JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Wave Optics

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In Young's double slit experiment the two slits are 0.6 mm distance apart. Interference pattern is observed on a screen at a distance 80 cm from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be ____ nm .

Answer: 450

Solution

The first dark fringe position is

y_{1} = \frac{D λ}{2 d}

. Given that this position equals

\frac{d}{2}

(directly opposite one slit), we have

\frac{D λ}{2 d} = \frac{d}{2}

. Solving:

λ = \frac{d ^{2}}{D} = \frac{( 0.6 \times 1 0 ^{- 3} ) ^{2}}{0.8} = 450 \times 1 0^{- 9}

m = 450 nm.

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Integer answer

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