1
$\begingroup$

We had an experiment in our physics lab where we used two grating materials one with 2500 lpi(say A) and the other with 15000lpi(say B).

Now when we shine white light on both of them we didn't get any spectrum in the transmitted light from A but we did get the pattern from B and my instructor told me that it had to do with the fact that the spatial coherence of the white light is smaller than the grooves distances for grating A (hence no diffraction in the transmitted light) while it is bigger than the distances in B.

But when I saw the light being reflected from both the gratings I could see spectrum in both of them !! Shouldn't the coherence condition be applicable here as well implying no pattern for the grating with 2500lpi ?

Can someone explain why I was able to see the spectrum in the reflected light but not with the transmitted one from grating A ?

Here's a picture of reflected pattern from both the gratings :-

One with 2500lpi :-

2500lpi

Here's the reflected pattern from 15000 lpi grating:- enter image description here

$\endgroup$
6
  • $\begingroup$ You can see "spectrum" in both, but clearly, the amount of dispersion in the one image is much greater than in the other. I can't answer how that relates to "coherence" though. $\endgroup$ Commented Mar 21 at 13:42
  • $\begingroup$ I don't understand why the 2500 lpi grating disperses light less than the 1500 lpi grating. Something is off. $\endgroup$
    – mmesser314
    Commented Mar 21 at 14:25
  • $\begingroup$ @mmesser314 it's 15000lpi not 1500lpi $\endgroup$
    – Ankit
    Commented Mar 21 at 15:23
  • 1
    $\begingroup$ What is meant by 'reflected pattern from both the gratings'. Don't you have four pictures, namely transmission and reflection from both gratings? What are the angles of the orders? $\endgroup$
    – my2cts
    Commented Apr 15 at 13:57
  • $\begingroup$ @Ankit Why did you remove the image of the second grating? $\endgroup$ Commented Apr 15 at 14:01

3 Answers 3

2
+50
$\begingroup$

Without more information on your experimental setup and the exact type of grating used, it is difficult to know exactly why. However, based on the pictures, it is fairly certain that the reflected patterns are indeed diffraction because the angle increases with wavelength, which is consistent with the formula $$\sin\theta = \frac{\lambda}{d}$$ and they are also consistent with other examples I could find.

Your gratings have line spacings of around $0.0106\,\text{mm}$ and $0.00169\,\text{mm}$ respectively, which correspond to average first-order diffraction angles of $3°$ and $18°$ respectively for visible light.

The lack of incident light coherence does not necessarily mean that there is no diffraction pattern. Incoherent white light diffracts into colors as well cf e.g. this post.

If the two gratings are of the same type, my hypothesis is that the diffraction angle of the first grating is too small and the colors within each order were too close to be distinguishable, resulting in a mostly white patch for each order.

You seem to be using transmission gratings which are designed to transmit light. Therefore, they have much less reflectivity and I suspect that it is only with this weaker reflectivity (combined with the likely weaker and more distant ceiling light than the experimental light) that the closely-spaced colors within each order can be visually distinguished as in your first picture. Therefore, you can make out the colors in the reflected pattern but not in the transmitted pattern.

On the other hand, your second grating does not suffer from the closeness problem and therefore produces distinguishable colors in both cases. So, in summary, I think this is an intensity issue rather than a coherence issue. Again, without further details on the experimental setup, it isn't really possible to pinpoint the exact reason.

$\endgroup$
1
  • $\begingroup$ It makes me wonder if that filter has the gratings facing downward so the light passes through the substrate before being refracted by the gratings. $\endgroup$ Commented Apr 21 at 23:31
1
$\begingroup$

What I think is happening here, that in A the diffraction is not of the type of diffraction that the experiment is about. The pattern that you are seeing is more like that of a dispersive diffraction, i.e. diffraction from scattering of light by a uneven surface. It is very similar to what you see from the reflection of white light on CD/DVD or any such reflective surface (even a mirror does this but is difficult to observe because scattering is much less, you can actually see this pattern in periphery of a bright light).

This explanation is also supported by the fact that in A, the different wavelengths are completely overlapping while in B where actually spatial diffraction is occurring due to the slits of the grating. In A, it is being reflected by the material of the slits itself and not the slits which also does diffraction upon reflection because it has very tiny unevenness.

$\endgroup$
1
  • 1
    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Apr 19 at 2:50
0
$\begingroup$

It is not quite clear exactly what the experimental conditions of the experiment was. For that, the incident angles of all the different experiments would need to be specified. One can then also record the observed angles of all the diffraction orders.

Nevertheless, one possible explanation can be that the diffraction angle for the grating with the higher number lines per inch is too large to produce a first diffraction order for normal incidence. So when the incidence angle is increased to allow for a reflection, the diffraction grating can produce a visible diffraction order.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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