Is there such a thing as a "hobbiest grade" spectrometer?

[rhd]

Oh my god, this is hilarious! Why did I just spend $200 on a spectrometer?

Ok, so I have to admit that I thought you were full of crap when you said you did this with a ruler! So I cracked out Photoshop. I wasn't sure exactly how to measure, so what I did was crop each image by dragging a crop box from the dead center of one circle, to the dead center of the other. It produced a long thin rectangle with 1/4 of each dot remaining. Here they are:

532 cropped to a 2141 pixel wide box

UNKNOWN cropped to a 2391 pixel wide box

Now, the first time around I actually corrected for the fact that this wasn't a perfectly straight photo, and used the Pythagorean theorem to calculate the distance between top-left and bottom right. That wasn't necessary, the width was enough. Correcting for 33 pixels of vertical offset produced almost no benefit.

Well, the math was simple:

Calibration Shot:
532 / 2141 = 0.248482018

Unknown:
? / 2391 = 0.248482018
? = 0.248482018 x 2391
? = 594.1205044

So.... the unknown wavelength was calculated at 594.12 nm, and it was actually 593.5 nm. That's an error rate of 0.620504437 nm.

There was so much room for improvement here too. For one, I just scotch taped a small flexible diffraction grating to my lens. With a larger grating, I could have used almost twice as many pixels to get the same scale. I'm sure that an automated piece of software could also do a better job of identifying the dead center of the dot. My estimation could have been off by a few pixels.

Lol - it didn't take us long

 

[Cyparagon]

I've been trying all day, and I can't get consistent results. The distance between the dots changes slightly with their position in the view. Since the distance for each wavelength is different, this further complicates matters. laser_freak used a linear approach, and it shouldn't be linear - at least not if you want accuracy. At times, I found an approach that worked for two wavelengths, like here, but never the whole spectrum. Not yet anyway.

 

[rhd]

I don't think this should have worked as well as it did. The pixel to wavelength ratio isn't constant throughout the spectrum.

 

[rhd]

I've been trying all day, and I can't get consistent results. The distance between the dots changes slightly with their position in the view. Since the distance for each wavelength is different, this further complicates matters. laser_freak used a linear approach, and it shouldn't be linear - at least not if you want accuracy. At times, I found an approach that worked for two wavelengths, like here, but never the whole spectrum. Not yet anyway.

That's what I was trying to convey also. Doesn't the equation you sent me for the web tool work better?

EDIT: How would I get the "B" variable out and onto the same side of this equation as the "A" ?

A = 532 / sin(arctan(2141/B))

 

[Cyparagon]

I tried an adaptation. I started with what I know to work: λ = dsin(arctan(S/L)) where:

d is the distance between the slits
λ is wavelength
S is the distance between orders
L is the distance to the screen

d is 1µm, and with a known wavelength and pixel count (thrown in where S would be), I solved for L, which is a constant. It gets close, but not nearly close enough. It's off by 20nm in many cases. There must be something in the camera optics or elsewhere that I need to take into account.

Oh, and don't forget units. All in meters in the original, but if you want λ in nm, It can be simplified to this assuming a 1000line/mm grating:

λ = 1000sin(arctan(S/L))

And this assuming a 500line/mm grating:

λ = 2000sin(arctan(S/L))

 

[rhd]

Do you know how to re-write your initial equation with "d" and "L" (and only those variables) on one side?

 

[rhd]

Given L and d as constants, we should be able to solve for UNKNOWN wavelength with the following two equations:

(#1)
arctan(2141/L) = arcsin (532/d)

(#2)
arctan(2391/L) = arcsin (UNKNOWN/d)

My problem is that I don't know enough trig to be able to isolate "d" and "L" on one side of the equations, so that we can connect the two and solve.

 

[laser_freak]

When the angle is small, only a few degrees, the approximation of x ~ m(lambda)L/d can be used. Therefore x is approximately proportional to lambda.

 

[Cyparagon]

Given L and d as constants, we should be able to solve for UNKNOWN wavelength with the following two equations...

I don't know either. It might not be possible, or you might end up with complex numbers. You can skip a lot of the busywork with Wolfram Alpha.

When the angle is small, only a few degrees...

It's not only a few degrees though - not with a diffraction grating. It's more like 10-30 degrees.

 

[laser_freak]

Ok, let's not use the small angle approximation.

d * b / ((b^2) + ((a^2)((d^2)-(x^2)))/(x^2))^(1/2) = y

plug the equation into wolfram alpha to get a pretty picture of it.

a = distance between orders of known laser
b = distance between orders of unknown laser
d = distance between slits
x = wavelength of known laser
y = wavelength of unknown laser

 

[rhd]

1000 * 2391 / ((2391^2) + ((2141^2)((1000^2)-(532^2)))/(532^2))^(1/2)
= 574

(not close)

I feel like there has to be a way to get that "d" figure out of the equation.

My theory is that it's a constant, so you should be able to reduce it out of both the known and unknown wavelength's equation.

 

[qumefox]

Ok, with the direction this thread has gone.. I can't help but think about this video.

Heh. You have to watch the whole thing though.. and remember to set your calculator to maths.

 

[lasersbee]

BTW... if anyone is interested I contacted the
eBay seller of the $200.00 Unaligned Spectrometer
and asked how much they charge to have it aligned....

Hold onto your hats......
an additional $300.00....

Either they are trying to gouge or they are not that
simple to align.


Jerry

 

[icecruncher]

BTW... if anyone is interested I contacted the
eBay seller of the $200.00 Unaligned Spectrometer
and asked how much they charge to have it aligned....

Hold onto your hats......
an additional $300.00....

Either that are trying to gouge or they are not that
simple to align.


Jerry

I posted that in reply #13 btw.

 

[lasersbee]

I posted that in reply #13 btw.

Thanks for reminding me .. I must have missed it..:beer:

Jerry

 

[icecruncher]

Thanks for reminding me .. I must have missed it..:beer:

Jerry

Not problem, just thought maybe you were skimming and missed it.

I was surprised also. From $200 to $500 for just alignment seemed steep. I still would really like one, so it's on a wish list.

Maybe my wife will buy for me on our anniversary. that would be nice. (I have a very nice wife :wave