Momentum and radiation pressure

[nemzy]

A plane electromagnetic wave of intensity 6.00 W/m2 strikes a small pocket mirror, of area 30.0 cm2, held perpendicular to the approaching wave.

(a) What momentum does the wave transfer to the mirror each second?
kg · m/s


Ok, well for a perfect reflector, the formula is given as total momentum = 2U (total energy) / c (speed of light = 3e8)

Well from the given info i can find the radation force, which is Force = Radiation Pressure * Area

Radiaton Pressure = 2*Wave intensity / speeed of light (2* since it is a perfect reflector)

But, i have NO absolutely no freaking idea how to solve for U??

 

[Q_Goest]

absolutely no freaking idea how to solve for U??

Just a wild guess, but isn't that simply the energy per unit area (energy flux) multiplied by the area of the mirror?

6.00 W/m2 * 30.0 cm2 * 1 m^2/100^2 cm2 = 0.018 W

 

[Curious3141]

Why do you need to bring in the pressure formula ?

Energy transferred over an interval = power*time, correct ? And the power is given by intensity*area, yes ? So find the light energy incident upon the mirror in one second. Let's call that W. W is the total energy of all the incident photons hitting the mirror in one second.

For a single photon, E = pc where E is the energy of the photon and p is the momentum. Since the photon is reflected perfectly the momentum transferred from one collision and reflection event is 2p. Since W should be proportional to E (related by the number of photons incident upon the mirror in unit time), the total momentum should be proportional to 2p with the same factor. So the answer is just \frac{2W}{c}