1
\$\begingroup\$

This is a weird set up but I'm trying to figure out at what angle of deflection a pendulum has based on using a laser point reflecting off an attached mirror. I've worked out the math but I'll need to know the height from the laser origin the beam is deflected at in order to figure out all the relevant angles.

This beam cannot be directly observed, however it is possible to observe after a few seconds so I need a way to 'preserve' the last position of the laser pointer for a few seconds.

Here is my current idea:

Take a glow in the dark paint and shine a UV light on it to 'charge' it and make it glow. Then use a 5mW red laser pointer in order to cause the point where the laser hits to lose it's charge so it looks like a dark dot surrounded by glowing green material.

Is there any alternative ways to doing what I'm trying to do? I know that by using a 405nm blue laser it will instead cause the point to glow so I can just make one specific point glow - but I do not have a blue laser pointer on hand so that's why I'm doing it in reverse with this red one.

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Aside from my answer, another way would be to use a laser powerful enough to burn a hole in a piece of paper, and just use paper as the "sensor material". \$\endgroup\$
    – The Photon
    Commented Nov 6, 2015 at 17:36
  • \$\begingroup\$ Have you tried looking at a piece of white paper through a digital camera? Of course this depends on the wavelength of your IR laser. Please remember that even a low power IR source can damage eyesight. Some classifications of visible light lasers are called eye safe because eyelids shut (blink). This isn't true for invisible IR. \$\endgroup\$
    – gbulmer
    Commented Nov 6, 2015 at 18:31
  • \$\begingroup\$ I am kind of confused. Is it a red laser or IR? Everyone is saying "IR" but the OP said red, I thought. Anyway, CMOS image sensors (digital cameras) are VERY sensitive to IR light. But this makes things look funny, so they usually employ IR blocking filters in the optics. If you remove the IR blocking filter, they are fantastic IR detectors. \$\endgroup\$
    – user57037
    Commented Nov 6, 2015 at 18:42
  • \$\begingroup\$ How about temperature sensitive (thermochromic) paint or pigment? The laser probably transfers some heat to the painted surface, and it might cause the paint to change colors, especially near the apex of the pendulum, where speed is slowest. This is just speculative. I have certainly not done such a thing. \$\endgroup\$
    – user57037
    Commented Nov 6, 2015 at 18:48

1 Answer 1

Reset to default
1
\$\begingroup\$

The usual way to detect an IR beam is to use an "laser viewing card" or "IR sensor card".

These are sheets of paper or plastic coated with a phosphor material that can be "charged" by multiple IR photons (I don't know if this works through a 2-photon process or by two charging steps through an intermediate state), and then emit a visible photon. They're sold by the major optical components distributors like ThorLabs, Newport, etc. The size of the sensor region is typically between 25 and 75 mm square. If you're working in an organization with some IR lasers already in use, look around and there should already be a few of these available.

I'm not sure if these will maintain a glow for "a few seconds", but if you observe them with a sensitive camera in a dark room, they may work.

\$\endgroup\$
3
  • \$\begingroup\$ An interesting idea but this is more for home experimentation and I'm mostly limited on budget which is why my focus is on cheap materials and the papers run $80. But thank you for the suggestion. \$\endgroup\$
    – FrankerZ
    Commented Nov 6, 2015 at 17:48
  • 1
    \$\begingroup\$ @FrankerZ, the more you tell us about your requirements in your question (like you have a budget of $XX) the better will be the answers you get. \$\endgroup\$
    – The Photon
    Commented Nov 6, 2015 at 19:02
  • \$\begingroup\$ cheaper alternative: "IR test card" from MCM electronics. marketed as a way to test consumer tv remote control, worth a look. \$\endgroup\$
    – MarkU
    Commented Nov 6, 2015 at 20:25

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.