Optic theory

[AilSnail]

Doug, I like it too - Ra's last post has me confused though.

The green filtered picture illustrates how lux at 1m gets skewed in favor of small reflectors.

Ra, I'm not sure I am following you. You seem to be opposing your earlier post. A short focal length will not compete with a long focal length lens when it comes to illuminate something far away, especially if f/# is kept constant.

I'm going to think a bit more on what you are saying.

 

[AilSnail]

This one illustrates a long and a short fl, with the same diameter.
It shows that the light is spread over a larger area with the shorter fl, like we concluded earlier. It also illustrates that in a perfect lens, the only light beams which are going straight ahead, are the ones coming from the direction of the infinately small focal point.

Now, I happen to have a 21.5mm aperture x 6.9mm bfl aspheric, and another lens with a focal like 50mm or so, which I masked to to get the apertures the same.
The latter is much better at lighting up distant targets.

So if I understand correctly what you are saying, Ra, either your theory is flawed, or my equipment, method or eyes skews my observation in favor of the longer fl.

 

[Ra]

That is what I meant with "This will be the hard part of this post !"

On these forums there is a lot of confusion and mis understanding about what determines the throw of a light !

But there is hope: Take a good look at the last picture AilSnail posted:

Now concentrate only on the green beams: That beam, comming out of the lens, towards the object is exactly the same with both lenses !!

That means that the center of each beam has exactly the same amount of light (lux at 1m), This also means: THE THROW OF THE CENTER BEAM IS THE SAME WITH BOTH LENSES !! And that is that I've been trying to tell you..

For the entire beam, I hope the earlier posts were clear enough about that: If the lens "sees" a larger surface, it projects a larger surface, but if the lens diameter stayes the same, the surface brightness at the object stayes the same!

So, back to the pic AilSnail posted: The lens with the short focus projects a bigger spot at a sertain distance compared to the lens with the longer focal length. But in the middle of both spots the lux-reading will be the same, so they'll have the same throw.

Now what happens is: The lens with the short focal length grabes much more lumens from the emitter, thats why the beam is wider. Now if you illuminate an object with this wider beam objects in the neighborhood light up as well by the wider beam, That is the main reason the narrow beam seems to throw further: You have a clearer wiew to the object, not disturbed by illuminated objects nearby !!

AilSnail,, Try to do the same experiment, but now measuring the lux at a distance of about 10 metres, at the center of each beam. The brightness of a illuminated object is directly related to the amount of lux it receives, so the amount of lux you measure at a distance is directly related to throw !!

Hope this helps you to understand..


Regards,

Ra.

 

[AilSnail]

Thanks for explaining more. I don't have a lux meter, but maybe I can set up something with a photoresistor.

Meanwhile I need to think some more about whether your theory holds up, in theory.

What a cliffhanger.

Imagine that you put a reflector behind a star, a reflector so immensely large that the star is almost not measurable in comparison - I'm talking about a sincerely huge construction, almost half the universe. The reflector is parabolic, and focused towards the earth. That was just for fun, I'll get back to the theory shortly.

 

[Ra]

The green filtered picture illustrates how lux at 1m gets skewed in favor of small reflectors.

Sorry, but no.. It does not !! the size of the reflector isn't a issue here, only the difference in surface brightness is: If I would put a halogen bulb in the smaller reflector the image would look like this:

The other thing you can see is that the surface brightness reflected by the reflector is mostly of the same intensity, no matter the source-reflector surface distance: The outer rim of the reflector is further away from the filament than the central part, yes ?? But it has the same intensity !! So no matter how small or big or deep or shallow your reflector is,, if its parabolic (with the source in focus..) it would look the same when you look into it from a distance, evenly lit over its entire surface !

The HID-arc has a much higher surface brightness than halogen: Thats why the picture with the small HID-torch looks like this: Not because of the smaller reflector!!:

Regards,

Ra.

 

[AilSnail]

The HID-arc has a much higher surface brightness than halogen: Thats why the picture with the small HID-torch looks like this: Not because of the smaller reflector!!:

But the smaller one with the short arc reads a higher lux at 1m than the large halogen doesn't it? I guess it would depend on which sort of measurement equipment you use?

 

[Doug S]

Here is the same data presented differently; It shows how much of the light of the XR-E that will hit a lens of a given F/#. The F/# is the same as focal length divided by diameter.

A very useful graph! Optics is a bit far from my formal training but I can offer one practical comment on using this graph. It becomes very difficult to design lenses with very low f numbers because at some point the rays from the source will strike the lens material at less than the critical angle and thus never enter the lens.

 

[AilSnail]

Nice of you to point that out, we should delve into that a bit also.

 

[Ra]

Every time I try to quote someone the link to CPF crashes!!!

So I'll try it this way:

QUOTE: (AilSnail But the smaller one with the short arc reads a higher lux at 1m than the large halogen doesn't it? I guess it would depend on which sort of measurement equipment you use? END QUOTE.


No, it doesn't !! That is, not if the entire reflector of the halogen torch is lit by the filament!

The amount of lux you measure comming from a torch is a combination of the surface brightness and the dimensions of that surface. So with a lower surface brightness (halogen) you need a bigger reflector to get the same lux-reading at a distance, than with higher surface brightness (HID).

Every lightsource has its typical surface brightness (within a sertain range)

You can overdrive a halogen bulb to get a higher surface brightness, but physics of nature determine the limits: Going too far will melt the filament!
With halogen you'll never reach the surface brightness of HID!!

As a result: With a halogen torch with a 4 inch reflector, you'll never reach the throw of a HID-torch with the same reflector diameter!

AND: You'll never see military searchlights with a CCFL-tube as a lightsource !!


Regards,

Ra.

 

[Ra]

And...

QUOTE (Doug_S): A very useful graph! Optics is a bit far from my formal training but I can offer one practical comment on using this graph. It becomes very difficult to design lenses with very low f numbers because at some point the rays from the source will strike the lens material at less than the critical angle and thus never enter the lens. END QUOTE.

You are absolutely right Doug..

With short focal lenghts the lens absolutely needs to be aspheric, and even then there indeed is a limit of angle of the incomming rays at the edge of the lens.

F/0.7 is about the shortest possible focal length of a single aspheric lens!

A lens is the best way to grab light from an emitter that emits at the front.
A conventional reflector is the best way to grab light from emitters that emit to the side and towards the back.

The acrylic optics for use with emitters like Cree and Luxeon are a combination of the workings of a conventional reflector with a lens: They grab almost 95% of the lumens output of the emitter: more than ever is possible with only a reflector or a single lens!


Regards,

ra.

 

[AilSnail]

I have a photoresistor provided by doug owen a few years ago, that thread is probably here somewhere. It decreases resistance when light hits. It is probably about 5mm dia.
I hung it ~2.5m from where I sat with the lenses. I also got a filter from Doug, which was supposed to stop 50% of the light, with some rather good accuracy IIRC. Was a bit scratched though. I used a white plastic tube of 22.5mm ID and 30mm long, to "shutter" the 27mm on the front side, and the 38mm on the back (flat) side - this was the way they could be easily held, in front of the luxeon3. So effectively three lenses of different fl, two with the same D. Here is a link to the 27mm. The focal lengths of the two larger lenses are not accurately measured. The 27mm is brightest a bit further away from the LED than the fl, which I also wrote about here. I suspect the lens is the same as Newbie has, which I mentioned in the first post, and can be seen with many nice pics here, here and here. Here are the resistor readings.

27mm - F/0.31 - 22.5D x 6.9bfl - 1.77K- filter 2.55K
65mm - F/0.49 - 65.0D x 35bfl - 0.32K - filter 0.5K
Occ38 - F/2.44 - 22.5D x 55bfl - 0.81K - filter 1.25K
Ambient: 1.3Meg min.

 

[AilSnail]

Without mapping the resistor, and making corrections to get a linear scale, the most I will draw from this is that there is a high probablility that the 22.5x55 is more than twice as bright as the 22.5x6.9, and that the 65x35 is more than twice as bright as the 22.5x55.

Ra, these results are not supporting the theory you presented. Perhaps you could help me understand why?

What were the lux readings of the two lights in the first green picture, at one meter?

 

[McGizmo]

If you have a deep reflector, you collect more lumens from the source, that results in a wider beam (more sidespill) But you'll have the same throw with a shallow reflector, but because you send less lumens towards the object, the beam will be tighter (more laserlike if visible)

So, bottom line: With a lens diameter of 50mm and a focal length of 1 metre you'll have exactly the same throw as with a 50mm lens with a focal length of 30mm !! Only the beam of the first will be useless because of the pathetic amount of lumens its made of..

Hi Ra,

By side spill, are you refering to the divergence of the beam and the fact that it is no truly colimated? Since our image or source is not infinitely small, most of our image surface is not in focus, right? I have come to think that the greater the focal length of the reflector (parabolic) the better the collimation or less divergence one gets.

I consider spill to be that portion of the beam that is beyond the spot and is composed of light exiting the flashlight with no contact with the reflector along with light that is hitting the reflector in such a manner that it does not assume a collimated path. This latter would be from bumps on the surface of the reflector not conforming to the parabolic surface as well as from light originating well away from the focal point.

If we take the same LED driven at the same current and put it behind two 2 25 mm OD reflectors, one being very deep and of a short focal length and one being very shallow and of a long focal length, it has been my experience that the two will NOT have the same lux measure or throw. If I follow you, you are saying they will? The short focal length gathers more lumens but has more divergence and will have a larger spot. The long focal length gathers less lumens but has tighter collimation and will have a smaller spot but measuring the same lux as the short focal length?

Are your comments independent of source (image) size and independent of deviation from specular surface (orange peel)?

 

[AilSnail]

deleted

 

[AilSnail]

Luxeon3 added to the same old graphs. Intensity graph stolen from lumiled datasheet. I should like to know how to change the colour of the curves in Oo calc, and how to lay the intensity graphs of lux3 and xr-e over each other.

 

[AilSnail]

The yellow lines converge at the reflector axis. The focal point of the parabolas are at the bottom of the filament.

 

[greenlight]

I read this thread with my inova X1 in mind. It makes a good beam with its 5mm led, and the aspherical lens design seems very well thought out.

 

[Ra]

AilSnail: Try to figure out if the short focal lens (F/0.31) is fully lit over the entire 22.5mm diameter surface: That is very difficult with focal lengths that short !!

The best way to do this: Take a heavy filter to protect your eyes, make a tiny hole of about 3mm diameter in piece of paper and place the hole exactly on the place of the photoresistor, look through the hole during operation of the lightsource: The entire 22.5mm diametre must be evenly lit by the source! If not, the lens needs more collimating, till it does!

It is best to do this for each lens before you take a measurement !!!

The best and most reliable way to diaphragm lenses is with a piece of paper with a circular hole at the center ! Take three papers together, fold them once and cut half a circle out of the fold: you'll have three identical holes, one for each lens (they don't need to be exactly round, as long as they are identical! Ofcource the other option is to use only one paper with hole for the three measurements !

Only following the instructions above will provide honest measurements !!

Regards,

Ra.

 

[Ra]

And McGizmo:

You are right: Certainly a deep reflector (short focal length) has a very tiny focal spot! The part of the emitter that is not in that focal point causes sidespill !

Take a look at the drawings of post #22.. They exacty show what is happening: with a shorter focal length the source is closer to the lens so the outcomming beam will be wider. But at the center of the beam the surface brightness of the source is the same, no matter how long the focal length !

So with the longer focal length the beam will be narrow, but the surface brightness at the center of that beam will be the same !

Example: Take a piece of white paper, put some light on it sideways so that it is evenly lit. Now take a magnifying-glass or lens and magnify the surface of the paper.. You''ll see the details on the paper magnified, yes.. BUT, do you see more light through the magnifying glass??? NO !! If you magnify the piece of paper, the surface brightness is not magnified !! And that is what this is all about.. But now you have a piece of paper 1 square mm small, put that on a black surface: looking at it with the magnifying-glass makes it bigger! AND.. It seems brighter, but thats only because the surface seems bigger. Now look through the magnifying glass from a distance of about 1 metre.

Now try to move the lens till the little piece of paper seems to occupy the entire surface of te lens.. Thats what a torch-lens or -reflector is supposed to do: magnify the surface of a light-source. But like you (hopefully) have seen before: surface brightness is not magnified !!

Conclusion: A point at infinity "sees" a surface with sertain dimensions, and a sertain surface brightness, only that will determine the amount of lux (throw) it receives.

So if we have perfect optics: (aspheric lens or parabolic and mirror-finished!! reflector):

With the same source: Bigger reflector or lens means more throw.. Same diameter reflector or lens, no matter the deepness or focal length means the same throw.
Higher surface brightness with the same reflector or lens means more throw.


Regards,

Ra.

 

[Doug S]

The best and most reliable way to diaphragm lenses is with a piece of paper with a circular hole at the center ! Take three papers together, fold them once and cut half a circle out of the fold: you'll have three identical holes, one for each lens (they don't need to be exactly round, as long as they are identical!
Regards,

Ra.

Actually, splitting hairs here, only the outer layer sheet will have a round hole. The other two will have holes with the same major axis but the minor axes (for you non-native English speakers, axes is the plural of axis, one of the many goofy English plural constructions that must drive you folks crazy) will differ by 1X and 2X the paper thickness. I like your second option better:
Of cource the other option is to use only one paper with hole for the three measurements !

If you magnify the piece of paper, the surface brightness is not magnified !!Ra.

In fact, if I am thinking correctly, the apparent surface brightness is decreased by the factor of the area magnification, that is magnification squared.