Fresnel lens tests

My daughter’s boy friend suggested that with a soft plastic [such as vinyl] such a light could be folded and put into a pocket.

focal lenght to diameter ratio is one thing, and important to capture as much light as possible.
there is however another reason why the 50mm focal length lens has a brighter spot. The reason is the longer focal length.
have a look at the ‘thin lens’ lens formula:
1/F=1/s+1/s’

s is the distance from the LED die to the lens, while s’ is the distance between the lens and the image of the LED on the wall (assuming you have the image clearly in focus as seems to be your case).

If you know the distance between the lens and the wall (s’), and the focal length F, you can calculate s as follows:
s=1/(1/F-1/s’) (use excel :wink: )

with s and s’, you can calculate the magnification:
M=-s’/s

Now, for the shorter focal length, you’ll get a bigger magnification that for the shorter focal length, because the lens is closer to the LED, but the distance to the wall stays the same.
If you know the LED die size and multiply it with the magnification, you get the size of the image on the wall.

Since the lens really forms an image of the LED die, reducing the size will get you a brighter image.
if you would find 2 lenses with different focal lengths, but with the same diameter to focal length ratio, the lens with the longest focal length will result in the brightests image since it captures the same amount of light, but projects it in a smaller area.

I hope this is not too technical, and helps you understand more about the things you’re seeing.
Here you can find some more fresnel lenses if anyone is interested: short focal length Fresnel lenses for sale

Best regards,
Johan

Perhaps you didn’t understand my thoughts. If I hold the two lights at different distances from the wall, so that the spots come out about the same size, then I can see the total amount of focused light in the relative brightnesses. The effect of the focal length is compensated by the different distances from the wall, so all I see is the part of the led’s output that is focused by each lens. With similar aspheric lenses, the two lights look similar, so if one is brighter in this test that means that more of the light is captured. That appears to be accounted for the the difference in the ratios of diameter to focal length, which determines (through trigonometry) the solid angle subtended. The led’s pattern must be wide enough (XM-L with dome intact) that substantial light hits that extra outer part of the lens. It also shows that, in this application, a Fresnel lens has comparable efficiency to that of the aspheric, that is the steps between the segments don’t cause much light to be lost.
I fully understand the first order optics, but in this case I have compensated for the difference in angular spread by adjusting the distances, rather than by including it in a calculation.

Hi Fritz,

If you get a chance can you take a measurement of the die image on the wall at a set distance?

For example image= 16" @ distance of 3 feet from wall.

Thanks

116” distance.
Click for full size.

Thanks