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MHT CET 2026 (13 May, Shift 1) · Photons: Energy, Momentum & Radiation Pressure

Photons: Energy, Momentum & Radiation Pressure question from MHT CET 2026 (13 May, Shift 1)

Q5MHT CET 2026 · 13 MAY · SHIFT 1Medium

Photons: Energy, Momentum & Radiation Pressure

A parallel beam of light is incident normally on a plane surface absorbing 40%40\% of light and reflecting the rest. If the incident beam carries 6060 W of power the force exerted by it on the surface is (C = velocity of light = 3×1083 \times 10^{8} m/s)

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Answer: (b)

  1. Absorbed part gives P/cP/c per unit fraction, reflected part gives 2P/c2P/c: F=Pc(a+2r)F = \dfrac{P}{c}(a + 2r)
  2. a=0.4, r=0.6a = 0.4,\ r = 0.6: F=603×108(0.4+1.2)=2×10−7×1.6F = \dfrac{60}{3\times10^{8}}(0.4 + 1.2) = 2\times10^{-7}\times1.6
  3. F=3.2×10−7 NF = 3.2\times10^{-7}\text{ N}
  4. So the answer is (b) 3.2×10−7 N3.2\times10^{-7}\text{ N}.
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