JEE Main 2026 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 2 April 2026, 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, the intensity at some point on the screen is found to be

\frac{3}{4}

times of the maximum of the interference pattern. The path difference between the interfering waves at this point is

\frac{\lambda}{x}

where

\lambda

is wavelength of the incident light. The value of

x

Official answer

NTA final key.

Chapter: Wave Optics
Topic: Young's Double Slit
Idea tested: Intensity at Fractional Path Differences

Same idea in other shifts

Asked 5× in all

6 Apr 2026, Shift 1 · Q36In interference experiment the path difference between two interfering waves at a point $A$ on the screen is $\lambda/3$ , where $\lambda$ …EasySingle correct
6 Apr 2026, Shift 2 · Q45In a Young double slit experiment, the wavelength of incident light is 6000 Å, the separation between slits $S_1$ and $S_2$ is 5 cm and…MediumDiagram basedHas a figure
9 Apr 2024, Shift 1 · Q59In a Young's double slit experiment, the intensity at a point is $\left(\frac{1}{4}\right)^{th}$ of the maximum intensity, the minimum…MediumNumerical value
6 Sep 2020, Shift 2 · Q24A Young's double-slit experiment is performed using monochromatic light of wavelength $\lambda$ . The intensity of light at a point on the…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 42 of the 174 JEE Main shifts held from 2017 to 2026.