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Momentum and radiation pressure

  1. Feb 1, 2005 #1
    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. jcsd
  3. Feb 1, 2005 #2

    Q_Goest

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    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
     
  4. Feb 1, 2005 #3

    Curious3141

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    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, [tex]E = pc[/tex] 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 [tex]\frac{2W}{c}[/tex]
     
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