Verified Solution • 2025–26 Session moderate

QPages 365-366: Q7: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q7 Solution: A beam of white light is incident normally on a plane surface…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often calculate the momentum transfer for the reflected portion using only $P/c$, forgetting that reflection reverses the photon's direction, resulting in a momentum change of $2P/c$. Always apply the factor of 2 to the reflected power component to correctly account for the reversal in momentum vector.

A beam of white light is incident normally on a plane surface absorbing 70% of the light and reflecting the rest. If the incident beam carries 10 W of power, find the force exerted by it on the surface. Use h=6.63×10−34 J-s=4.14×10−15 eV-sh = 6.63 \times 10^{-34} \text{ J-s} = 4.14 \times 10^{-15} \text{ eV-s}, c=3×108 m/sc = 3 \times 10^8 \text{ m/s} and me=9.1×10−31 kgm_e = 9.1 \times 10^{-31} \text{ kg})

Step-by-Step Verified Solution

**Answer:** Force F=4.3×10−8F = 4.3 \times 10^{-8} N. **Step-by-Step Explanation:** Power = 10 W. Momentum per second for absorbed light: pa=0.7×10c=73×108=2.33×10−8p_a = \frac{0.7 \times 10}{c} = \frac{7}{3 \times 10^8} = 2.33 \times 10^{-8}. For reflected light: momentum change = 2×0.3×10c=63×108=2.0×10−82 \times \frac{0.3 \times 10}{c} = \frac{6}{3 \times 10^8} = 2.0 \times 10^{-8}. Total force: F=2.33×10−8+2.0×10−8=4.33×10−8F = 2.33 \times 10^{-8} + 2.0 \times 10^{-8} = 4.33 \times 10^{-8} N.

All Questions in this Chapter

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