- #1
Niles
- 1,866
- 0
Hi
Say I am looking at an AOM working in the Bragg regime (i.e., only a single diffracted beam). It is easy to show using Bragg's law that the frequency-shift Ω of the diffracted wave is given by
[tex]
\Omega = 2n\omega \frac{v}{c}\sin(\theta)
[/tex]
Here Ω is *also* the frequency the AOM is driven with, in other words the LHS is constant in the sense that in does not depend on the incoming light (so the frequency-shift imparted on the wave is constant). However, the RHS does depend on the incoming light, since the angle θ of the diffracted beam is equal to the angle of incidence of the incoming beam, so I can change it easily by e.g. turning the AOM.
In my book it says that the shift Ω is zero for forward scattering and maximum for backscattering. This is what I don't understand: The shift Ω is the same as the frequency of the phonons in the material, which is *constant*. So how can I change the frequency shift of the diffracted wave by changing the angle on incidence? Niles.
Say I am looking at an AOM working in the Bragg regime (i.e., only a single diffracted beam). It is easy to show using Bragg's law that the frequency-shift Ω of the diffracted wave is given by
[tex]
\Omega = 2n\omega \frac{v}{c}\sin(\theta)
[/tex]
Here Ω is *also* the frequency the AOM is driven with, in other words the LHS is constant in the sense that in does not depend on the incoming light (so the frequency-shift imparted on the wave is constant). However, the RHS does depend on the incoming light, since the angle θ of the diffracted beam is equal to the angle of incidence of the incoming beam, so I can change it easily by e.g. turning the AOM.
In my book it says that the shift Ω is zero for forward scattering and maximum for backscattering. This is what I don't understand: The shift Ω is the same as the frequency of the phonons in the material, which is *constant*. So how can I change the frequency shift of the diffracted wave by changing the angle on incidence? Niles.
Last edited: