wave concept question

leeyiren

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on: September 19, 2010, 01:26:52 pm »

hello, everyone.

We all know that electromagnetic waves travel more slowly in a transparent medium than in a vacuum.

Their frequency stays the same but their wavelength gets less.

v=f X lambda

My question is why "their frequency stays the same but their wavelength gets less." Huh

Can it be that the wavelength stays the same but their frequency gets less?

In that case, the velocity still decreases.

Are there any simulations to show that the frequency stays the same and the wavelength decreases when light enters glass from air?

Remember frequency has the element of time.

Thanks.


« Last Edit: September 19, 2010, 01:29:47 pm by leeyiren »	Logged

Fu-Kwun Hwang

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Reply #1 on: September 19, 2010, 04:52:49 pm »

You would not asked the above question if you have understood the meaning of wave.

Wave is the same change of some physics quantity, however, there is a time delay due to distance in space.

You are welcomed to check out formation of wave

The source determine the frequency of the wave. The wave propagate with the same frequency.
The same phase will produce the same amplitude. $\phi=\vec{k}\cdot\vec{r}-\omega t$
when wave move from $\vec{r}$ to $\vec{r}+d\vec{r}$ , there is a time delay $dt=\frac{\vec{k}\cdot d\vec{r}}{\omega}$ .
Since it is always the same change in time (however, different time delay due to distance from the source),
wave always keep the same frequency when propagate. (otherwise, it is not a wave).

You might want to check out refraction of wave from one media to another media,too!


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leeyiren

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Reply #2 on: September 19, 2010, 05:14:03 pm »

hi prof. thx for ur prompt reply. I hope u can help me clear this doubt.

The textbook says "the shorter the wavelength, the higher the frequency. As the frequency gets higher, the energy increases".

I read about Planck–Einstein equation from wikipedia which is not taught in the textbook or during lessons.
The following equations show how energy, wavelength and frequency is related.
λν = c
E=hv
E=(hc)/λ
h=Planck's constant = 6.62606896(33)×10−34 J•s
c=3x10^8 m/s.

so I can conclude that
increase frequency or decrease wavelength -> E increases.
decrease frequency or increase wavelength -> E decreases.

Am i right?

On the other hand
E=kA^2
The amount of energy carried by a wave is related to the amplitude of the wave. A high energy wave is characterized by a high amplitude; a low energy wave is characterized by a low amplitude.

It is taught to us that

"Putting a lot of energy into a transverse pulse will not effect the wavelength, the frequency or the speed of the pulse. The energy imparted to a pulse will only effect the amplitude of that pulse."

Doesn't this contradict the Planck–Einstein equation?


« Last Edit: September 19, 2010, 05:17:38 pm by leeyiren »	Logged

Fu-Kwun Hwang

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Reply #3 on: September 19, 2010, 09:33:27 pm »

For mechanical wave : e.g. wave on a string propagate in x direction (displacement in y direction)
$y(t)=A \sin (kx-\omega t)$
So $vy(t)=\frac{dy(t)}{dt}=-A\omega \cos(kx-\omega t)$
The kinetic energy per unit length $K=\frac{1}{2}m v^2=\frac{1}{2} \rho dx A^2\omega^2 \cos^2(kx-\omega t)$ where $\rho$ is the mass per unit length
So the energy of the wave proportional to amplitude $A^2$ and frequency $f^2$ , where $\omega=2\pi f$ .

However, the wave speed is determined by the wave media (string) $v=\sqrt{\frac{T}{\rho}}$
The frequency of the wave is determined by source. And the wavelength $\lambda =v/f$
If you did not change the frequency of the wave source, putting more energy means increase the amplitude of the wave.
Actually, it is also possible to change the frequency of the wave source, however, more power is required to generate wave with the same amplitude.

$E=hc/\lambda$ is the relation for light. It means that the power of light is proportional to the frequency (assume electric field is the same).
Light source determined the frequency of the generated light.


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leeyiren

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Reply #4 on: September 19, 2010, 09:49:57 pm »

I am grateful for your time. However I dun get your equations and explanations. Is there any other way you can explain?


« Last Edit: September 19, 2010, 09:51:39 pm by leeyiren »	Logged

Fu-Kwun Hwang

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Reply #5 on: September 19, 2010, 10:17:57 pm »

Quote

"Putting a lot of energy into a transverse pulse will not effect the wavelength, the frequency or the speed of the pulse. The energy imparted to a pulse will only effect the amplitude of that pulse."

The above statement is under the assumption that frequency of the wave source is not changed!


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