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A parallel beam of light is incident normally on a plane surface absorbing 40 % of the light and reflecting the rest. If the incident beam carries 60 W of power, the force exerted by it on the surface is

Options

  1. A3.2 × 10 - 8 N
  2. B3.2 × 10 - 7 N
  3. C5.12 × 10 - 7 N
  4. D5.12 × 10 - 8 N

Correct answer

B. 3.2 × 10 - 7 N

Step-by-step solution

Momentum of incident light per second p 1 = E c = 60 3 × 10 8 = 2 × 10 - 7 Momentum of reflected light per second p 2 = 60 100 × E c = 60 3 × 10 8 = 1 .2 × 10 - 7 Force on the surface = change in momentum per second = p 2 - - p 1 = p 2 + p 1 = 2 + 1 .2 × 10 - 7 = 3 .2 × 10 - 7 N

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