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ArcticMyst Security by Avery

Single Slit Diffraction Spectroscopy - My Attempt to Calculate Wavelengths

Joined
Oct 24, 2008
Messages
1,057
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48
Well here is my crack at it and I actually got pretty good results, at least they were more accurate than I though they would be.

The Goal:
To calculate/estimate the wavelength of my Red and Blu-Ray lasers using the known wavelength of my green 532nm Laser with stuff I've got around the house. :)

The wavelength of laser diodes can shift but DPSS lasers don't shift wavelengths.

The Set Up:
I didn't have a suitable prism or diffraction grating so I decided to use single slit diffraction. Single slit diffraction causes the beam to interfere with itself and an interference pattern is generated. An interference pattern has a central maxima and diminishing maxima on either sides. The angle between the maxima changes as the wavelenght changes. So by measuring the distance between them I could use that information to extrapolate the wavelength.

So First I needed to create my slit. I made it by positioning two X-acto blades next to each other while they are stuck on a magnet. By using the neodymium magnet I can position the blades so they create an extremely narrow slit (I'll calculate just how narrow later). The narrower the slit the farther apart the maxima are spaced, making taking measurements more accurate.

DSC09042.jpg

DSC09062.jpg

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DSC09044.jpg


To further improve the accuracy of the measurements, I let the pattern diverge for 33ft.

I used my 160mW 532nm CNI PGL-III-A , my 250mW 660nm LOC, and my 240mW 405nm 6x lasers.

I know that my 532nm DPSS is exactly 532nm. So I'll use that and the relative measurements to work out the true wavelength for my other two lasers are. I suspect that my 405nm 6x laser is lower than 405nm and I really don't know what my 660nm LOC truly is, but I have no reason to suspect it's far from that 660nm mark.

DSC09043.jpg

DSC09057.jpg

DSC09053.jpg

DSC09045.jpg

DSC09051.jpg

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The measurements:

This was a pain in the a$$ but I think I got a pretty good guestimation of it. There are alot of variables and it's diffucult to see the exact center of each maxima but I think my measurements work accurate to within a 1/8 in or so. I measured in inches and converted to mm for the math.

The maxima are labeled as K values.
k=0 - is the center maxima
k=1 - are the maxima to the immediate left and right
k=2 - are the 2nd from the left and right
k=3 - are the 3rd from the left and right
and so on

Notice how the longer the wavelength is the farther spaced the maxima are. That is an artifact of wavelength because I did not change the distance or the gap on the slit. I only switched lasers and different pattern emerge. Using that difference in the patterns is the basis of this experiment.

DSC09052.jpg

DSC09047.jpg

DSC09049.jpg


At a distance of 33ft, ~10,058mm the distances from the center maxiam to the k values are as fallows, and (by the small angle approximation) the angles are calculated as well.

532nm Laser
K=1 121mm .689 deg
K=2 187mm 1.056 deg
K=3 273mm 1.555 deg
660nm Laser
K=1 148mm .834 deg
K=2 227mm 1.295 deg
K=3 326mm 1.856 deg
405nm Laser
K=1 89mm .507 deg
K=2 143mm .815 deg
K=3 200mm 1.139deg

The Math:

So now I had my data and I could put the lasers away and crunch the numbers.

Now the whole point of this was to see if I could figure out the wavelength of my 660nm Laser and my 405nm Laser relative to my 532nm Laser, but I was curious about my slit too.

Using the equation that describes maxima interference patterns...

(B)sin(A) = (k + 1/2)(W)

B - width of slit (nm)
A - angle to k
K - which maxima (k=1,2,3...)
W - wavelenght

... I could calculate the width of the slit for all 9 of my measurements. Believe it or not all of my results were within 6um of one another. I got an average width of 69000nm or .069mm! Less than a 1/10th of a mm!

Mmmkay, now on to the main results.

I used a formula called Snell's Law to compare the wavelengths and angles.

(sin(A)/sin(b)) = w1/w2

A - angle of 1st laser
B - angle of 2nd laser
w1 - wavelength of 1st laser
w2 - wavelength of 2nd laser

This says that the ratio of the sine of the angle between corresponding maxima of two different lasers is proportional to the ratio of their wavelengths.

For example, I know that angle between the center and the 1st maxima of my 532nm Laser is .689 degrees. And I know that my red Laser made an angle of .843 degrees with it's first maxima. So I can figure out what the ture wavelenght of my red Laser is by using the accepted valise of 532nm ....

651nm

I was stoked to see it come that close! But that was just one. I measured when k=2 and k=3 too. So I can actually do the math two more times and compare and average the results.

Using 532nm to guess the wavelength of my red Laser
k=1 651nm
k=2 646nm
k=3 635nm

644nm average

The same can be done for 405nm too.

Using 532nm to guess the wavelength of my Blu-Ray Laser
k=1 391nm
k=2 407nm
k=3 389nm

395nm average

Ha! I suspected it was a bit lower than 405nm!

Now I'm not about to consider these numbers to be super accurate. I know that there were many variables, especially with measuring the maxima. I want to get a better diffraction grating and try this again and with more accurate measurements and with some kind of marked scale to project the pattern onto. If I wanted to compare two lasers of the same type of diode this set up won't work at all. The margin of error is too big. But if you think about what I just did...

I just calculated the wavelength of a laser, to and accuracy of +-20nm, with a ruler and a couple of razorblades. :)

-Tony
 





aXit

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Sep 15, 2009
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Nice work on the slit width, a 10th of a millimetre, that's about the thickness of printer paper; or was that the accuracy of your thickness measurements from 532? Would the results improve with a smaller slit? Maybe placing a thin material between the blades then removing it could get a smaller slit

Great job on the estimations. +-20nm isn't bad for the setup.

Enjoyed reading it, thanks.
 
Joined
Sep 12, 2007
Messages
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My eye can probably guess a wavelength ±20nm, but nice work and proof of concept. You should have far better results with a plastic diffraction grating. They're less than a dollar if you look around.
 
Joined
Oct 24, 2008
Messages
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Points
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Nice work on the slit width, a 10th of a millimetre, that's about the thickness of printer paper; or was that the accuracy of your thickness measurements from 532? Would the results improve with a smaller slit? Maybe placing a thin material between the blades then removing it could get a smaller slit

Great job on the estimations. +-20nm isn't bad for the setup.

Enjoyed reading it, thanks.

Thanks!

The results would only improve with more accurate measurements. But a smaller slit would spread out the interference pattern even more, so yeah, the smaller the slit the better. Calculating the width of the slit gave me an idea of the accuracy of my measuring. The width doesn't change, well temp could do it, but generally it's the same. I used all 9 of my measurements to figure out the width. All of them came to with in 6um of each other. So that tells me that my measurements, although not perfectly accurate, were at least very consistent.

When I get a better diffraction grating, than the one I have, I will do this again. My ultimate goal would be to compare my two 405nm builds and confirm my suspicion that one of them is indeed a shorter wavelength than the other.

My Ehgemus 6x build is about 200-240mW (I'm guessing) and it "appears" to be just as bright as my 90mW (confirmed output) when shined on a nonfluorescing surface. And I could figure out where my LOC is (660-635nm I figure)

But it just goes to show that something as intangible and abstract as the wavelength of electromagnetic waves can be measured with stuff you've got around the house! That too me is astounding!

-Tony
 

Morgan

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Feb 5, 2009
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Any further developments a couple of weeks on Xplorer877?

Are the individual maxima of a single diffraction the same length? If they were, then you could measure from the end of one to the end of the next. This would remove the variable of trying to guestimate the middle?

I may be guestimating myself there but measuring this way on all the samples will give a similar proportional measurement wouldn't it?

Just thought I'd throw that out there.

Really interesting thread.

Stuart and I agree +1

M
:)
 
Joined
Oct 24, 2008
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Thanks!

The individual maxima do not technically have a "length" because they decay and fade rather than abruptly end. Plus I don't know of any formula that would relate the length to the wavelength.

-Tony
 

Morgan

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Thanks!

The individual maxima do not technically have a "length" because they decay and fade rather than abruptly end. Plus I don't know of any formula that would relate the length to the wavelength.

-Tony

Ah, I see so defining the darkest point is just as difficult as defining the brightest!

I meant though, rather than being able to define wavelength from just a single measurement of a single laser, the ratio of the measurement from ends of any maximas on one laser compared to another, (obviously this isn't as easy as I thought from the pics but just clarifying...), would still be the same as the ratio of the measurement from the middle of any maximas on one laser compared to another, right?

M
:)
 
Joined
Oct 24, 2008
Messages
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Points
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Yeah it would, but again it's hard to make accurate measurements. I think what I'll do next time is to use a real diffraction grating and project the pattern onto a non-fluorescing marked scale. Then it will be much easier to see or I could take a picture and get the measurements there.

-Tony
 




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