Lens calculation problem (Optics)

  • Thread starter mhjerde
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  • #1
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I'm building a light source for photography and I'm trying to calculate the focal length for a lens that will give me a 40 degree (full width) cone beam from a LED emitter.
See attached image.
[PLAIN]http://dl.dropbox.com/u/2301/LEDOpticDrawing.jpg [Broken]

The emitter itself measures about 0.5 x 0.5 mm and sits inside a 5.5 mm x 5.5 mm package. The plan is to mount plano convex lenses on top of an array of these emitters. I'm out of my element here and clearly in need of help :-). This is what I've tried...

First I tried to calculate using angle of view:
Edim is the emitter size: 0.5 mm
f is focal length
Beam angle = 2 * arctan(Edim/(2 * f) * rad
Solving for f, this gives a focal length of 0.7 mm. The result looks unreasonable. If I try to calculate what such a lens would look like using the formula 1/f = (n – 1)/R, I get a 9 mm thick 5.5 mm dia lens.


Next I tried to get to focal length via numerical aperture:
D is lens diameter: 5.5 mm
f is focal length
n is refractive index: 1.585 (polycarbonate)
Beam angle = arcsine( (2 * (D/f)) / n) * rad
Solving for f, this gives a focal length of 10.8 mm. Can this be correct? Frankly, it seems larger than I expected.
Does the NA formula assume that the emitter is positioned at the focus point? How can I take the distance from the emitter to the lens into the formula? I will need to mount the lens very close to the emitter, ideally 1.4 mm from the actual emitter surface.

M
 
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Answers and Replies

  • #2
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I think I made a silly mistake. Looks like the right answer to the above is a focal length of 1.4 mm.
 
  • #3
Andy Resnick
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I guess I'm a little confused by your question- what is the cone angle illumination from the emitter without your lens on top- most LEDs come with a lens already mounted to the chip? And how even does your illumination need to be? Can you simply get some LED lights and stick a diffuser plate on them (or have it bounce off a diffuser plate)? If it's an array of LEDs, can't you simply mount them on a curved surface to generate a spreading beam?
 
  • #4
sophiecentaur
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Are LEDs bright enough? They are pretty disappointing in the home for reading, even.
 
  • #5
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The emitter I plan to use has a 'viewing angle' of 120 degrees. The viewing angle is defined as 2[tex]\theta[/tex]1/2, or 50% power angle. This is basically a lambertian emitter.

120 degrees is too wide. I want to narrow it into a 40 degree viewing angle. Light outside the camera field of view is lost in the sense that it is not contributing the the illumination of the subject or scene. 40 degrees is a working theory, it may be that it is better to have a slightly wider light.
(Ideally, I would want to be able to adjust the angle.)

I'll need to experiment a bit, but I wanted to have a sensible starting point. Buying small lenses for testing seems very expensive. I have not found a source for small inexpensive plastic lenses.

BTW, there are several reasons why I'm using this particular type of emitter instead of using the typical narrow beam 3 or 5 mm round LEDs. These are RGB emitters so I'm able to adjust the color temperature or the emitted light. If I don't use a multi-die emitter, I get problems with the color mixing. This type of emitters also have lower thermal resistance so they can be driven at higher power and brightness.
 
  • #6
Andy Resnick
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Most likely, you can get a large Fresnel lens (like 5" x 7" or even larger) and play with the spacing between the lens and LEDs until you are happy with the result.

http://www.andrewdavidson.com/articles/led/

By holding the Fresnel lens close to the LED, you will "tighten" the beam divergence rather than focus it.
 

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