In Young's double slit experiment, monochromatic light of wavelength… | JEE Main 2024 (1 Feb, Shift 2)

Q55JEE Main 2024 · 1 FEB · SHIFT 2Difficult

Superposition, Coherence & Intensity

In Young's double slit experiment, monochromatic light of wavelength 5000 Å is used. The slits are $1.0\text{ mm}$ apart and screen is placed at $1.0\text{ m}$ away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is _________ $\times 10^{-6}\text{ m}$ .

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

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Answer: 125

$I = \frac{I_{max}}{2} \Rightarrow \cos^2\frac{\phi}{2} = \frac{1}{2} \Rightarrow \phi = \frac{\pi}{2} \Rightarrow \Delta x = \frac{\lambda}{4}$
$y = \frac{\lambda D}{4d} = \frac{5\times10^{-7}\times 1}{4\times10^{-3}}$
$y = 1.25\times10^{-4}\text{ m} = 125\times10^{-6}\text{ m}$ , so the answer is $125$ .

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