Optical Power Standard

[Jubathoph]

I have often wondered this: Why don't we measure the "optical efficiency" for lenses, i.e. a "lens power derate" which would be measured by (optical output with lens/ optical output no lens)?

It would make power comparisons between lenses very simple. Right now, all threads I see are on "AAA lens is X% more power than BBB lens, and BBB lens is Y% less power than CCC lens." It is very confusing. Especially because there is no baseline information on actually how efficient any single lens is. We are making relative, not absolute, comparisons. (Please direct me to this if I am wrong, there is a lot of information on this forum.)

I could see it possibly being a challenge to measure raw diode output, due to the divergence of the diode as well as heat radiation. (Well put by RHD in this post: http://laserpointerforums.com/f54/calculating-efficiency-questions-65642.html#post942835)

However, I think the heat could be treated as a "systematic error" and eliminated if the experiment is conducted properly.

Any thoughts??

Regards,

-J

 

[Cyparagon]

Excellent idea.

I could see it possibly being a challenge to measure raw diode output, due to the divergence of the diode as well as heat radiation.

No need to measure it right after the diode.

 

[Jubathoph]

Awesome!

In other words:

1). Colimate the diode
2). Measure the output
3). Pass the collmated beam through the test lens
4). Measure the new output

Efficiency = step 4/ step 2

Why aren't we doing this again? Guess I need a LPM.
The only problem I see with this is the beam is diverging after step 4 so it will need to be close to the LPM. some of light will be converted to heat, and due to the close proximity to the sensor the heat will artificially register as optical power. If we introduce 2 test lenses we can get it back to a collimated beam and keep a good distance from the LPM. Then, the efficiency will be (Y/X)^1/2

What do you think?

-J

 

[Cyparagon]

some of light will be converted to heat

The heat will first go to heating the lens, and only after it is sufficiently above room temperature, will itself start to radiate enough to effect the readings. It probably won't get that high because of convective losses. Even if it did... there are too many other things in the room that give off heat for it to be your primary concern.

Unless you're sending enough power through the lens to melt it, I wouldn't worry.

 

[Jubathoph]

So mainly its the heat from the diode/driver/battery assembly that is the biggest concern. Got it.

 

[Bluefan]

The heat irrandiance falls of with a square of the distance, so only when you keep a diode very close to a thermal power meter you catch a lot of it.

To measure the absolute efficiency of a lens I think it's best to take a stabilised collimated laser and pass it through 2 of the lenses under test, one to focus it and one that collimates it again, like Jubathoph posted. The heat radiation of the diode is far away AND absorbed by the lenses.
But because the beam before the two lenses and after the two lenses is collimated a photodiode based power meter can be used which are usually more precise (and insensitive to heat radiation). Because the beam passes two lenses the total efficiency measured are the single lens efficiency squared.

I'm at my parents at the moment or I'd build the setup.