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Light waves

  1. Sep 30, 2009 #1
    1. The problem statement, all variables and given/known data
    The laser used in a typical DVD player has a wavelength λ = 650 nm and power of
    240 mW.
    (a) Look up the speed of light in empty space and calculate the frequency f of this laser light.

    (b) Light of f carries energy in bundles of hf, called “photons”, where h is Planck’s
    constant. Calculate the energy of a photon using the value of f you found in
    part (a).

    (c) Calculate the number of photons striking the DVD disk every second.

    (d) A photon also carries momentum h/λ. Momentum for an object with mass m is defined as mv, where v is the speed (so long as v is much, much less than c). In a collision between a photon and a stationary massive object, in which the photon bounces straight backwards, the momentum of the object after the collision is twice the photon’s momentum before the collision. Estimate the speed of a typical DVD of mass 16 g (grams) – if it were not held stationary by the DVD player – after one second of bombardment by the laser.
    Assume the same momentum gain 2h/λ for each collision.


    2. Relevant equations
    Planck's constant = h = 9.80665m/s
    Speed of light = 300,000km/s = 3e8m/s

    3. The attempt at a solution

    (A) v=λf
    f=v/λ
    f=(3e8m/s)/(6.7e-7m)
    f=4e14s (is this Hz?)

    (B) energy=hf
    = (9.80665m/s)(4e14s)
    = 4e15m/s2

    (C) I know power is energy transmitted per unit time, so can I divide the energy by 1s? If so, then:
    (4e15m/s2)/(1s)=4e15m/s
    Where does the 240 mW come in?

    (D)
     
  2. jcsd
  3. Sep 30, 2009 #2

    Kurdt

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    a.) Yes the units should be in Hertz. Be careful with significant figures.

    b.) I would look up Planck's constant and try again.

    c.) If power is the energy per unit time and you want to know how many photons are transmitted in unit time then you need to find out how many photons make up the power of the laser.
     
  4. Oct 1, 2009 #3
    Whoops, I clearly used gravity instead of h. I've been neck deep in physics the last few hours. Thanks for catching that.

    (A) v=λf
    f=v/λ
    f=(3e8m/s)/(6.5e-7m)
    f=4e14Hz

    You said to watch my significant figs. But I am only sure of the speed of light to one sig fig, so shouldn't v/λ be to one sig fig?

    (B) energy=hf
    = (6.62606896e-34J/s)(4e14s)
    = 2.7e-19J/s2 = 2.7e-16mW/s

    (C) To find the amount of photons that make up the power of the laser,
    240mW = 0.24W = 0.24J/s
    (0.24J/s)/(2.7e-19J/s2) = 8.9e17 photons/s

    (D) Photons momentum = h/λ = (6.62606896e-34N/m)/(6.5e-7m) = 1e-27N
    next I need the momentum for the DVD.
    If I understand, it says the momentum would be twice as much as the photons momentum p. So I need to multiply the photons momentum by 2.
    p2 = mv
    (p2)/(m) = v
    v = (2e-27N)/(.016kg)
    v = 1.25e-25m/s
    The DVD's speed after one second of bombardment would be 1.25e-25m/s
     
  5. Oct 2, 2009 #4

    Kurdt

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    What i actually meant for a.) is check your rounding. For b.) the units of energy are not watts per second. c) you have the correct method, but check the units again. For d.) you need to work out the momentum imparted by all the photons that hit the cd in one second. You hve only worked out the momentum of one photon.
     
  6. Oct 2, 2009 #5
    (A) v=λf
    f=v/λ
    f=(3e8m/s)/(6.5e-7m)
    f=5e14Hz

    (B) energy=hf
    = (6.62606896e-34J/s)(5e14Hz)
    = 3e-19J/s2


    (C) To find the amount of photons that make up the power of the laser,
    240mW = 0.24W = 0.24J/s
    (0.24J/s)/(3e-19J/s2) = 8e17s
    So this number does mean 8e17 photons per second, right?

    (D) Photons momentum = h/λ = (6.62606896e-34N/m)/(6.5e-7m) = 1e-27N
    next I need the momentum for the DVD.
    p2 = mv
    (p2)/(m) = v
    v = (2e-27N)/(.016kg)
    v = 1.25e-25m/s

    If this is the momentum for one photon striking the DVD, then I can just multiply it by 8e17 photons a second:
    (1.25e-25m/s)(8e17) = 7e-8m/s

    I still don't understand why my old answer was only for one photon. Can you please explain that to me?

    The DVD's speed after one second of bombardment would be 7e-8m/s
     
  7. Oct 3, 2009 #6
    Does that look right?
     
  8. Oct 4, 2009 #7

    Redbelly98

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    To 3 sig figs, c is 3.00e8 m/s.

    Now you know c to 3 sig figs. :smile:

    Okay, but since we know λ to 2 sig figs, your answer should be to 2 sig figs.

    Okay, but watch the sig figs and units.
    Hint: check the units of h, it's not J/s

    Yes and no. It should mean that. But your answer of "8e17s" means a time duration of 8e17 seconds, which is different than saying 8e17 per second.

    The units should have worked out to "per s" or 1/s here.

    Correct number, wrong units. Again look up the units for h.

    This result looks correct.

    Not quite, try doing that multiplication again.

    Because you got the old answer by calculating the momentum of ONE photon:
     
  9. Oct 4, 2009 #8
    (A) v=λf
    f=v/λ
    f=(3.00e8m/s)/(6.5e-7m)
    f=4.62e14Hz
    (B) energy=hf
    = (6.62606896e-34J s)( 4.62e14 1/s)
    = 3.06e-19J s2
    (C) To find the amount of photons that make up the power of the laser,
    240mW = 0.24W = 0.24J/s
    (0.24J/s)/( 3.06e-19J s2) = 8e17 1/s
    (D) Photons momentum = h/λ = (6.62606896e-34J s)/(6.5e-7m) = 1.0e-27J s/m
    If I’m not mistaken, 1 Newton = 1 joule/meter
    So, can’t I convert to get 1.0e-27N/m?

    Next I need the momentum for the DVD.
    p2 = mv
    (p2)/(m) = v
    v = (1.0e-27N/m)(2)/(.016kg)
    v = 1.25e25m2/s2
    My units get pretty ugly in step (D). But I think they’re right. But I’m notorious for butchering units.
     
  10. Oct 5, 2009 #9

    Redbelly98

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    For units, look back to step (B).

    What should the units be, given that you are calculating an energy? Do you see the error now?

    Yes.
    No.

    J s/m = (N*m) * s/m = ?
     
  11. Oct 5, 2009 #10
    (B) energy=hf
    = (6.62606896e-34J s)( 4.62e14 1/s)
    = 3.06e-19J
    (C) To find the amount of photons that make up the power of the laser,
    240mW = 0.24W = 0.24J/s
    (0.24J/s)/( 3.06e-19J) = 8e17 1/s
    (D) Photons momentum = h/λ = (6.62606896e-34J s)/(6.5e-7m) = 1.0e-27N s
    J s/m = (N*m) * s/m = N s because the meters cancel out
    I think that straightened out my units up to this point.

    p2 = mv
    (p2)/(m) = v
    v = (1.0e-27N s)(2)/(.016kg)
    This is the other place where I'm worried about my units. Should I convert N back to (kg)(m)/s2? Then I can divide out the kg. But it seems I should divide out another s, because as of now my velocity is an acceleration (m/s2). Unless my units are incorrect, then I can just divide by 1/s to see how fast it is going after only 1s.
    v = 1.25e25m/s2/(1s)
    v = 1.25e25m/s
     
    Last edited: Oct 5, 2009
  12. Oct 5, 2009 #11

    Redbelly98

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    Good job on cleaning up the units. And yes, the units do get tricky for this last part. One way to think about it is:

    You've actually calculated the velocity change per sec here, i.e., the acceleration in m/s^2. So multiply this acceleration by the time interval, 1 sec, to get the velocity after 1 sec. The units become m/s then, as required.

    (And in 2 sec, the DVD would have twice that velocity.)
     
  13. Oct 5, 2009 #12
    v = 1.25e25m/s2
    So I should multiply this by 1s rather than divide?
    v = 1.25e25m/s2*1s
    But won't that give me
    v = 1.25e25m/s3
    ?
     
  14. Oct 5, 2009 #13

    Redbelly98

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    Think about it. Carefully.
     
  15. Oct 5, 2009 #14
    (A) v=λf
    f=v/λ
    f=(3.00e8m/s)/(6.5e-7m)
    f=4.62e14Hz
    (B) energy=hf
    = (6.62606896e-34J s)( 4.62e14 1/s)
    = 3.06e-19J s2
    (C) To find the amount of photons that make up the power of the laser,
    240mW = 0.24W = 0.24J/s
    (0.24J/s)/( 3.06e-19J s2) = 7.8e17 1/s
    (D) Photons momentum = h/λ = (6.62606896e-34J s)/(6.5e-7m) = 1.0e-27J s/m
    1 Newton = 1 joule/meter
    1.0e-27J s/m = 1.0e-27N s

    p2 = mv
    (p2)/(m) = v
    v = (1.0e-27N s)(2)/(.016kg)
    v = 1.25e-25m/s2
    v = 1.25e-25m/s2*1s
    v = 1.25e-25m/s
    This is the momentum for one photon striking the DVD, so I will multiply it by 7.8e17 photons a second:
    (1.25e-25m/s)(7.8e17) = 9.8e-8m/s
    The DVD's speed after one second of bombardment would be 9.8e-8m/s
     
    Last edited: Oct 6, 2009
  16. Oct 6, 2009 #15

    Redbelly98

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    Looks good.
     
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