1. Feb 1, 2005

### 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??

2. Feb 1, 2005

### Q_Goest

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

3. Feb 1, 2005

### 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 $$\frac{2W}{c}$$