3 Answers
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The answer isn’t as simple as the previous two posts suggests. If you are using a laser in the visible part of the spectrum, then at low power – think laser pointers – you will get a very bright spot in the image. At powers up to 100x or so more than a laser pointer, you get saturation of the sensor – the whole image will be white. This is due to the creation of more photo-electrons in the CCD than the circuitry can process so the charges bleed over into adjacent detectors. Saturation goes away after the laser is turned off. Physical damage doesn’t happen until you get into kilowatt class lasers (about 10^6 time more powerful than a laser pointer). But now you are looking at very big laser systems with big power demands and thermal control systems.
If you are using an IR laser, like Nd:YAG, then things are different. IR lasers have photon energies that are lower than the the energy required to generate photo-electrons in silicon (CCDs are silicon detectors, as are CMOS devices). Also, the optics of the camera, which are designed to focus visible band light, probably don’t focus the IR light to the image plane where the CCDs are located. The net result is you may warm up the CCDs but unless you are looking at weapons grade lasers, you won’t hurt or disturb the camera.
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You could ruin the camera. Why be so silly?
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A low enough power one would just be a bright light.
A higher power well-focussed one would burn the sensor; no great difference to the effect on someone eye.
(And remember that doing it to anyone else’s camera would be criminal damage; wrecking your own gear your choice).