Q99 / 125 · BITSAT 2019 · Physics

In Young's double-slit experiment, the distance between slits and screen is 2 m and distance between slits is 0.25 mm. A light of wavelength 800 nm is used to find fringes on the screen. If the screen moves with a speed of 5 m s-1, then first maxima will move with a speed of:

  • Correct answer
    A
    16 mm s-1
  • B
    8 mm s-1
  • C
    20 mm s-1
  • D
    50 mm s-1

Solution

Given, D=2 m, d=0.25 mm=2.5×10-4 m

Velocity of screen, dDdt=5 m s-1

Speed of first maxima =dβdt=?

In Young's double-slit experiment, fringe width β=Dλd

On differentiating with respect to t, we get

dβdt=λd·dDdt

=800×10-92.5×10-4×5

=16×10-3 m s-1=16 mm s-1

Hence, the correct answer is option: 'A'.

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