A radiation of energy E is normally incident on a perfectly reflecting surface. The momentum transferred to the surface will be: (Given : speed of light 'c')
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2\,E\,c^{-1} \)
Radiation of energy \( E \) carries momentum \( p = E/c \). By the conservation of momentum the surface receives the change in the momentum of the radiation.
For a perfectly reflecting surface the radiation is turned back along the same line, so its momentum changes from \( +E/c \) to \( -E/c \).
\( \displaystyle \Delta p = \frac{E}{c} - \left(-\frac{E}{c}\right) = \frac{2E}{c} \)
The momentum transferred to the surface is \( 2E/c \), twice the value for a perfectly absorbing surface.