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UltravioletPhotography

Infrared with a solar panel


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So, I discovered that a diffraction grating with a spacing of d can only diffract light with a wavelength λ smaller than d. The angle θ with λ=d is 90°. I also found a function to calculate x for a given λ:

post-284-0-91595100-1577987004.png

 

Where L is the distance between the diffraction grating and the wall, and d is the spacing of the grating. X becomes infinite as λ approaches d. In my case I have a 1000 lines/mm grating, which means that I can not see a spectrum above 1000 nm with that grating. If I even want to try I must get a grating with a spacing bigger that 1 μm. A 500 lines/mm grating would allow me to reach 2 μm teoretically.

 

Learning is always a good thing.

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All the sources I've seen say that the largest wavelength you can resolve with a diffraction grating is TWICE the line spacing. For example, read the ThorLabs guide here: https://www.thorlabs...ogpages/802.pdf

post-94-0-17587500-1577992781.png

 

But that's the theoretical maximum. It actually degrades more quickly than that, apparently.

 

Also, check your formula, that doesn't look right:

post-94-0-01654700-1577993507.jpg

 

I'm not sure why there is a discrepancy between the "twice the line spacing" criterion that I see online and this formula (or yours). Incidentally the units on your formula don't work out, so I am pretty sure there is an error. :smile:

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This is the error in my formula:

post-284-0-59547300-1577995705.jpg

 

when you write "" like that, it can magically become a "-". You have no idea how many times I made errors like this. In the bottom of my formula, I put an absolute value because I was getting negative numbers when I tried with real ones.

 

Even using your formula, post-284-0-80738700-1577997498.png, because L^2 becomes negligible as x gets big. Maybe Thorlabs has another definition for "groove period". Experimentally this is what I see.

 

As an example, I tried with the laser on my infrared thermometer:

post-284-0-85542300-1577999153.jpg

 

Grating: 1000 lines/mm, wavelength: probably 650 nm (even if the laser is labeled as 630-670 nm. That's a very wide range for a laser, did they take into account temperature drifting?)

post-284-0-00592900-1577999283.jpg

 

L = 0.112 m

x = 0.1025 m and 0.0955 m (yes, I have two different values on the left side and on the right side, I made sure the grating was parallel to the wall. I don't know what is going on). Average=0.099 m.

Calculated wavelength: 662.29 nm. Also notice that you can not see 2nd and greater order diffractions, as they would be equivalent to more than 1000 nm.

 

Also, this is the shape of the graph of my (corrected) function, without constants.. For wavelengths below about half a spacing, x is approximately linearly dependent on λ. When it approaches d (1 in this case) it shoots to infinity very quickly.

 

post-284-0-10742900-1578000287.jpg

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when you write "" like that, it can magically become a "-". You have no idea how many times I made errors like this. In the bottom of my formula, I put an absolute value because I was getting negative numbers when I tried with real ones.

On the second line you also dropped a minus sign. If you fix that you will not need to add absolute values.

 

The laser experiment was nifty, but in the 660nm range it's clearly well within what the 1000 line/mm grating is capable of, so it should be giving good results there.

 

Also, this is the shape of the graph of my (corrected) function, without constants.. For wavelengths below about half a spacing, x is approximately linearly dependent on λ. When it approaches d (1 in this case) it shoots to infinity very quickly.

 

Yes, and this is what my intuition was saying should happen -- that was why I originally asked if you had accounted for the non-linearity. Anyhow, it definitely seems that this setup cannot answer the original question about the limits of the sensor. You need a different grating for that.

 

Maybe Thorlabs has another definition for "groove period". Experimentally this is what I see

Yeah, it's possible. I would like to know how they arrive at their conclusion.

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I put some data that I received while trying to determine the limit of my sensor. Nikon D7100.

This is a test with water. Testing the stack RG1000_2mm + S8612_1mm.

 

1. RG1000 = 1 / 80S, ISO200, f8;

2. RG1000_2mm + S8612_1mm = 61.3 S, ISO200, f8;

3. Visible light (S8612_1mm).

Light: One 300W halogen floodlight.

 

post-242-0-82836900-1578041824.jpg

 

Filters:

1. Visible light (S8612_1mm)

2. RG1000

3. RG1000_2mm + S8612_1mm.

 

Light: One 300W halogen spotlight, + another 150W.

 

Total:

RG1000 = 1 / 20S, ISO200, f8;

RG1000_2mm + S8612_1mm = 59.7S, ISO200, f8;

 

In the frame black glass: ИКС7, ИКС3, TС3.

 

post-242-0-92753300-1578041897.jpg

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This is the result of the comparison:

On the diffraction grating 1000 lines (from Uviroptics lab.) Light: one 60W incandescent light bulb. Total darkness.

RG1000 vs RG1000_2mm + S8612_1mm

 

post-242-0-43905900-1578042703.jpg

 

Another comparison:

 

post-242-0-83717100-1578042813.jpg

 

 

Here is a comparison with different IR glasses.

All equipment is absolutely motionless fixed - any movement is excluded.

 

This result is stable, I tried several times.

I'm sure anyone can repeat this and get the same or very similar result.

 

So I don’t know what exactly the RG1000_2mm + S8612_1mm stack shows on the grating, but I'm sure, if this is a some further than 1000 nm?

 

s8612 -> RG610 -> KC19 -> RG715 -> RG9 720 -> IR80N -> RG830 ->

RG850 -> RG1000 -> Икс3 -> Икс7-> Тс3 -> Тс3+Икс7-> RG1000_2mm + S8612_1mm.

 

post-242-0-43173500-1578043891.jpg

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The last possible IR image for my camera RG1000_2mm + S8612_1.75 mm. Need exposure compensation for to see something.

This is the limit. The 2 mm S8612 shield blocks completely and sees nothing.

White balance NX-D on the left. Same + Photoshop auto levels on the right.

 

post-242-0-84801500-1578044894.jpg

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Nice photos! Water really darkens with the RG1000 + S8612 stack. Similar to Cadmium's experiment. I think that with that stack you are seeing mostly beyond 1000 nm, and maybe the beginning of the second big absorption peak of water at around 1200 nm. Surely 1200 nm is beyond the capabilites of a silicon sensor, no doubts there, but since this peaks are quite broad you can start seeing them.

Also, playing with math we discovered that a diffraction grating can not diffract light with a wavelength greater than d. This means that you should not be able to see >1000 nm in your spectrums, if you used a 1000 lines/mm grating, and the longer wavelengths are stretched faster and faster when approaching 1000 nm (you lose linearity). It seems that you can see something with the RG1000 + S8612 stack, and this means that there is something in the 900+ range.

 

Water has an attenuation length of around 0.26 cm^-1 at 1000 nm. This means that there are 0.26 attenuation lengths in one cm. So a thickness of 5 cm should transmit 27% of the light, and at 7 cm the transmission drops at 16%. Don't take this numbers as accurate, since they are only true for 1000 nm monochromatic light, and the attenuation length of water varies significantly in that region.

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Stephano - I don’t think Ninjin is seeing the 1200 peak any more than you were. What happens is that those absorption glass filters like RG1000 do not have a sharp cut off, it is fairly gradual. So you are still seeing the 975 peak because the camera gain falls off very fast past 1000nm. (The 975 peak is also broad.)
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Stephano - I don’t think Ninjin is seeing the 1200 peak any more than you were. What happens is that those absorption glass filters like RG1000 do not have a sharp cut off, it is fairly gradual. So you are still seeing the 975 peak because the camera gain falls off very fast past 1000nm. (The 975 peak is also broad.)

If the slope of the camera sensitivity is greater than that of the spectrum of light transmitted by the filter, "the camera wins" and you see mostly the 975 nm peak. I didn't know that this peak was so strong. I can not find a chart with the values for the absorption length (which is indicated with λ, don't confuse it with the wavelength), but reading some graphs I found a value of approximately 40 m^-1, so 2.5 cm. In this case, in my original experiment with the solar panel, using a 7.5 cm water thickness, I have a depth of 3λ and a transmission of only 5%.

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Are you talking about the WATER absorption? Or the FILTER absorption?

 

The water curve is here, calculated directly by me from the raw numbers.

post-94-0-02931600-1578068805.png

 

This is over the 400-1200nm range:

post-94-0-31664600-1578069002.png

 

NOTE THE UNITS ARE CM^-1

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I can not find a chart with the values for the absorption length (which is indicated with λ, don't confuse it with the wavelength), but reading some graphs I found a value of approximately 40 m^-1, so 2.5 cm.

Here I meant the water absorption

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Thanks, I got pretty close, it seems it is slightly more than 40 m^-1. At 1453 nm what is the value, something like 35 cm^-1? That's impressive, it means 1% transmission at 1.32 mm thickness. Very interesting.

 

Edit: I wrote this before reading your value

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Your absorption curves are interesting. Many years ago I saw an article in Popular Science or some such magazine about a new automotive safety device that was being developed which detected the presence of actual ice on the road from it's infrared spectrum. Must have been too difficult to make it work reliably in a real world mass production environment. Probably exorbitantly expensive too, given the technology of the time.
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Here is the stack transmittance of 1.75mm S8612 with 2mm RG1000 (neglecting reflections). It is blocked to better than OD4 over the whole range, but without knowing the sensor gain curve, we can't figure out where the peak would be, since the sensor gain has the opposite slope.

 

Diabatic graph:

post-94-0-45817500-1578080412.jpg

 

Same graph but on a semilog instead:

post-94-0-17024200-1578080569.jpg

 

Looking at this, it seems like it's probably the region in between the 976nm peak and the 1205nm peak. The absorption of the water in that range is still rather high, even though not as high as either peak separately. I do think it's more likely to be closer to the 976nm side of things.

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Schott RG1000 2mm + Schott S8612 1mm.

Keep in mind, these graphs are based on the RG1000 OD5 @ 3mm reference thickness data,

so when transposed to 2mm the actual OD is lower than the visual range OD shown on the graph due to the limited RG1000 data provided.

First graph shows Internal Transmittance (Ti), however both Ti and T graphs cross the 1E-03 (OD3) line at 1160nm.

 

post-87-0-79533600-1578089106.jpg

 

post-87-0-06287800-1578089486.jpg

 

RG1000 reference thickness is 3mm thick, with maximum data of 1E-05 (OD5).

post-87-0-43718300-1578090335.jpg

 

post-87-0-73088000-1578089533.jpg

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Andy, That is good. Nice to see your graphs also. :-)

However, your graphs are for a different stack formula (using S8612 1.75mm thick).

The graphs I show are for the stack I used in the photo I posted (using S8612 1mm thick).

My post is more related to Ninjin's #30 post on page 2 of this topic:

https://www.ultravioletphotography.com/content/index.php/topic/3617-infrared-with-a-solar-panel/page__view__findpost__p__31771

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I did some measurements on my original photo, using Paint:

I cut a rectangle on the paper region and on the water region, not too big in size but enough to do what I needed to do;

I reduced the size of the rectangles to 1 x 1, to average all the pixels;

then I picked up the color obtained in RGB values.

 

For the paper region I obtained 34, 14, 49 (R, G, B) Total sum = 97;

For the water region I obtained 11, 10, 23 (R, G, B) Total sum = 44;

 

Ratio water/paper (water transmission) = 0.454.

 

Using the formula post-284-0-20649900-1578151511.png, where t is the transmission, x the depth of the water and λ is the attenuation length, I came up with this formula (hopefully correct) to calculate λ:post-284-0-06338400-1578151647.png.

 

Using the data I have (x = 0.075 m, t = 0.454), λ is 0.095 m, or 10.54 m^-1.

 

This were the colors I extracted from my image, in 200 x 200 squares:

 

Paper

post-284-0-76222400-1578151963.png

 

Water

post-284-0-44195300-1578151991.png

 

Paper, brightness 100

post-284-0-07759200-1578152047.jpg

 

Water, brightness 100

post-284-0-27130800-1578152069.jpg

 

Note that my camera saw a bluer color through the water. This is an artifact, since in my "solar panel and water without paper" photo the color is still magenta.

 

I don't think that my measurements are reliable, since there was a bit of noise and my sensor was at the limit (clearly seen by the bluer color through the water), but at least it should give an idea.

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Stefano, was that done using the RAW? If you do it on a JPEG, JPEGs have a nonlinear curve applied to them as part of the algorithm.
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Andy, That is good. Nice to see your graphs also. :-)

However, your graphs are for a different stack formula (using S8612 1.75mm thick).

The graphs I show are for the stack I used in the photo I posted (using S8612 1mm thick).

My post is more related to Ninjin's #30 post on page 2 of this topic:

https://www.ultravio...dpost__p__31771

Oh, I thought we were talking about post 32:

https://www.ultravio...dpost__p__31773

 

I thought the objective was to see where the sensor ends. The water seems like a bad way to determine that, since you can't tell which peak you are seeing, and the filters also let us down because they do not cut sharply enough for this purpose. The diffraction grating is beyond its specs. Maybe we need a prism??

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Andy, No, sorry, I have not done any tests with RG1000 2mm + S8612 1.75mm, which would be below OD4 at 1200nm.

I guess Ninjin tested that also, but I have not.

 

post-87-0-89933900-1578157119.jpg

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Stefano, was that done using the RAW? If you do it on a JPEG, JPEGs have a nonlinear curve applied to them as part of the algorithm.

Probably my little camera can only shoot JPG (they are JPG).
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Maybe we need a prism??

Probably. If they are linear (and they surely are for small angles) they can be used. Or maybe a diffraction grating with, say, 100 lines/mm would be very linear in that range. I saw and used gratings as low as 80 lines/mm in my school laboratory.

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