Wave Optics

JEE Main level, numerical answer, medium

53 of 191

Question 53 of 191

Two slits are 1 mm1 \mathrm{~mm} apart and the screen is located 1 m1 \mathrm{~m} away from the slits. A light of wavelength 500 nm500 \mathrm{~nm} is used. The width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern is _____×10−4 m\times 10^{-4} \mathrm{~m}

Show answer and solution

Answer: 2

d=1 mm,D=1 m,λ=500 nm10(λDd)=2λDaa=d5=10×10−4 m5=2×10−4\begin{aligned} & \mathrm{d}=1 \mathrm{~mm}, \mathrm{D}=1 \mathrm{~m}, \lambda=500 \mathrm{~nm} \\ & 10\left(\frac{\lambda \mathrm{D}}{\mathrm{d}}\right)=\frac{2 \lambda \mathrm{D}}{\mathrm{a}} \\ & \mathbf{a}=\frac{\mathrm{d}}{5} \\ &=\frac{10 \times 10^{-4} \mathrm{~m}}{5} \\ &=2 \times 10^{-4}\end{aligned}

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.