Wave Optics

JEE Main level, single correct, medium

59 of 191

Question 59 of 191

In Young's double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is 7λ4\dfrac{7\lambda }{4}. The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is:

Show answer and solution

Answer: (A) 12\dfrac{1}{2}

Given the path difference is ∆x=7λ4∆x=\dfrac{7\lambda }{4}
Hence, the phase difference can be found as follows:
ϕ=2πλ∆x=2πλ×7λ4=7π2\begin{matrix}\phi & = & \dfrac{2\pi }{\lambda }∆x \\ & = & \dfrac{2\pi }{\lambda }\times \dfrac{7\lambda }{4} \\ & = & \dfrac{7\pi }{2}\end{matrix}
The formula to calculate the intensity is given by
I=Imax⁡cos⁡2(ϕ2)...(1)I={I}_{\max }{\cos }^{2}\left(\dfrac{\phi }{2}\right)...\left(1\right)
From equation (1), it follows that
IImax⁡=cos⁡2(ϕ2)=cos⁡2(7π2×2)=cos⁡2(7π4)=cos⁡2(2π−π4)=cos⁡2π4=12\begin{matrix}\dfrac{I}{{I}_{\max }} & = & {\cos }^{2}\left(\dfrac{\phi }{2}\right) \\ & = & {\cos }^{2}\left(\dfrac{7\pi }{2\times 2}\right) \\ & = & {\cos }^{2}\left(\dfrac{7\pi }{4}\right) \\ & = & {\cos }^{2}\left(2\pi -\dfrac{\pi }{4}\right) \\ & = & {\cos }^{2}\dfrac{\pi }{4} \\ & = & \dfrac{1}{2}\end{matrix}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Wave optics: YDSE, diffraction and polarisation simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Wave Optics (Youngs Double Slit Experiment), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.