Question

A monochromatic source of light emits $10^9$ number of photons at a perfectly plane mirror perpendicularly. The force exerted by the light on the surface of the mirror is 6 $pN$, when the power of the source of light is $x \times 10^{-15}W$. Find the value of $x$.

Answer 1

$9 \times 10^{11}$

Answer 2

The force exerted by a beam of light on a perfectly reflecting surface at normal incidence is related to its power. Newton's second law states that force is the rate of change of momentum. For a photon, momentum $p = \frac{h}{\lambda}$ and energy $E = \frac{hc}{\lambda} = pc$. When light is incident perpendicularly on a perfectly reflecting surface, the change in momentum of a photon is $2p$. If $n$ is the number of photons emitted per second, the force is $F = n \times 2p$. The power of the source is $P = n \times E = n \times pc$.

From $P = n \times p \times c$, we get $n \times p = \frac{P}{c}$. Substituting this into the force equation, $F = 2 \times (n \times p) = 2 \times \frac{P}{c}$.

Given: Force, $F = 6 \text{ pN} = 6 \times 10^{-12} \text{ N}$. Power of the source, $P = x \times 10^{-15} \text{ W}$. Speed of light, $c \approx 3 \times 10^8 \text{ m/s}$.

Using the formula $F = \frac{2P}{c}$: $6 \times 10^{-12} = \frac{2 \times (x \times 10^{-15})}{3 \times 10^8}$

Now, we solve for $x$: $6 \times 10^{-12} \times (3 \times 10^8) = 2x \times 10^{-15}$ $18 \times 10^{-4} = 2x \times 10^{-15}$ $x = \frac{18 \times 10^{-4}}{2 \times 10^{-15}} = 9 \times 10^{-4 - (-15)} = 9 \times 10^{11}$.

The number of photons ($10^9$) is implicitly assumed to be the rate of photons per second for the power and force to be related in this manner.