Wave Optics

JEE Main level, single correct, medium

113 of 191

Question 113 of 191

The maximum number of possible interference maxima for slit separation equal to 1.8λ1.8 \lambda, where λ\lambda is the wavelength of light used, in a Young's double slit experiment is

Show answer and solution

Answer: (B) 3

As sin⁡θ=nλd\sin \theta=\frac{n \lambda}{d} and sin⁡θ\sin \theta cannot be ≯1\ngtr 1
∴1=nλ1.8λ\therefore 1=\frac{n \lambda}{1.8 \lambda}
or n=1.8n=1.8
Hence maximum number of possible interference maximas, 0,±10, \pm 1 i.e. 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.