Wave Optics

JEE Main level, numerical answer, medium

77 of 191

Question 77 of 191

In a Young's double slit experiment, an angular width of the fringe is 0.35∘0.35^{\circ} on a screen placed at 2m2m away for particular wavelength of 450nm450nm. The angular width of the fringe, when whole system is immersed in a medium of refractive index 75\dfrac{7}{5}, is 1α\dfrac{1}{\alpha }. The value of α\alpha is _____ .

Show answer and solution

Answer: 4

Angular fringe width is given by θ=λD\theta =\dfrac{\lambda }{D}, where, DD is the distance between the source and the screen and λ\lambda is the wavelength of the light.
We have, θ1θ2=λ1λ2\dfrac{{\theta }_{1}}{{\theta }_{2}}=\dfrac{{\lambda }_{1}}{{\lambda }_{2}}
Then, θ2=0.35∘450nm×450nm75=0.25∘=14{\theta }_{2}=\dfrac{{0.35}^{^{\circ}}}{450nm}\times \dfrac{450nm}{\dfrac{7}{5}}={0.25}^{^{\circ}}=\dfrac{1}{4} (∵λ2=λ1μ)\left(∵{\lambda }_{2}=\dfrac{{\lambda }_{1}}{\mu }\right)
Thus, value of α=4\alpha =4.

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.