A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{2E}{c} \)
Electromagnetic radiation of energy E carries momentum p = E/c. At normal incidence on a perfect reflector the radiation comes back with the same momentum in the opposite direction, so the change in its momentum is twice its initial momentum. This change is transferred to the surface.
\( \displaystyle \Delta p = \frac{E}{c} - \left(-\frac{E}{c}\right) = \frac{2E}{c} \)
Shortcut: Units: E/c has the unit J s m−1 = kg m s−1, the unit of momentum, while \( E/c^{2} \) has the unit of mass. This removes options 2 and 3.