from JEE Main 2025 (28 Jan, Shift 1)

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