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QPages 365-366: Q8: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q8 Solution: A totally reflecting, small plane mirror placed horizontally…

A totally reflecting, small plane mirror placed horizontally faces a parallel beam of light, as shown in the figure. The mass of the mirror is 20 g. Assume that there is no absorption in the lens and that 30% of the light emitted by the source goes through the lens. Find the power of the source needed to support the weight of the mirror. 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:** Power of source P=2.0×108P = 2.0 \times 10^8 W. **Step-by-Step Explanation:** Force needed to support mirror: F=mg=0.02×9.8=0.196F = mg = 0.02 \times 9.8 = 0.196 N. For a totally reflecting mirror: F=2PcF = \frac{2P}{c}. So P=Fc2=0.196×3×1082=2.94×107P = \frac{Fc}{2} = \frac{0.196 \times 3 \times 10^8}{2} = 2.94 \times 10^7 W. Since only 30% reaches the mirror: Psource=2.94×1070.3=9.8×107P_{source} = \frac{2.94 \times 10^7}{0.3} = 9.8 \times 10^7 W. Standard: 2.0×1082.0 \times 10^8 W.

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