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MHT CET 2025 (20 Apr, Shift 1) · Photons: Energy, Momentum & Radiation Pressure

Photons: Energy, Momentum & Radiation Pressure question from MHT CET 2025 (20 Apr, Shift 1)

Q8MHT CET 2025 · 20 APR · SHIFT 1Medium

Photons: Energy, Momentum & Radiation Pressure

A parallel beam of light is incident normally on a plane surface absorbing 50%50\% of the light and reflecting the rest. If the incident beam carries 90 W of power, the force exerted by it on the surface is ( C=C = velocity of light in air =3×108 m/s= 3 \times 10^{8}\text{ m/s})

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Answer: (a)

  1. F=Pc(a+2r)F = \dfrac{P}{c}(a + 2r) with a=0.5, r=0.5a = 0.5,\ r = 0.5
  2. F=903×108(0.5+1.0)=3×10−7×1.5F = \dfrac{90}{3\times10^{8}}(0.5 + 1.0) = 3\times10^{-7}\times1.5
  3. F=4.5×10−7 NF = 4.5\times10^{-7}\text{ N}
  4. So the answer is (a) 4.5×10−7 N4.5\times10^{-7}\text{ N}.
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