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Saturday, 31 October

JEE Main 2025 · PhysicsNumerical answerNumerical valueMediumMulti-step

JEE Main 4 April 2025, Shift 2, Physics Q46

Question 46 of 75 in this shift, Physics question 21 of 25, Section B.

In a Young's double slit experiment, two slits are located 1.5 mm apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm. If the 20 maxima of the double slit pattern are contained within the central maximum of the single slit diffraction pattern, then the width of each slit is x×10−3x\times10^{-3} cm, where x-value is ______.

Official answer

15

NTA final key.

Chapter
Wave Optics
Topic
Single Slit Diffraction
Idea tested
Diffraction Envelope in YDSE

Same idea in other shifts

Asked 2× in all
  1. 8 Apr 2024, Shift 2 · Q59Two slits are 1 mm apart and the screen is located 1 m away from the slits. A light of wavelength 500 nm is used. The width of each slit…MediumNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.