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Doppler shift of Light

  • Thread starter Cheezay
  • Start date
  • #1
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Homework Statement


How fast would a motorist have to be traveling for a yellow (l = 595 nm) traffic light to appear green (l = 550 nm) because of the Doppler shift?


Homework Equations



v=[(c)(f0/fs)2-c] / [(f0/fs)2+1]

The Attempt at a Solution


v= Speed of motorist
c= Speed of light
f0= Observed wavelength (green, (5.5x10^-7 m)
fs= Source wavelength (yellow, (5.95x10^-7 m)

First of all, am I using the correct equation? Because i'm almost certain my work is correct, since after plugging in the above known numbers, i get 2.35x10^7 m/s for an answer, but this answer is not correct. Any help would be greatly appreciated!
 

Answers and Replies

  • #2
Doc Al
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First of all, am I using the correct equation?
Looks OK. (Except for some reason you use f to represent wavelength. f usually stands for frequency.)

Because i'm almost certain my work is correct, since after plugging in the above known numbers, i get 2.35x10^7 m/s for an answer, but this answer is not correct.
That answer looks good to me.
 
  • #3
76
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oooh i love doppler shift =]

ive always used F(observed)= [c/(c+v)]*F(actual)
 
  • #4
26
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Thanks RoryP. I don't know what was wrong with the equation that i posted, but i tried yours out and it worked!
 
  • #5
Doc Al
Mentor
44,882
1,129
ive always used F(observed)= [c/(c+v)]*F(actual)
That equation isn't quite right; it should be:

[tex]f_{obs} = f \sqrt{\frac{c + v}{c - v}}[/tex]
 
  • #6
76
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No worries Cheezay, yeah ive never seen the equation you started with, but then again ive only been doing physics for 2 years now so i might bump into it soon!

Yeah i just checked my notes from 6th form and the equation i used is for c>>v, so dont konw if that makes any difference =]
 

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