A perfectly reflecting mirror has an area of 1 cm2 Light energy is allowed to fall on it for 1h at the rate of 10 Wcm-2. The force that acts on the mirror is
Answer & explanation
Correct answer: option 2
Let E = Energy falling on the surface per second = 10 J
Momentum of photons
$ p =\frac{h}{λ}=\frac{h}{(c_1v)}$
On reflection,
Change in momentum per second $= 2p = \frac{2E}{c}$
We know that,
Change in momentum per second = force
$F=\frac{2E}{c}=\frac{2×10}{3×10^8}$
$= 6.7 × 10^{-8} $ N