JEE Main 2018PhysicsWave OpticsHardMCQ

JEE Main 2018 — Wave Optics Question with Solution

JEE Main 2018 (08 Apr)

Question

The angular width of the central maximum in a single slit diffraction pattern is 60° . The width of the slit is 1 μm. The slit is illuminated by monochromatic plane waves. If another slit of the same width is made near it, Young's fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e., the distance between the centres of each slit.)

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Show full solutionCorrect option: B
Correct answer
B25 μm

Step-by-step explanation

In diffraction, dsin30°=λ, λ=d2



Young’s fringe width [ d'- the separation between two slits] 

β=λ×Dd'⇒10-2=d2×50×10-2d'
⇒10-2=10-6×50×10-22×d'⇒d'=25 μm.

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About this question

This is a previous-year question from JEE Main 2018, covering the Wave Optics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.