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Doppler Effect of a laser

  1. Nov 30, 2013 #1
    1. The problem statement, all variables and given/known data

    A laser emits a monchromic beam of wavelength λ, which is reflected normally from a plane mirror, receding at a speed v. What is the beat frequency between the incident and reflected light?

    2. Relevant equations



    3. The attempt at a solution

    The solutions starts off with this

    [itex]f_{1} = \frac{f_{0}}{1 + \frac{v}{c}}[/itex]

    But I'm not exactly sure where this equation came from. The solution uses

    [itex]f_{0}[/itex] frequency of source
    [itex]f_{1}[/itex] frequency of incident light (source) as measured by moving mirror
    [itex]f_{2}[/itex] frequency of reflected light as measured by the moving mirror

    I know that the Doppler effect is often stated as

    [itex]\frac{λ^{'}}{λ} = \sqrt{\frac{1 - \frac{v}{c}}{1 + \frac{v}{c}}}[/itex]

    So I'm not exactly sure where the first equation came from. Thanks for any help.
     
  2. jcsd
  3. Nov 30, 2013 #2
    Are you sure about your equation? In my book it is the same, but without the square root.

    If you replace the wavelengths by the frequency and isolate for f1, you will get the equation, they start off with.

    Remember that the two velocities are of the observer and reciever. So one of them is zero. The top velocity in your bracket, that is.

    The lower velocity is positive, when the mirror moves away from a stationary source. So you get, what they have.
     
  4. Nov 30, 2013 #3
    Ya. I just read underneath that part and it says that when [itex]v << c[/itex] Eq. (4-44) (the equation I posted in the first post for the Doppler Effect) is approximated by

    [itex]\frac{λ^{'}}{λ} = 1 - \frac{v}{c}[/itex]

    Which I didn't realize. I remember reading it earlier but it left my mind because I would rather use the equation in post one. It's interesting that your book doesn't mention it being a approximation, but that is probably because [itex]v << c[/itex] occurs a lot.

    Thanks for the help.
     
  5. Nov 30, 2013 #4
    Wait hold on. What allows us to simply replace the wavelengths with the frequencies and just interchange them? I thought that the equation applied to wavelengths. Oh ok I know why lol.
     
  6. Nov 30, 2013 #5
    Yeah, well the "approximation" is the most commonly used equation, which probably explains it. Anyway, all equations are based on approximations to a certain degree :-)
     
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