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I would like to get a wide laser beam but a high quality one.

That's why I'll make it go through a high-end Canon lens.

But I would like to know if that is likely to damage my lens !

Maybe it will make holes in the anti-glare sections ?

Maybe the lens will just burn !

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    \$\begingroup\$ I would say it is probably fairly safe, laser lens assemblies are made with the same AR coatings. If all you want is a wider beam angle a single lens should be pretty good, the only reason camera lenses have so many elements is to get a flat focal plane and rectangular projection. \$\endgroup\$
    – Phil
    Jan 11, 2013 at 17:42
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    \$\begingroup\$ Is this about photography at all? \$\endgroup\$
    – mattdm
    Jan 11, 2013 at 19:45
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    \$\begingroup\$ The usual method is to bounce it off of a convex surface (specifically, a drop of mercury). It's "high-quality" enough for holograms. With any lens, you're likely to get some phase dispersion. \$\endgroup\$
    – user2719
    Jan 12, 2013 at 4:15
  • \$\begingroup\$ @mattdm : Yes it is. It is about the lighthing used to take pictures. I want to use a laser beam to light a little object. \$\endgroup\$
    – Skippy Fastol
    Jan 12, 2013 at 10:24
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    \$\begingroup\$ It's not a "witty technique"; it's the standard means by which field illumination for holograms is done. Mercury doesn't evaporate at quite the rate of, say, water or alcohol, and its surface tension creates a very stable surface (very nearly a precise oblate spheroid if the drop is large enough relative to the support). \$\endgroup\$
    – user2719
    Jan 12, 2013 at 10:40

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The question is simply one of how much power per area on the lens surface. This is called irradiance, defined as power of electromagnetic radiation per unit area which is measured in watt per square meter.

To calculate this, take total power of laser but divide by the size of the cross sectional area of the beam.

A typical lens will be designed to handle the irradiance of the sun at the very least which is about 120 watts per square meter. Of course, interior components of the lens should be able to handle much higher levels as it focuses the sunlight.

What you need to find, is whether the lens can dissipate the excess heat from the inefficiency of transmission of the light. A good lens transmits quite a bit of the light, but there is always some loss. That loss can be measured in watts, and it will be concentrated in the area of the beam.

The consider the ability for glass to conduct the heat away. You must be able to guarantee that the glass can conduct the generated heat away before the glass gets to the melting temperature. Regular glass has a thermal conductivity of 1.05 but you might want to contact Canon for the specifics of the the glass they use.

Given the power of laser, the size of the beam, and thermal conductivity of the glass, you should be able to calculate whether the lens will melt or not!

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    \$\begingroup\$ That value of 120 watts per square meter corresponds to the definition of 'sunshine', as opposed to 'stormy'. About 1350 watts per square meter hits the upper atmosphere, about 1000 (what a beautifully nice number) makes it down to the surface of the earth. \$\endgroup\$
    – Phil
    Jan 13, 2013 at 7:19
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It is unlikely to damage your lens as a result of passing through it, your sensor on the other hand might just die. The key thing when working with lasers is to avoid having it focused on anything you don't want damaged. It is possible to etch and cut glass with sufficiently powered lasers, however, unless you have the laser focused on that specific point for a long enough time it is unlikely.

If you want to just photograph a laser in action then there should be no issues.

Also! When playing with lasers, remember to wear appropriate safety gear on your eyes!

Hope that helps!

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    \$\begingroup\$ I'm pretty sure that when cutting glass with a laser, they specifically choose a laser with a wavelength that's mostly absorbed by the type of glass they're cutting. \$\endgroup\$
    – Fake Name
    Jan 12, 2013 at 3:15

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