X-Ray Diffraction: Explaining Visible Light Reflection

  • Thread starter Thread starter manofphysics
  • Start date Start date
  • Tags Tags
    Diffraction Ray
manofphysics
Messages
41
Reaction score
0
"When a wavefront of X-rays strikes an atom, the electrons in that atom interact with the X-rays and immediately re-emit the X-radiation, normally without change of wavelength, and the X-radiation that is emitted by the atom is emitted as a spherical wavefront "

Now my question is : visible light is also an electromagnetic wave, but we only experience Reflection in a straight line(if the object is well polished)Why does the visible light also not produce spherical wavefronts after striking the electron on the surface of the polished object, instead of showing proper straight line reflection?

I know this may sound like a elementary concept to some, but still I would be grateful if someone could clear up this doubt.Thanks,
 
Physics news on Phys.org
manofphysics said:
"When a wavefront of X-rays strikes an atom, the electrons in that atom interact with the X-rays and immediately re-emit the X-radiation, normally without change of wavelength, and the X-radiation that is emitted by the atom is emitted as a spherical wavefront "

Now my question is : visible light is also an electromagnetic wave, but we only experience Reflection in a straight line(if the object is well polished)Why does the visible light also not produce spherical wavefronts after striking the electron on the surface of the polished object, instead of showing proper straight line reflection?

I know this may sound like a elementary concept to some, but still I would be grateful if someone could clear up this doubt.Thanks,

Visible light has a very long wavelength compared to interatomic distances in a solid, so the interaction of visible light with a solid can be described using the macroscopic electrodynamics of continuous materials (as described, e.g., in the book by Landau and Lifgarbagez). For this case one can consider the system characterized entirely by the dielectric function \epsilon or equivalently the index of refraction n.

The usual procedure of equating the (incident, reflected, and refracted) macroscopic electric fields at the boundry leads to the kinematic equations
<br /> \sin(\theta_{\tt inc.})n_{\tt inc.}=\sin(\theta_{\tt reflect.})n_{\tt reflect.}=\sin(\theta_{\tt refract.})n_{\tt refract.}\;,<br />
but, since the incident and reflected have the same index of refraction the incident and reflected waves make the same angle w.r.t. the normal which means that the light reflects off the mirror in just the way one believes from geometrical optics. The final equality in the above equation is, of course, Snell's law.

For x-rays the situation is entirely different since the wavelength of x-rays is about the same as the interatomic spacing or less and macroscopic electrodynamics does not apply. Thus we think about the x-rays interacting with each atom individually not en mass (i.e., macroscopically). So, when the x-ray polarizes the atom there will generally be different angular momentum components of emitted light due to the resulting wiggling of the atomic electrons, but often the zeroth (i.e., spherical) component dominates and this gives rise the the idea you have mentioned about spherical waves being emitted from the position of the atom.
 
Hi. I have got question as in title. How can idea of instantaneous dipole moment for atoms like, for example hydrogen be consistent with idea of orbitals? At my level of knowledge London dispersion forces are derived taking into account Bohr model of atom. But we know today that this model is not correct. If it would be correct I understand that at each time electron is at some point at radius at some angle and there is dipole moment at this time from nucleus to electron at orbit. But how...
Back
Top