How to measure light wavelenght?
Please, don't say 'with a waveleghtometer' . This is a Physics 'practical problem', and I'm thinking interference.
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Weren't you the answerer who commented that a CD makes a spectrometer......
You could build a basic one quite easily, and calibrate it against a LEDs.
Interference is exactly how it works of course.
You could build a basic one quite easily, and calibrate it against a LEDs.
Interference is exactly how it works of course.
Well, the problem is that it is IR. I've asked about seeing a IR LED glow (blink) in a remote control about a week ago, and now the problem is to actually measure how far is its wavelength (here, I got it right finally), to visible light. Also I think that mistake would be to large with a diffraction grating (this was about the first thing I thought of).
actually measure how far is its wavelength (here, I got it right finally),
To be accurate, the English would be "how LONG is its wavelength"
using a diffraction grating would still work, all you need to do is look at the image WITH a mobile camera.
Steve
I wanted to say FROM visible light. I've got terrible problems with any written language, probably from being left-handed and having the tendency of writing backwards or mirror write (learning Hebrew doesn't help much), so I can either concentrate on spelling, or punctuation, or what I actually mean. Sorry :(
I'll try the diffraction grating, by the way.
I'll try the diffraction grating, by the way.
Definitely try the diffraction grating! Within reason, it's not limited to any particular wavelength (which is why you can make a rainbow with one :-), and the IR produced by LED is close enough to visible that it'll work. Just use a digital camera to view the result.
Hey, just thought of a great trick. Use a small incandescent white-light source (not a fake "white" LED) adjacent to your IR source, and take a single picture of both of them going through (or reflecting off) the grating. You should get the normal visible-light rainbow, along with the IR spectrum. Now you can calibrate the IR directly against the known red and violet, and get an estimate of the wavelength.
Hey, just thought of a great trick. Use a small incandescent white-light source (not a fake "white" LED) adjacent to your IR source, and take a single picture of both of them going through (or reflecting off) the grating. You should get the normal visible-light rainbow, along with the IR spectrum. Now you can calibrate the IR directly against the known red and violet, and get an estimate of the wavelength.
This is a class problem, or your own little idea ? If its a class problem for 16 year olds, your education system is light years ahead of ours.
Steve
Steve
It could be both, Steve. Some of the advanced science classes here assign the students to complete a "term project" -- they choose the topic themselves, and have to write it up and usually give an oral report at the end.
There is no way that anyone in a public school (in your sense) student in the UK would get assigned a project like this. Its really tricky. Only someone operating at a VERY high physics level here would assigned something like this.
Kudos to Gruffalo.
Steve
Kudos to Gruffalo.
Steve
Yes, exactly (the student chooses the topic, which usually needs to be approved), and definitely kudos to Gruffalo for choosing something challenging but not impossible!
Well, it is a Maths specialized school, and a problem for 16-17-18 (first year undergraduates too) year olds who said they liked Physics. And it isn't a term project, it's a by-the-way thing before they think of some half-year projects and problems which are somewhat more challenging and longer to do, so I bet you'll hear some more questions from me after they are out.
Anyway, it has very little to do with the education system here, because it is an "advanced Maths class" and a JC school.
Anyway, it has very little to do with the education system here, because it is an "advanced Maths class" and a JC school.
Is that selected by examinations ? We have this weird idea here that you can't select by ability, unless its "sports"
Yes, its examination after grade 7 (out of 11 staring with 0), with 13-14 year olds. The problems at the exam are Olympiad-like and are supposed to require no knowledge whatsoever, only 'talent'.
Exams like this are not usually done in 'normal' state schools, but I don't think there is any law prohibiting it, and , once again, a JC school has a bit more freedom, for example, our headteacher even got us out of writing a general certificate Maths exam last year (GCSE analogue, written at the age of 15) and we did our own Algebra exam instead
Exams like this are not usually done in 'normal' state schools, but I don't think there is any law prohibiting it, and , once again, a JC school has a bit more freedom, for example, our headteacher even got us out of writing a general certificate Maths exam last year (GCSE analogue, written at the age of 15) and we did our own Algebra exam instead
Does your department have a double-beam IR spectrophotometer? they're the sort of thing that get given away to physics lab's when they're old.
L
The usual trick is to use a diffraction grating. You send your light beam through the grating, and different wavelengths of light emerge at different angles. You measure the angle and infer that the light detected at that angle must be such and such a wavelength. More here:
http://en.wikipedia.org/wiki/Diffraction_grating
Also here:
http://en.wikipedia.org/wiki/Spectrometer
http://en.wikipedia.org/wiki/Diffraction_grating
Also here:
http://en.wikipedia.org/wiki/Spectrometer
Also I think this is a nice picture. It's a CFL seen through a diffraction grating. So you can see how this wavelenght-to-angle trick works. Source: http://en.wikipedia.org/wiki/File:CFLspectrum.agr.jpg
Did you try Googling "measuring wavelength of light"? It helps to spell correctly when doing a search. You can also search on "diffraction grating" or "spectrograph," as Ork* said.
Sep 14, 2010. 9:39 PMorksecurity
says:
Spectrograph would be a traditional solution, comparing it to known wavelengths.
(Tried websearching this yet?)
(Tried websearching this yet?)
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