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Need some help with Doppler effect problem! Exam Monday!

  1. Aug 21, 2004 #1
    Hi everyone,

    really would love it if someone could help me with this. I think it's probably a very easy problem, only I can't seem to find the right formlula, or understand the only one I found...

    This is the problem:

    I'm in my car, driving towards a crossing with a red traffic light (lambda = 700 nm). How fast do I need to drive to make the traffic light appear green (lambda = 500 nm) to me?

    (I suppose this must have to do with the Doppler effect, but the only formula I can find on it says: v/c = (lambda - lambda0) / lambda0

    v = speed of the source of radiation, c = speed of light, lambda = measured wavelength, lambda0 = something that strictly translated from Swedish to English means "resting" wavelength, don't know if that makes any sense? I just have no idea what lambda0 is!?)

    Hope I explained it ok! Does anyone have any suggestions on how to solve this problem?

    Thanks,
    Linda, Sweden
     
  2. jcsd
  3. Aug 22, 2004 #2

    Galileo

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    The formula for the Doppler effect for light is:

    [tex]\frac{\lambda'}{\lambda}=\sqrt{\frac{1-v/c}{1+v/c}}[/tex]
    where [itex]\lambda'[/itex] is the Doppler-shifted wavelenght and [itex]v[/itex] is the relative velocity between source and observer. (v is positive if they are approaching each other)

    For v<<c, this equation is approximated by
    [tex]\frac{\lambda'}{\lambda}=1-\frac{v}{c}[/tex]

    You know the wavelenghts, so you can solve for v.
     
  4. Aug 22, 2004 #3
    Thanks a lot for that!!
    Was easy to solve with your help :smile: , now lets hope I pass my exam tomorrow as well!
    Thanks again,
    Linda, Sweden
     
  5. Oct 16, 2011 #4
    I want to ask that will this phenomenon hold true for a celestial object like sun? Does Doppler Effect hold true also for the stationary light source? I mean, that if you are traveling to the signal (in above problem) with velocity Vp (19.44m/s) and the velocity of signal light is Vs (3x10^8 m/s) and if you apply the relative velocity concept, it gives:

    Vr = Vp + Vr (In case of making the light source at 0 velocity)
    = 300000019.44 m/s (Speed of object?!?!)
    OR
    Vr = Vp - Vr (In case the observer is stationary)
    = 299999980.56 m/s (Speed of light?!?!)

    Do clear this to me. I know there can be my misconception but please help me to solve this!
    Thanks in advance
    :)
     
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