In a Young's double slit experiment, each of the two slits A and B , as shown in the figure, are oscillating about their fixed center and with a mean separation of 0.8 mm . The distance between the slits at time t is given by d=(0.8+0.04sinωt)mm, where ω=0.08rads−1. The distance of the screen from the slits is 1 m and the wavelength of the light used to illuminate the slits is 6000A˚. The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point O.
The maximum speed in μm/s at which the 8th bright fringe will move is ____.
Answer: 24
Solution
The position of the nth fringe is y=ndλD.
The speed of the fringe is:
v=dtdy=−nd2λDdtdd
Given d=0.8+0.04sinωt:
dtdd=0.04ωcosωt
For maximum speed, ∣cosωt∣=1, which occurs when sinωt=0, i.e., d=0.8 mm.
Maximum rate of change:
dtddmax=0.04×0.08=0.0032 mm/s
Maximum speed of the 8th fringe:
vmax=8×(0.8×10−3)26000×10−10×1×0.0032×10−3
=0.64×10−68×6000×10−10×0.0032×10−3=24 μm/s
JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Numerical answer
