1
$\begingroup$

I would like to display an image onto a screen using a digital light projector and somehow record the amount of light the screen is receiving from each pixel.

I have thought about using a visible light filter over an array of very small photovolatic cells and recording the voltage output by each cell, but the logistics of the wiring would be immense, not to mention I can't seem to find cells small enough.

I have searched for 'UV screens' and many other terms but I can't seem to find anything. There are many results for UV diodes but they are too large to get the resolution I need.

Does anyone know if such screens exist?


Let me explain a common problem where this is an issue:

This frequently occurs with 3D printers that use ultraviolet-cured resin: enter image description here

The theory is: projector displays black&white (binary, not grayscale; uniform regions of 100% brightness and 0% brightness) image onto the bottom of a transparent container of the resin, curing one "layer" of the item, then the layer is lifted (using a mechanical actuator), the space fills with more uncured resin, and next layer is displayed and cured.

In reality the projector's light is non-uniform - some areas brighter, other dimmer. This depends on a lot of hard to determine factors, like light source construction, lens alignment errors etc. As effect, rhe resin will cure to a different degree in different areas of the image.

Independently, the projector is capable of displaying full grayscale (or even color) images (for no other reason that that's what's cheapest due to demand and economy of scale).

Therefore, if you know the brightness layout of the projected image, you can compensate in software - adjust the projected image of item layer, by dimming areas that were brightest; display a grayscale image instead of 1bpp, that results in actual uniform exposure of the "screen" surface, and uniform curing of the resin.

But in order to generate these adjustments, you must obtain the individual map of brightness of the projected image, as generated with this specific projector. With visible light, that wouldn't be very hard, just take a photo of the screen displaying a white sheet using a normal camera. This being ultraviolet though makes it quite tricky.

$\endgroup$
5
  • $\begingroup$ If you are projecting a digital image then the image file itself already contains the information about the alternative brightness of each pixel, in fact that is what a digital image is. $\endgroup$ Commented Jan 17, 2016 at 12:16
  • 1
    $\begingroup$ The light if from a digital light projector which has several hotspots, all different each time you turn it on. I need to test the screen under a thin layer of liquid to see which areas of the liquid is receiving too much light, and which is getting to little. The refraction and diffusion through the liquid also means that images with a greater surface areas are receiving more combined light than areas with smaller areas. Come to think of it, I am wondering if a simple digital camera will work with a a visible light filter applied to it. I could just read the opacity values of the image files. $\endgroup$
    – user88720
    Commented Jan 17, 2016 at 12:28
  • $\begingroup$ What does "UV" mean in this context? Ultraviolet, or something else? $\endgroup$
    – Dave Tweed
    Commented Jan 17, 2016 at 14:51
  • $\begingroup$ I would use a fluorescing surface (liquid or sheet) and record the image with a high resolution machine vision camera (or some other gain characterised camera). If you need a bit more resolution you could pan the camera and take 4 or 9 overlapping images of the projected area. $\endgroup$
    – KalleMP
    Commented Feb 2, 2017 at 8:21
  • $\begingroup$ You should be able to get 'static' measurements using B&W photo paper. It will get exposed to UV light just like to regular. $\endgroup$
    – SF.
    Commented Nov 26, 2021 at 12:38

2 Answers 2

1
$\begingroup$

If I am reading your question correctly, you want to calibrate the output of a UV emitter, by measuring the intensity of each part of the display.

However - why do you need an array of pixels to do the calibration? Can you not use a single UV sensor, and scan that slowly over your output? Or go part-way and use a 1D array and take the image as a strip? https://en.wikipedia.org/wiki/Strip_photography

Pixel-by-pixel scanning will take longer, but might be more reliable as you don't have to worry about non-uniformities in your detectors!

Something similar is used in microscopy (but in reverse), as a "gain reference" image for the camera, which helps to mitigate the different sensitivities of the CCD (such as due to microchannel preamplification non-uniformities)

$\endgroup$
-2
$\begingroup$

This is a bit like a perpetual motion machine. You have a digital image. You project that image onto a screen. Then you want to capture that image from the screen to make a digital image.

The obvious answer is that this process is silly. Just use your original digital image.

$\endgroup$
3
  • 1
    $\begingroup$ The light is from a projector which has several hotspots, all different each time you turn it on. I need to test the screen under a thin layer of liquid to see which areas of the liquid is receiving too much light, and which is getting to little. The refraction and diffusion through the liquid also means that images with a greater surface areas are receiving more combined light than areas with smaller areas. So I need to display an image and then record it so I can calibrate the projector by applying a mask. $\endgroup$
    – user88720
    Commented Jan 17, 2016 at 14:10
  • $\begingroup$ @user: This is totally different from what you asked about. I answered the question you actually asked. $\endgroup$ Commented Jan 17, 2016 at 14:12
  • 3
    $\begingroup$ I'm sorry Olin but I was asking for advice with possible solutions for a UV screen, and you said the process was 'silly', which was not a useful answer, hence why it wasn't accepted. What is sent to the projector, and the amount of light an individual area on the screen receives are two different things. $\endgroup$
    – user88720
    Commented Jan 17, 2016 at 14:34

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.