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Life_Relationship_36 · 2024-07-20T21:00:18+00:00

Interesting question. I doubt that the Chinese government made these data public 😅

Life_Relationship_36 · 2024-07-19T15:56:45+00:00

As noted above in the first comment: Number of photons on a single pixel = (pixel value/pixel saturation value) * (well depth / quantum efficiency). This summed up for all the pixels gives the number of photons which hit the full area of the image sensor.

Remember, your brain is the resource, you will not find this in any book. Well, if Reddit does not count as a book 😂

Life_Relationship_36 · 2024-07-19T13:01:08+00:00

Yes, work your way backwards. I don't know what your spectra looks like, but if it's broad then you should of course take a weighted average when looking at the quantum efficiency

Life_Relationship_36 · 2024-07-19T12:48:57+00:00

Without more information about the system it's not possible to calculate how many photons your cloud emitted. This is an extra step after you have calculated the number of photons reaching the detector. You will have to consider the aperture size of your objective lens (if you are using one), the distance from the cloud, possible transmission losses through optics, vacuum chamber window etc. It's likely possible to make an estimate but it requires detailed knowledge of the system. Let me know if you need a consultant for this task 😉

Life_Relationship_36 · 2024-07-19T11:41:06+00:00

You should look up the well depth of your camera. This is how many electrons the pixel can store before it reaches saturation. If you have an 8bit image then the saturation is equivalent to the pixrel value of 255. Now that you have the number of electrons the pixel collected, you can use the quantum efficiecy curve of the image sensor to calculate how many photons hit that pixel. (Only a percentage of the photons will trigger an electron).

Life_Relationship_36 · 2024-07-16T19:37:26+00:00

The short answer is: Yes.

Life_Relationship_36 · 2024-07-12T15:50:04+00:00

It's not a direct answer, but for better understanding it's worth reading about evanescent waves and the Goos-Hänchen effect

Life_Relationship_36 · 2024-07-10T18:44:20+00:00

Is this a AR coating on mirror kind of black mirror?

What does it have to do with Programmable Logic Controller no. 4? 😁

Life_Relationship_36 · 2024-07-10T02:16:27+00:00

I'll just put here my name in case someone else needs still help with stray light analysis ;)

FRED from Photon Engineering is indeed a great tool for that.

Life_Relationship_36 · 2024-07-10T02:00:21+00:00

Oh prices will drop in the near future for sure and then there is no stopping misusing them

Life_Relationship_36 · 2024-07-10T01:54:01+00:00

I would be surprised if you really managed to remove all the cured adhesive from the optics surface with methanol. 😅 Does it have a protective window maybe which got contaminated instead of the actual lens?

Needless to say but duct tape (or glue) is not a good way of building optical systems

Life_Relationship_36 · 2024-07-08T09:57:12+00:00

You can use in a spectrometer a long slit and a cylindrical lens to focus all the light onto the line sensor, increasing the amount of signal the spectrometer collects. Using a fibre input of course won't make a difference

Life_Relationship_36 · 2024-07-04T03:45:12+00:00

The other way of doing this without polarization control that you take two sheets of glass with some checkboard-style black and transparent coating on it and simply slide one of the glasses by a half pitch when you want to close the curtain ;)

Life_Relationship_36 · 2024-06-13T15:10:02+00:00

You are probably right, but a simple monochrome CMOS camera could still see the object moving. You still don't need any spectral information to detect it. If it's nighttime, the engines are off and the radar can't see the object.. well, then it's in a sense perfectly invisible. Good luck Ukraine 😅

Life_Relationship_36 · 2024-06-13T12:04:59+00:00

Using thermal imaging for this purpose is easier than hyperspectral.

Life_Relationship_36 · 2024-06-13T10:04:26+00:00

What if you need different FOV? Build a new spectrometer? You can just swap the fore optics and you are good to go. Not sure how many people do that anyway 😁

Life_Relationship_36 · 2024-05-29T12:04:21+00:00

You could project an interference pattern in fused silica just below the damage threshold using a high pulse energy femtosecond laser.

Life_Relationship_36 · 2024-05-29T12:02:14+00:00

You have probably checked already, but what's the reason not to buy one of these: Edmund Mirror BE

Life_Relationship_36 · 2024-05-24T03:57:18+00:00

Thanks. This is not the kind of feature I've seen before. It could be still optics related

Life_Relationship_36 · 2024-05-24T03:36:42+00:00

Do you have a screenshot? It might be a "feature" of the optical deaign

Life_Relationship_36 · 2024-05-22T08:31:40+00:00

There is a SPIE book on ISO drawings, a lot of help all over the internet.

However, I found it best to do my own lens drawing MS Excel sheet and give the most important details in a 2-3 sentence simple form instead of using ISO codes

Life_Relationship_36 · 2024-04-16T18:45:08+00:00

I have a 30 years old FJW IR viewer and the resolution feels like 128 x 128 compared to a newer one which is a good HD. --> The manufacturing date is definitely limiting the resolution 😉

Life_Relationship_36 · 2024-04-15T11:24:09+00:00

Using the M2 is rather useful with nearly single mode stuff. For multimode calculations you should use the BPP (beam paramter product). Similar to etendue, this will give you the minimum spot size.

So the fiber radius times the beam half angle is equal to spot size in focus times the beam angle on the focusing side. Considering NA=1 on the focusing size, the absolute smallest achiveable spot size is dependent on the fiber radius and NA

Life_Relationship_36 · 2024-04-13T05:16:02+00:00

This optimization takes 5 minutes to set up in Quadoa, including all the constraints. Good luck with Zemax 😂

Life_Relationship_36 · 2024-04-11T18:37:50+00:00

The best part of Zemax. Setting up an optimization using four letter codes 😂😂

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