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I am looking for an answer to this question all over the net and can't find any answers.

It is clear that a strong laser beam will damage CCD while it is in use(filming), but what about if the camera is not taking pictures? Will the laser beam damage a turned off CCD?

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  • what does CCD mean?
    – ides
    Jul 27, 2013 at 1:56
  • 3
    @ides A CCD is a type of image sensor. See photo.stackexchange.com/questions/9120/…. While most digital cameras these days now use CMOS sensors (also explained in that link), CCDs are still common in technical applications.
    – mattdm
    Jul 27, 2013 at 2:04
  • Bear in mind that many cameras protect the CCD (or CMOS) sensor in some physical way when it is not active. DSLRs have a mirror which redirects the light through the viewfinder rather than onto the sensor (though some have a semi-transparent mirror like Sony SLT so in those cases not a complete protection), and many small cameras have built-in, automatic lens caps. Oct 17, 2013 at 6:11
  • @drfrogsplat They also have mechanical shutters covering the sensor even when the mirror is locked up. The shutter is the real source of protection.
    – Michael C
    Feb 15, 2017 at 21:27
  • can bullets kill people when they are sleeping? ;-)
    – szulat
    Apr 17, 2019 at 15:48

3 Answers 3

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It would likely depend on the power of the laser, but yes, the damage is done by harming the light sensitive portions of the sensor. A sufficiently powerful laser would cause damage even if the sensor was off. The only thing I'm not sure about is if it would take more power to damage it while it is off then while it is on.

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If a laser beam is strong enough to cause thermal damage (ie burn or melt the sensor, the filter in front of it, or even the shutter or aperture), it will be almost irrelevant if an image sensor (of whatever type) is powered on or not (it being powered on could worsen the thermal damage a bit by causing an already elevated temperature).

Damage from an optical overload that would in itself be thermally harmless, in a powered on state, is well known for photomultipliers but not for normal image sensors.

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Lasers are nothing more than light, so you essentially have to calculate the number of photons that are falling upon an particular pixel, which you can then translate to a number of electrons in the pixel. "Damage" only occurs when the number of electrons exceeds the well-depth of the pixel (measured in electrons).

But the damage that occurs really depends on how much over the well depth you go. If you generate 55,000 electrons in a single pixel that has a well-depth of 50,000; you're pretty alright, but if it was more like 400,000 you would start to have a problem.

Also, it depends on how long this occurs for and how often. If you do it once for a millisecond, you're fine, if you do it for an hour, you probably messed up your sensor fairly well, or at least that column in the sensor.

Shining a laser on a sensor that is not turned on will not do any damage, assuming you don't generate enough heat to wreck the sensor.

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