Calculating Speed from Light Wavelengths

In summary: relativistic situation is involved, so the classical answer will yield a smaller result than v in the formula.
  • #1
burgerkin
35
0

Homework Statement



I drive through a traffic light. When I was pulled over, I tell the police that the red light of wavelength 675 nm appear green to me, with a wavelength of 525 nm cause of doppler effect. How fast was I traveling?

Homework Equations


[tex]\Delta[/tex] [tex]\lambda[/tex] / [tex]\lambda[/tex] = v/c

v is the speed that the source is receding, isn't it the same as the speed I am traveling?

The Attempt at a Solution



v = c (675-525) /675

I just followed the equation to find v, but it was incorrect.

please help.
 
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  • #2
burgerkin said:

Homework Statement



I drive through a traffic light. When I was pulled over, I tell the police that the red light of wavelength 675 nm appear green to me, with a wavelength of 525 nm cause of doppler effect. How fast was I traveling?

Homework Equations


[tex]\Delta \lambda / \lambda = v/c [/tex]
v is the speed that the source is receding, isn't it the same as the speed I am traveling?

The Attempt at a Solution



v = c (675-525) /675

I just followed the equation to find v, but it was incorrect.

please help.

You should get rid of the superfluous "tex" starts and stops; what's between each pair ends up on its own line. I've fixed the tex code in the quote.

In your formula, the denominator should have the observed wavelength, not the unshifted wavelength.
 
  • #3
Thanks for fixing it!

I did try to use the observed wavelength when I first tried it, but it still gave me wrong answer!
 
  • #4
burgerkin said:
Thanks for fixing it!

I did try to use the observed wavelength when I first tried it, but it still gave me wrong answer!

What value did you calculate?
 
  • #5
I calculated v in the formula, I am assuming that is the speed I am traveling, but I think that is not the case. It says that it is V source, I am understanding it as the speed at which the light source is leaving me? so it is also the speed at which I am moving?
 
  • #6
burgerkin said:
I calculated v in the formula, I am assuming that is the speed I am traveling, but I think that is not the case. It says that it is V source, I am understanding it as the speed at which the light source is leaving me? so it is also the speed at which I am moving?

For light, v is the relative velocity of the source and observer. If the source and observer are getting closer together over time, then v is negative. If they are growing further apart, then v is positive.

In this case it is you, the observer in the car, that is taken to be moving, but what really counts is the closing speed between the car and traffic light. That speed is just the speed of the car as it approaches the light.
 
  • #7
So if the question is asking me to calculate the speed the car is moving, it is the v in the formula, isn't it? Why did i get the wrong answer. I don't think I fully understand this..
 
  • #8
burgerkin said:
So if the question is asking me to calculate the speed the car is moving, it is the v in the formula, isn't it? Why did i get the wrong answer. I don't think I fully understand this..

Yes, that is the v. At this point I can't say why you got the wrong answer, because you haven't shown any numbers! What result did you get?

Also, just to make sure, I should ask if you've covered the relativistic version of the Doppler shift yet. It will yield a slightly different result due to relativistic correction factors. If you haven't covered it, don't worry about it and proceed with the "classical" formulas you've been using.
 
  • #9
^^ I think that is the problem.. I used the formula and my answer is slightly smaller than the correct answer. I don't know what exactly is the correct answer, but I know it is smaller than v. Yes, relativistic situation is involved i guess, we studied a little relativity, but teacher did not cover much.
 
  • #10
burgerkin said:
^^ I think that is the problem.. I used the formula and my answer is slightly smaller than the correct answer. I don't know what exactly is the correct answer, but I know it is smaller than v. Yes, relativistic situation is involved i guess, we studied a little relativity, but teacher did not cover much.

Well, not to dash your hopes, but the relativistic version will yield a smaller result than the classical one.

Perhaps you should post your calculation in all its numerical glory.
 
  • #11
oh yes, I made a mistake, the answer is slightly smaller..

let us use red light wavelength 685nm, then light shifted to green to observer to 555 nm,

i DID, 685-555=130

130 / 555=0.234

0.234 x 3 x10e+8 =7.03e+7 m/s (wrong answer)

the correct answer given is 6.22e+7
 
  • #12
Okay! So it looks as though you will want the relativistic version.

No doubt you have the formula in your book or notes, but to save time here it is:

[tex] \frac{\lambda_o}{\lambda_s} = \sqrt{\frac{1 + \beta}{1 - \beta}} [/tex]

where: β = v/c, and v is the relative velocity between observer and source.

Square both sides and solve for β, then find v from the definition of β. You should obtain the desired result.
 
  • #13
I know it is odd, but I do not have the formula in book..and my teacher barely covered relativity..

yes, it worked out right finally! thanks so very much!
 

Related to Calculating Speed from Light Wavelengths

1. How do you calculate speed from light wavelengths?

The speed of light is a fundamental constant in physics, and it is equal to approximately 3.00 x 10^8 meters per second. To calculate speed from light wavelengths, you can use the formula: speed = wavelength x frequency. The wavelength is the distance between two consecutive peaks in a light wave, and the frequency is the number of waves that pass a given point in one second.

2. Why is the speed of light used in this calculation?

The speed of light is used in this calculation because it is a universal constant that is the same for all observers regardless of their relative motion. This means that the speed of light is a reliable and consistent measurement to use in scientific calculations and theories.

3. Can this calculation be used for all types of light?

Yes, this calculation can be used for all types of light, including visible light, radio waves, microwaves, and X-rays. This is because the speed of light is the same for all types of electromagnetic radiation.

4. What units should be used for the wavelength and frequency in this calculation?

The units for wavelength are typically measured in meters (m), nanometers (nm), or angstroms (Å). The units for frequency are typically measured in hertz (Hz), which is equivalent to 1/second. It is important to ensure that the units are consistent when using this calculation.

5. How is this calculation used in real-world applications?

This calculation is used in various real-world applications, such as in telecommunications, astronomy, and spectroscopy. For example, the speed of light is used to measure distances in space and to transmit information through fiber optic cables. In spectroscopy, the speed of light is used to identify the elements present in a sample by analyzing their unique light wavelengths.

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