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It is known that you can see IR with digital camera. I have personally tested with camera of phone and webcam with a RC of A/C. It is seen as mostly white light with red tint. I haven't tested with DSLR but according to WP that shouldn't work as it should be equipped with IR-cutoff filter. People surgically remove them to capture full-spectrum image. My question is how is it possible? Firstly, the CCD are set to capture RGB which doesn't include gamut to see IR. Assuming it can capture more wider spectrum of light, how the display render it? It also have limited gamut without IR (ok, displays do emit IR and UV because heat and stuff but not to be seen). Even if this, too is a false limit, why it is visible? Viewing IR through a lens or mirror doesn't make it visible. It is different from false-colour image of 'heat-signature'. It is also different from photochemical method how we take X-ray, where the image show presence and intensity of the ray, not the ray itself. We can also view UV and IR with such photochemical method where chemical gives a grayscale image depending on intensity of the light or a false-colour image with different colour for different frequency of light (none happens in normal films because AgX aren't much sensitive to IR) but how can photographically see IR and UV as visible light?

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  • $\begingroup$ "where the image show presence and intensity of the ray, [...] a grayscale image depending on intensity of the light" - to provide some context for the answer bellow; a digital color image is pretty much 3 grayscale images, one showing the intensity coming from a distribution of frequencies around the R part of the spectrum, one around the G part, and one around the B part. So what's captured is a simplified version of the input; the monitor then reproduces that simplified version back. $\endgroup$ Commented Oct 7, 2021 at 8:16
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    $\begingroup$ The silicon bandgap (1.12eV) corresponds to a wavelength of 1.1 microns. $\endgroup$
    – Jon Custer
    Commented Oct 7, 2021 at 13:09
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    $\begingroup$ Because a silicon CMOS detector is sensitive to light in the near IR region. $\endgroup$
    – Jon Custer
    Commented Oct 7, 2021 at 13:13
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    $\begingroup$ Re, "full spectrum image" FYI, Photographers sometimes use that phrase to mean one thing. Astronomers often use it to mean something different. $\endgroup$ Commented Oct 7, 2021 at 15:04
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    $\begingroup$ the camera on the phone is not transmitting light nor color: it is transmitting encoded numbers. The CCD is reacting to the em-source by registering a voltage, and this voltage is then encoded as a number, this number is processed with others and then treated as if it is one of the 3 primary colors when constructing an image matrix to present to you via the phone's display. $\endgroup$
    – Yorik
    Commented Oct 7, 2021 at 16:23

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How cameras see infrared light

A camerasensor consists of an array of photodiodes, all sensitive to, more or less, the wavelengths of visible light. That is not completely true, as some part of the infrared spectrum is detected as well. Now this does not give you any color resolution. There are different methods to implement this, but the most common is to use different colorfilters over the array. So some sub-pixels are used to capture light that passes through the green-blue filter and represent the intensity of the red color spectrum, some capture light that passes through the red-blue filter, and so on.

Now the reason you see infrared as more or less white is that those colorfilters do a poor job at filtering infrared light. So every sub-pixel picks up the infrared light, which is why the display shows the infrared source as white. The red-ish tint comes from the fact that infrared is closest to the red spectrum and hence gets absorbed slightly less by the green-blue filter than by the others.

DSLR Cameras and infrared light

DSLR cameras have an additional infrared filter over the whole camera sensor, so the filtering process is done before the conversion of light.

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  • $\begingroup$ The reddish (actually, purple) tint comes not from difference in sensitivity to IR. Instead, all subpixels are approximately the same way sensitive to IR, but contributions of R, G and B are weighted by a color matrix that converts from camera color space to linear-sRGB color space. This results in red being more pronounced than if you naively mapped RGB photosites to sRGB subpixel values. Particularly, if you try doing color conversion on a blown-out highlight, you'll get purple, while all subpixels register the same value (the maximum possible value). $\endgroup$
    – Ruslan
    Commented Oct 7, 2021 at 8:25
  • $\begingroup$ So, camera sensors work in exclusive manner (all colurs execpt). $\endgroup$ Commented Oct 7, 2021 at 8:44
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the CCD are set to capture RGB which doesn't include gamut to see IR.

That is not strictly true. The photosites of a CCD image sensor or a CMOS image sensor are sensitive to a broad range of wavelengths. Selective sensitivity to "red," "green," and "blue" bands is accomplished by covering photosites with color filters. But the transmittance spectrum of a color filter can be complex.

Don't think of a "red" filter as transmitting only red light. Think of it as blocking blue and green light. It blocks blue and green, it looks red to our eyes, but that does not necessarily mean that it blocks all infrared or ultraviolet wavelengths.

In fact, the standard R, G, and B filters used in cameras do not block much infrared, which is why color cameras always incorporate an infrared cutoff filter that covers the whole array. That "cutoff" filter is what some photographers are able to remove from their cameras when they want to do infrared photography.

Blocking ultraviolet, on the other hand, is easy. Ordinary window glass does that, and so do most of the glasses and/or plastics that are used to make modern camera lenses. If you want to do ultraviolet photography, you need special lenses made from more exotic materials.


how can I see the result ?

Doctors use X-rays to make pictures of our bones. Our eyes aren't receptive to X-rays, so how can we see those pictures?

It's because the pictures made by an X-ray machine or an infrared or ultraviolet camera are false color images. The wavelengths of light that are reflected off the paper, transmitted by the film, or emitted by your computer monitor are not the same wavelengths that were captured by the film or the image sensor. The light that you see reflected/transmitted/emitted from a false-color picture is merely a representation of the intensities of wavelengths that you would not have been able to see if you were looking directly at the subject.

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A note on UV transmission: ordinary window glass does not block UV as stated in one of the previous posts. Instead it needs a special coating, or needs to be made from different material. Google "Trucker's tan" and you'll see what I'm on about. In UV photography, uncoated lenses tend to have better UV transmission. As you get farther into the UV, say UV-C, transmission becomes worse for ordinary glass.

See https://asgs-glass.org/optical-transmission/ for optical characteristics of various glass types, or https://kolarivision.com/full_spectrum_conversion/ for a brief intro to the glass filters used in DSLR sensors. The full-spectrum "filters" are really just clear window glass.

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    $\begingroup$ According to the links I found, "trucker's tan" is asymmetric tanning caused by driving for extended periods with the driver's side window down. So I don't think it supports your assertion that ordinary window glass does not block UV. $\endgroup$
    – gandalf61
    Commented Jan 21 at 17:34
  • $\begingroup$ Not so: nejm.org/doi/full/10.1056/NEJMicm1104059. While some window glass may be coated and/or tinted to block UV, in general, it does not. $\endgroup$
    – user152237
    Commented Jan 22 at 18:26
  • $\begingroup$ Here's a link to another set of spectral characteristics of various glass types: researchgate.net/publication/… .Again, these show transmission well into the near-UV range. $\endgroup$
    – user152237
    Commented Jan 22 at 18:37

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