JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Wave Optics

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

1 0^{th}

bright fringe having its centre at a distance of 10 mm from the central maximum. Distance of the centre of the same

1 0^{th}

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

Answer: 11

Solution

In Young's double-slit interference, the distance of the

n^{t h}

bright fringe from the central maximum is

y_{n} = \frac{nλ D}{d}

For fixed slit separation

d

, screen distance

D

, and fringe order

n = 10

y_{n}

is directly proportional to

λ

\frac{y _{2}}{y _{1}} = \frac{λ _{2}}{λ _{1}}

y_{2} = 10 mm \times \frac{660}{600} = 11 mm

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Integer answer

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