JEE Main 2022 · 27 Jun · Shift 1 · Physics Q53

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PhysicsInteger+4 / -1Question 53 of 90

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.

Your answer

Answer and worked solution

Official answer: 450

First dark fringe at $y=\dfrac{\lambda D}{2d}$ ; opposite a slit, $y=\frac d2$ .

So $\lambda=\dfrac{d^2}D=\dfrac{(0.6\times10^{-3})^2}{0.8}=4.5\times10^{-7}$ m $=450$ nm.

Answer: 450

Topics:Interference and Young's double slit

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JEE Main 2022 · 27 Jun · Shift 1, Physics question 53