# Cool Question about the doppler effect

If you are driving in a car up to a set of trafic lights that are red, how fast would you need to be going to make the lights appear green? (take the velocity of light to be 3x10^8 ms-1 and the wavelength of red light to be 620nm and the wavelength of green light to be 540nm)

Ive calculated it wrong... :s and found it to be 44 million ms-1.

Any help greatly apreciated.

Galileo
Homework Helper
It helps if you show what you've done so we can see where it went wrong.

The approach to the problem is simple: Take the formula for the doppler shift:

$$\frac{\lambda '}{\lambda}=\sqrt{\frac{c-v}{c+v}}$$
where $\lambda '$ is the Doppler-shifted wavelength and $v$ is the relative velocity between source and observer (positive when approaching).

The unknown is v, so solve your equation for v and just plug in the numbers.

hehe, yeah i know all that. Im just not sure what the differnce in wavelength is etc. I actually got 38 million metres / sec. Not 44. Just wondering what you got. okok this is what i did.

620nm - 540nm = 80nm

(80nm / 640nm) * 3x10^8

= v

v = 38.6 million ms-1

Galileo
Homework Helper
Difference in wavelengths? You are given the wavelengths and it's actually the ratio between the wavelengths that is important, not the difference.

The answer is higher that 38 Mm/s and lower than 44 Mm/s (Mm= Megameter :) ) so you probably did something wrong. What's your expression for v?

Galileo
Homework Helper
Sorry, I didn't see your last post.

Well, I`m not sure what your reasoning is. I don't why you took the difference in wavelengths and not just solve the expression for the doppler shift for v in terms of lamda and lambda'.

The answer comes quite close to the actual answer, because when the v<<c the approximation:
$$\frac{\lambda'}{\lambda}=1-v/c$$
can be used. This is easily solvable for v:

$$v=c(1-\frac{\lambda'}{\lambda})=c\frac{\lambda-\lambda'}{\lambda}$$