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A small mirror of mass m is suspended by a massless thread of length l . Then the small angle through which the thread will be deflected when a short pulse of laser of energy E falls normal on the mirror ( c = speed of light in vacuum and g = acceleration due to gravity)

Options

  1. A= 3 E 4 mc g l
  2. B= E mc g l
  3. C= E 2 mc gl
  4. D= 2 E mc g l

Correct answer

D. = 2 E mc g l

Step-by-step solution

Force due to beam assuming complete reflection F = 2 P C = 2 C dE dt ; P is power So change in momentum of mirror. m ( ~V -0)= Fdt = 2 C dE = 2 E C Now using work energy theorem ...(1) aligned & W _ g = k & - mg (1- )=0- 1 2 mv ^2 & ~g (2 ^2 2 )= v ^2 2 aligned as is small aligned & g 2 ( 2 )^2= 1 2 4 E ^2 ~m ^2 c ^2 & ~g ^2= 4 E ^2 ~m ^2 c ^2 & = 2 E mc g aligned

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