The radiation of energy ' $E$ ' falls normally on a perfectly reflecting surface. The momentum transferred…

The radiation of energy ' $E$ ' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is
  1. $\frac{E}{c}$
  2. $\frac{2 E}{c}$
  3. $\frac{E}{c^2}$
  4. $\frac{2 E}{c^2}$

Solution


Initial momentum, $\mathrm{P}_1=\frac{\mathrm{E}}{\mathrm{c}}$ Final momentum, $P_2=-\frac{E}{c}$ $\therefore \quad$ Momentum transfer to the surface is $\Delta P=P_1-P_2=\frac{E}{c}-\left(-\frac{E}{c}\right)=\frac{2 E}{c}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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