What frequency of light is recorded by a detector attached to the moving mirror

In summary, a stationary light source with a natural frequency of Fo is viewed in a mirror by a stationary observer. The mirror moves away from the observer with a velocity of Vrel<<c, recording a frequency of F1. When asked for the lowest average speed of atoms at some temperature T given some molar mass M, the equation for v_rms should be used, but the second term in that bracket goes to 0 when v_rel<<c.
  • #1
stunner5000pt
1,461
2
A stationary light source S wit ha natural frequency Fo is viewed in a mirror M by a stationary observer O. The mirror moves away from the observer wit ha velocty of Vrel << c

a) what frequency of light is recorded by a detector attached to the moving mirror

because Vrel << c classical may be used
let F1 be this observed frequency observed
then [tex] f_{1} = f_{0} (1 - \frac{v_{rel}}{c}) [/tex]
is this correct??

b) what frequency in terms of fo will the stationary observer measure for the light reflected off the mirror?
the mirror will now emit the f1 from above wouldn't it ??
sine this mirror is moving away wouldn't the doppler shift be [tex] f_{2} = f_{1} \frac{c}{c+v_{rel}} [/tex]
which would be [tex] f_{2} = f_{0} (1 - \frac{v_{rel}}{c}) \frac{c}{c+v_{rel}} = f_{0} \frac{c-v}{c+v} [/tex]

but i got the second part wrong! Whats wrong with it??

Also when asked for the lowest average speed of atom at some temperature T given some molar mass M
which formula should be used??
is it [tex] v_{rms} = \sqrt{\frac{3RT}{M}} [/tex] or [tex]v_{avg} = \sqrt{\frac{8RT}{\pi M}} [/tex]
 
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  • #2
Surely if v_rel << c, the second term in that bracket would go to 0?
 
  • #3
are you talking about part a) or b)? I did get the first part correct by the way
 
  • #4
Either! Assuming v_rel is << c in both cases..
 
  • #5
stunner,

Can you find a polynomial expression that approximates (c-v)/(c+v) when v<<c?
 
  • #6
k first of all the first one isn't wrong because i wasnt marked wrong thae fact that v<<c doesn't mean that the result in null so get off that!
Of course my approximation is lousy but I am trying to answer my prof's question properly according to him, at least


[tex] \frac{c-v}{c+v} = \frac{1-\frac{v}{c}}{1+\frac{v}{c}} = \frac{1-\beta}{1+\beta} = 1 - \beta + \frac{\beta^2}{2} + ... [/tex]

something like that? Doesnt that give the same answer as a) though??
 
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  • #7
stunner,

Try this. Divide the numerator and denominator of (c-v)/(c+v) by c. Then define x = v/c. The new denominator will be 1+x. Can you find a power series for 1/(1+x)?
 
  • #8
stunner,

I can't keep up with you!

You're very close to the right answer, but you're guessing on the power series. Figure it out. Write 1/(1+x) as (1+x)^-1, then it's easy to see all the derivatives: -(1+x)^-2, 2(1+x)^-3...
 
  • #9
jdavel said:
stunner,

Try this. Divide the numerator and denominator of (c-v)/(c+v) by c. Then define x = v/c. The new denominator will be 1+x. Can you find a power series for 1/(1+x)?

what you're saying is put v/c = x whihc would give

[tex] \frac{1 + x}{1 - x} [/tex] whiuch is clearly not [tex] \frac{1}{1-x} [/tex] and thus u cannt ot expand it out like the latter
 
  • #10
and power series for
[tex] \frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + ... [/tex] for abs (x) < 1 where abs means absolute value
 
  • #11
stunner,

But (1-x)/(1+x) = (1-x)*1/(1+x). And 1-x is already a power series. So just get the series for 1/1+x and you'll see the answer.
 
  • #12
are you sure that can be done??
 
  • #13
stunner,

A power series for 1/(1+x)? Why not?

y = (1+x)^-1 >> y(0) = 1
y' = -(1+x)^-2 >> y'(0) =-1
y'' = 2(1+x)^-3 >> y''(0) =2

etc...

So, y = 1/(1+x) = 1 - x + x^2...

So, what's 1/(1+x) when x << 1?
 

1. What is the relationship between the frequency of light and the movement of the mirror?

The frequency of light recorded by a detector attached to a moving mirror is directly related to the speed and direction of the mirror's movement. As the mirror moves, the frequency of light will shift either higher or lower depending on the direction of movement. This is known as the Doppler effect.

2. How does the frequency of light affect the data recorded by the detector?

The frequency of light recorded by the detector is crucial in determining the accuracy and reliability of the data collected. If the frequency of light is not properly accounted for, it can lead to incorrect measurements and skewed results.

3. Can the frequency of light be changed by adjusting the speed of the moving mirror?

Yes, the frequency of light can be altered by changing the speed of the moving mirror. As the mirror moves faster, the frequency of light will shift higher, and as it slows down, the frequency will shift lower. This is a fundamental principle used in many scientific experiments and technologies.

4. Are there other factors that can influence the frequency of light recorded by the detector?

Yes, there are other factors that can impact the frequency of light recorded by the detector, such as the distance between the light source and the mirror, the angle of reflection, and the properties of the mirror itself. These factors must be carefully considered and controlled in order to accurately measure the frequency of light.

5. How is the frequency of light measured by the detector?

The frequency of light is measured by the detector using a method called interferometry. This involves splitting the light beam into two beams and then recombining them, which creates an interference pattern. The frequency of the light can then be determined by analyzing this pattern and comparing it to a reference beam with a known frequency.

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