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**Laser diode chip aperture, lens aperture & wavelength relationships to divergence?**

From what I've been able to pull together over the last year of reading posts, and searching on google etc., is that the divergence of a laser is the product of:

1. Laser diode chip aperture (and if single mode?), the larger the laser chips aperture the lower the divergence?

2. Wavelength, i.e. 532nm, 405nm etc., I have searched for hours to find a graph which shows the relationship between wavelength and divergence but haven't been able to find one yet.

3. Clear aperture of the lens used to collimate the laser diode as well as wavelength being collimated.

Can someone please explain the relationships of the above in simple language from a big picture or top down understanding? I found some calculators and math to determine each, but haven't been able to pull it all together yet. If you look at my signature, you can find links to info as well as calculators about each of the above, but I don't have a full grasp of how the puzzle goes together yet to make the lowest divergence laser pointer possible from the available parts.

But here are my assumptions:

1. Single mode large aperture laser diode chip.

2. Shortest wavelength laser diode.

3. Largest diameter lens.

Perhaps using a large lens to reduce the divergence is simply the product of having an expanded beam? Please see this online lens divergence calculator: US-Lasers - Beam Divergence Calculator

Maybe I do understand all of the relationships well enough to buy what I need for a ultra low divergence laser pointer, if someone can confirm my above assumptions, I'd sure appreciate the help. Thanks!

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