4
$\begingroup$

When shining a white light source through a prism, one will see the spectrum of it—so a lot of colors depending on the type of light source. However, I would also like to see the invisible part of the spectrum like ultraviolet and infrared. So I thought using a special UV and a special IR camera to see that part of the spectrum. But of course, one needs a special light source emitting the whole light spectrum: UV + visible + IR.

Is this possible? If so, what kind of light source do I need? It's for a science show. So it does not need to be scientifically accurate.

$\endgroup$
4
  • 1
    $\begingroup$ Usually, when we say, "see the spectrum," we are talking about seeing how the light is dispersed. We can do that because our eyes can distinguish fine differences in colors of visible light. Cameras that can "see" UV or IR wavelengths that are invisible to us usually do not have that ability to distinguish fine differences. Unless you pay big \$\$\$ for a "multispctral" or "full spectrum" camera, you'll just get monochrome images of the invisible parts of the spectrum. $\endgroup$ Aug 30, 2023 at 13:47
  • 2
    $\begingroup$ Thanks everyone, you helped me a lot! I also tested something different: Halogen lamp going through a prisma: When looking with a normal webcam to that spectra, you can (obviously) only see the visible part. However if you remove the IR blocking filter in the webcam, you can see a huge part of light next to the visible. It looks so cool ! $\endgroup$ Aug 30, 2023 at 14:00
  • 2
    $\begingroup$ Yes you are right. But thats fine. I used a defiltered webcam and you can actually see the invisible part of IR while looking at the light going through a prisma from a halogen bulb! $\endgroup$ Aug 30, 2023 at 14:01
  • $\begingroup$ There are already cameras that do exactly this and are widely used in image recognition. Look for Hyperspectral Imaging (HSI) $\endgroup$
    – YPOC
    Aug 31, 2023 at 13:57

2 Answers 2

12
$\begingroup$

If you want something that can be "easy to find" in a lab, you can try a xenon lamp.

enter image description here

https://mmrc.caltech.edu/Stark/Xe%20lamp%20spectra.pdf

Xenon lamps are known as good broadband white-light sources.

If you want something cheaper but less accurate:

sun v. fluo lamp

Fluorescent light bulbs also have a relatively broad spectrum although it is less continuous. Nonetheless, you can also use some filters to show this principle for a science outreach event.

$\endgroup$
3
  • 1
    $\begingroup$ Excellent answer, but it leaves me a bit concerned about the transmissitivity of the prism (strictly, of the glass out of which it is made) to different wavelengths. $\endgroup$ Aug 31, 2023 at 5:57
  • $\begingroup$ Depends on the glass of which the glass is made of. Usually you have fused silica (SiO2) which is transparent up to 180 nm. Most glasses are transparent for visible and IR but absorb strongly in the UV, and usually very expensive glasses will have improved performance at the UV, but they don't come cheap. Do you know where the prism comes from to quickly check the transmittance? $\endgroup$
    – ondas
    Aug 31, 2023 at 7:33
  • $\begingroup$ Noting obviously that that's one for OP's consideration. I thought it worth raising since a lot of demonstration prisms will be glass of uncertain specification, or even (gasp) /plastic/. $\endgroup$ Aug 31, 2023 at 8:40
10
$\begingroup$

The demonstration can be done without sophisticated apparatus.

Place a (adjustable) slit in the slide holder of a projector which a tungsten filament halogen bulb.

Place a glass prism between the slide projector and a screen and adjust the focus of the projector to get a clear spectrum of the screen. In effect an image of the slit is being focused on the screen. Slant the screen so as to spread the spectrum out.

To demonstrate that infra red is present place the blacked (use Aquadag or soot produced by a candle) bulb of a mercury in glass thermometer first in the red part and then in the invisible to the eye infra-red part of the spectrum and note the larger temperature in the part beyond the red indicating the presence of infra-red.

Ultra violet requires the room to be darkened and the placement of some fluorescent paper over the bottom half of the blue end of the visible spectrum and beyond. It will be noted that the paper fluoresces (emits visible light) beyond the end of the visible spectrum showing the presence of ultra violet light.

Failing having a the projector, use a projector bulb with two parallel slits to collimate a beam but you will need some sort of cover over the bulb to observe the ultra-violet potion of the spectrum.

$\endgroup$
1
  • 1
    $\begingroup$ Wow thats awesome ! Thanks for the great ideas ! Will need to buy some fluorescent papers. Thanks a lot! $\endgroup$ Aug 30, 2023 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.