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JEE Main 20262 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
2 April
Shift
Evening Shift
Subject
Physics
Chapter
Wave Optics

JEE Main 2026 · 2 April · Evening Shift · Physics · Wave Optics

In a Young's double slit experiment, the intensity at some point on...

In a Young's double slit experiment, the intensity at some point on the screen is found to be 34\dfrac{3}{4} times of the maximum of the interference pattern. The path difference between the interfering waves at this point is λx\dfrac{\lambda}{x} where λ\lambda is wavelength of the incident light. The value of xx is _______.

Correct answer

0

Step-by-step solution

The intensity at a point in Young's double slit experiment is given by I = I_\max \cos^2\left(\dfrac{\phi}{2}\right)

Given I = \dfrac{3}{4} I_\max

\dfrac{3}{4} I_\max = I_\max \cos^2\left(\dfrac{\phi}{2}\right)

cos⁡(ϕ2)=32\cos\left(\dfrac{\phi}{2}\right) = \dfrac{\sqrt{3}}{2}

ϕ2=π6\dfrac{\phi}{2} = \dfrac{\pi}{6}

ϕ=π3\phi = \dfrac{\pi}{3}

The phase difference ϕ\phi is related to the path difference Δx\Delta x by ϕ=2πλΔx\phi = \dfrac{2\pi}{\lambda} \Delta x

2πλΔx=π3\dfrac{2\pi}{\lambda} \Delta x = \dfrac{\pi}{3}

Δx=λ6\Delta x = \dfrac{\lambda}{6}

Comparing with Δx=λx\Delta x = \dfrac{\lambda}{x}, we get x=6x = 6

Answer: 66
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