# Need some help with Doppler effect problem! Exam Monday!

1. Aug 21, 2004

### Linda

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. Aug 22, 2004

### Galileo

The formula for the Doppler effect for light is:

$$\frac{\lambda'}{\lambda}=\sqrt{\frac{1-v/c}{1+v/c}}$$
where $\lambda'$ is the Doppler-shifted wavelenght and $v$ 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
$$\frac{\lambda'}{\lambda}=1-\frac{v}{c}$$

You know the wavelenghts, so you can solve for v.

3. Aug 22, 2004

### Linda

Thanks a lot for that!!
Was easy to solve with your help , now lets hope I pass my exam tomorrow as well!
Thanks again,
Linda, Sweden

4. Oct 16, 2011

### chinmayrshah

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!