1
$\begingroup$

I know that when the slit width is less than that of wavelength then the slit will act as a point source and scatter the light in all directions.

But my question is that during the scattering of light does the photons get absorbed by the wall and then re-emmited as a scattered light or the photon rebounces without interaction with wall? Please give reason for your answer.

A fun way to test this will be to put coloured slit and observing the colour of light which is scattered. If the scattered light colour changes then we know that the photon is being absorbed before scattering. Please reply guys...

$\endgroup$
0
$\begingroup$

Whenever light is reflected/scattered from a surface, it is being absorbed and re-emitted. Reflection from a surface is interaction with the surface.

In the experiment you propose, you'll need a way to account for the fact that light of different wavelengths is always diffracted slightly differently. For example, the spacing of fringes in the double-slit interferometer depends on the wavelength of the light.

Perhaps you could simplify your proposed experiment. Instead of using a slit, just reflect white light directly off a polished surface made of the material the slit would have been cut from, and see how the spectrum of the light is altered. It's an experiment that can be done at home relatively easily.

If your question is about whether or not the diffracted light that emerges from a slit interacts with the material at the edges of the slit, I'm not sure. However, I suspect that a very small portion of it does. Measuring that effect would be much more of a challenge, because probably only a very tiny fraction of the light would interact with the slit material without being totally absorbed or back-reflected-- a percentage on the order of a few tens of nanometers divided by the width of the slit (which suggests that if there is such an effect, it should be more evident when the slit width is very small). I base that guess on the fact that when light is totally internally reflected, the evanescent wave extends on the order of a hundred nanometers beyond the reflecting surface. With grazing-incidence light and an opaque material, I imagine the penetration depth will be substantially less. Distinguishing the "interacted" light from the "merely diffracted" light would be tricky-- but your idea of looking for a color change is a good one, because "merely diffracted" light should have a very predictable pattern for each wavelength, and the "interacted" light would show up as a wavelength-dependent deviation from that pattern.

$\endgroup$
0
$\begingroup$

In diffraction, when wavelength is less than slit width, then does the scattered light gets absorbed & emitted from the wall or just rebound?

Some of the photons will hit the material beside the slit and will be absorbed by electrons of the wall. The re-emission will happens with different wavelengths, mostly in the infrared.

Photons will go through the slit uninfluenced beginning with some slits width. A simpler description of particle deflection is the case of a single edge. Here the slits width is infinite. But an intensity distribution behind the edge still appears.

For photons the first peak is equally distributed to the left and right of the geometrical shadow line. For electrons the first peak is right from a left edge, so electrons get only deflected from the edge meanwhile photons get deflected also behind the slit. This phenomena will be helpful to answer your questions.

H. Boersch FRESNELSCHE Elektronenbeugung 1940 (description of these experiments)

enter image description here

As you suggested, the interest should be focused to what you call rebounding.

... or the photon rebounces without interaction with wall?

A photon has an altering electric field component. It should be permissible to assume that by this field some photons interacts with the electric field of the surface electrons of the slits material. This are the photons which are not hitting the wall and not going uninfluenced through.

Since both the negative and the positive value of the photons electric field are involved, some photons arrive the observation screen behind the geometrical shadow and some arrive away from this line. Not so electrons, the all get deflected (in some range of distance to the edge) due to their always negative electrical potential away from this line.

This different behavior of photons and electrons at edges should be a very important argument for the interaction of the edges field and the particles field.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.