Wave Optics

JEE Main level, numerical answer, medium

67 of 191

Question 67 of 191

In a Young’s double slit experiment, the intensities at two points, for the path difference λ4\dfrac{\lambda }{4} and λ3\dfrac{\lambda }{3} (λ\lambda being the wavelength of light used) are I1{I}_{1} and I2{I}_{2} respectively. If I0{I}_{0} denotes the intensity produced by each one of the individual slits, then \dfrac{{I}_{1}+{I}_{2}}{{I}_{0}}=______.

Show answer and solution

Answer: 3

As we know, resultant intensity is given by,
I=4I0cos⁡2(Δϕ2)I=4{I}_{0}{\cos }^{2}\left(\dfrac{\Delta \phi }{2}\right) and phase difference can be written as ∆ϕ=2πλ∆x∆\phi =\dfrac{2\pi }{\lambda }∆x.
For first case:
I1=4I0cos⁡2(2πλ×λ4)=4I0cos⁡2(π4)=2I0{I}_{1}=4{I}_{0}{\cos }^{2}\left(\dfrac{2\pi }{\lambda }\times \dfrac{\lambda }{4}\right)=4{I}_{0}{\cos }^{2}\left(\dfrac{\pi }{4}\right)=2{I}_{0}
For second case:
I2=4I0cos⁡2(2πλ×λ3)=4I0cos⁡2(2π3)=I0{I}_{2}=4{I}_{0}{\cos }^{2}\left(\dfrac{2\pi }{\lambda }\times \dfrac{\lambda }{3}\right)=4{I}_{0}{\cos }^{2}\left(\dfrac{2\pi }{3}\right)={I}_{0}
Therefore, required value ofI1+I2I0=3\dfrac{{I}_{1}+{I}_{2}}{{I}_{0}}=3.

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.