Diffraction grating with monochromatic light vs. white light?

1. Dec 21, 2013

AFSteph

(A) What kind of pattern of would you get if you shone monochromatic light on a diffraction grating? What pattern of light would you get if you shone white light on a diffraction grating?

My answer: With monochromatic light, you would get a pattern of alternating light and dark bands. With white light you'll get dispersion patterns -- bands of color arranged according to their frequency.

(B) Is it correct to say that the difference in patterns produced by the monochromatic and white light is caused by the fact that monochromatic light waves interfere with each other when they pass through a diffraction granting, while the rays of white light do not? Explain.

My answer: I'm pretty sure this statement is incorrect, but I can't quite put it into words. Does it have something to iridescence -- that is, light waves canceling out certain frequencies so we see component colors?

Am I on the right track at lest? Help would greatly be appreciated! :D

2. Dec 21, 2013

Simon Bridge

(A) not bad - can you be more specific: i.e. what color is the central maximum? How are the colored bands ordered wrt the central maximum? Are there gaps between groups of colored bands?

(B) Nope - consider: what is white light made up of?

3. Dec 21, 2013

AFSteph

Hello again Simon! Thanks for your response.
Frankly, I don't think I need to be more specific as the words "maximum" are at no point mentioned in my textbook. The section on diffraction grating is about half a page long. What I'm trying to say is this course is not that in depth. (I wish it covered more -- honestly, answering these questions with such skimpy explanations is pretty annoying! :P)

So then the statement in (B) is correct? So... as white light is diffracted into its component colors/frequencies because the frequencies bend at different rates. They don't interfere with each other as they bend and separate, right? Although, theré's a quote in my textbook that bothers me.
That's basically the entire section on diffraction grating in my textbook. According to them, it is by interference but there's only a few sentences of explanation here and AUGH. Your thoughts? Is this a trick question? Sorry for the questions D:

4. Dec 22, 2013

Simon Bridge

The word "fringe" probably is ... each fringe is a maximum.
At the depth (shallowness) to which it covers diffraction grating, does it show you an equation for finding the range spacing or locations given the wavelength and the line-density? In which case you should be able to be more specific.

It's a good way to pick up extra marks ;)

It sounds like the book is not being clear about when diffraction happens.
White light is made up of many colors - each color being a single wavelength.
Each color, therefore, has it's own interference pattern.
The total pattern is the sum of all the individual patterns.

Here is white-light diffraction pattern, compared with that for green light alone:

Notice the central fringe is white - with rainbow fringes to either side.
The green in the rainbow corresponds to the location of the green-light fringes.
The rainbows are oriented with the blue end towards the center.

Last edited: Dec 22, 2013
5. Dec 22, 2013

AFSteph

Wow, thank you for the picture and the clear explanation! It's a bit late here but let's see if I can articulate this into something coherent. The answer for be would be:
"Is it correct to say that the difference in patterns produced by the monochromatic and white light is caused by the fact that monochromatic light waves interfere with each other when they pass through a diffraction granting, while the rays of white light do not? Explain."
Yes and no. Rays of white light do not interfere with each other as they pass through diffraction grating. After the white light disperses into its monochromatic component frequencies, these frequencies have their own individual interference patterns. The individual patterns make up the total spectrum pattern we see.

Is that good? I hope I got this. :D

EDIT: On the fringe/maximum front:
They do not give any equation (this is conceptual Physics -- they're pretty scant on the equations and mathematics side of things...) and the only talk of fringes is merely to describe fringes on shadows due to diffraction. Nothing deeper than that. None the less, thank you for the extra knowledge :)

Last edited: Dec 22, 2013
6. Dec 22, 2013

Simon Bridge

Doing well - waves of white light do not interfere because there is no such thing as waves of white light... white light is made up of waves of many different colors. Each color interferes separately.
I suspect that's all you need. You seem to be getting it.

Conceptual physics is pretty unhelpful, as you are discovering.
When you get the chance you should actually do the Young's Interference experiment.

Merry Xmas.

7. Dec 22, 2013

epenguin

Don't spectrometers and spectrophotometers usually use diffraction gratings rather than prisms? Though now you make me think about it, and after a quick google, I wonder about the multiple spectra at different angles - do they just select one and waste the rest?

8. Dec 22, 2013

hjelmgart

Though it may differ from instrument to instrument, I believe our spectrophotometer at the university uses a both.

I would guess, you can also use a single prism, and just move it while keeping the light beam fixed at one point. Then by moving the prism and having a fixed aperture, you could probably get through the entire visible spectrum by having either the lower or upper wavelengths enter first.

9. Dec 22, 2013

Simon Bridge

The different sets of rainbows on the spectrum pic above are called "orders". The center one is usually order 0, with the others numbered 1,2,3... to each side.

You can set up the line spacing so that there is only one order on each side of the central one.

Real optical spectrometers vary a great deal in the components they use. A common one consists of two telescoped bolted to a turntable arrangement marked out in degrees-minutes-seconds. You put a diffraction grating (or a prism or whatever) in the center, point one telescope at the light, and turn the other to see the fringes. Crosshairs let you pinpoint the fringes quite well.

Below are two common setups.

10. Dec 22, 2013

hjelmgart

You wouldn't by any chance know, how it works in an ellipsometer as well? I mean when you measure the zeroth order reflectance of, say some periodic structure? I know the overall theory of the the ellipsometer works, but this thread made me wonder, when I use it to measure reflectance versus wavelength.

Is it simply a monochromator like above, which is positioned in the detector, which measures the intensity of the beam after it has been reflected from my sample?

Also, if I use lenses to focus my beam to a smaller spot size, could this not cause me some error margins for my measurements?

11. Dec 22, 2013

epenguin

So I'm right that you only take from one fringe and throw the rest away and it doesn't matter? (It just went through my head a minute you could recombine beans of a given wavelength to get more intensity but difficulties would probably make it not worth it.) Then of course in a spectrophotometer you also throw away most of the one fringe at any one time because you have to pass the light through a narrow slit to select one narrow range of wavelengths.

Can you comment on the following train of thought? Basically what are the advantages of diffraction against refraction for an instrument? With a prism you can get nearly all the light of a certain wavelength. However there is nothing in refraction that tells you what the wavelength is, it would have to be externally calibrated somehow. Whereas in diffraction the wavelength can be known from fundamentals, the grid spacing and the measurable angle. But the turntable I guess would have to be well machined and stable. No wonder they are serviced from time to time, now I seem to recall engineers using some standard absorbers to do this. But then you could calibrate a prism in the same way, so I wonder again what is the advantage of diffraction against refraction?

12. Dec 22, 2013

Simon Bridge

I had to use wikipedia like everyone else ... does not look related.
You set up the incoming light to have a particular polarization and measure the change in polarization after reflection.

http://en.wikipedia.org/wiki/Ellipsometry

You don't always throw away the extra ones ... depends what you want to do with them.
Of course, if there are no higher order fringes, then there is nothing to throw away.

13. Dec 22, 2013

Simon Bridge

You use the known properties of the material you made the prism out of just like, with the diffraction grating, you use the known properties of the grating.

Bottom line? It's much much cheaper.

14. Dec 22, 2013

epenguin

But why is it much cheaper? With a diffraction grating you have to be scratching thousands of accurately enough-defined lines some tiny distance apart. A prism is just a block of glass. I suppose it is difficult to get it uniform density and composition?

15. Dec 22, 2013

SammyS

Staff Emeritus
Diffraction gratings can be made very cheaply by a method known as replication.

See this link: http://www.optometrics.com/diffraction_gratings.html#THE REPLICATION PROCESS

16. Dec 22, 2013

Simon Bridge

By comparison - getting high-enough quality optically clear glass with precise edges to get a similar quality spectra is quite difficult.

I used to make slits with razor blades (the kind we used to shave with - look up "double edges safety razor blade") - I could stack a number of them and draw them across a soot-coated glass slide. The separation was the width of the blade, manufacturers specs: 0.004" (about 0.1mm). The slit width itself could be something less than half that iirc. These days you just buy them to whatever specs you want.

Try comparing prices.
It can be quite counter-intuitive what ends up costing you money.