Flashlight Optics - Dome, Dedoming and Throw

I wanted to write a longish text about flashlight optics for quite a while now, starting with all the optics and photometry basics, up to the details of what throw depends on and why, including how dedoming increases throw, all with a coherent line of argumentation and some experiments to try. I even started writing that document, but it seems I just don't find the time to finish it in reasonable time.

I now decided to take a shortcut and just do a very short version of it, especially focusing to the dome chapter because dedoming is a relevant topic currently. Unfortunately that means I can't 'prove' every detail, but I'll try to be concise and conclusive anyway.


I'll for now consider ideal optics here most time, e.g. perfectly shaped reflectors with 100% reflectivity or lenses with 100% transmission etc.

I have to introduce some physical quantities (with quite some simplifications):

Solid angle W [sr]: It's symbol is actually Omega, but I'll resort to "W" here. It's sort of the beam divergence; a big solid angle means a wide beam, a small solid angle a thin beam. It's unit is steradian (sr). Its sort of a two-dimensional angle.
Luminous flux F [lumen]: the amout of light emitted, measured in lumen.
Luminous intensity I [cd]: For throw, this is what you want your flashlight to have. It's the flux per solid angle: I=F/W. It's often improperly referred to as "lux@1m", but the correct unit is candela (cd). Even the flashlight manufacturers got it right meanwhile (ANSI/NEMA FL1).
Illuminance E [lx]: Tells you how brightly your flashlight illuminates the target. It's flux per area: E=F/A, with lumen/m² = lux (lx) as unit. Illuminance on a target and intensity are linked by the inverse-square-law E=I/d², d being the distance to the target.
Throw d [m]: According to ANSI/NEMA this is the distance at which the illuminance onto a target falls below a threshold of 0.25 lx, which is about the full moon's illuminance. Can be directly calculated using the inverse-square law E=I/d² => d=sqrt(I/E0) with E0=0.25lx.
Luminance L [cd/m²]: This is the most important ingredient to understand throw. It's often referred to as the surface brightness; it's the emitted intensity I per emitter area A, L=I/A, candela per mm², some kind of intensity density :) I's also Flux per solid angle and area: L=F/AW. It has a couple of counter-intuitive properties covered below. It's also the brightness seen by the eye.

Again we'll for now simplify things and consider a lambertian emitter, i.e. a completely diffuse emitter. A lambertian emitter has a certain beam profile (cosine law for emitted intensity) with a beam angle of 120° (FWHM), so you see most LEDs actually are approximately lambertian emitters; the dies themselves as well as the LED with dome (with the exception of the XR-E for example).

With a lambertian (diffuse) extensive emitter, the luminance does not change with the distance: A white wall has the same brightness regardless of your distance to it (as long as you still see it as an area).
With a lambertian (diffuse) extensive emitter, the luminance does not change with the viewing angle: A white wall has the same brightness regardless at what angle you look at it (as long as it's <90°). The same holds for the LED (a very small wall). The intensity I decreases to the side of course (cosine law), but not because of a change in luminance L, but because of a reduced apparent emitter area seen under an angle.
A lens or reflector will not change the luminance of the beam! A lens for example may collimate the beam and thus produce a thinner beam (smaller solid angle W) but it will also show a magnified virtual image of the die, i.e. a bigger apparent emitter area A. The product A*W (called etendue) stays the same, and so does L=F/(A*W) (actually that's what my avatar means). The beam behaves as if it came from a bigger emitter, but only under a smaller angle - but with the same perceived brightness (in terms of luminance).

A flashlight's aspheric lens works just that way: It increases the effective emitter area A at the expense of a smaller solid angle. If you look at the die through the lens, it will look equally bright (same luminance L), with a much bigger area A that actually equals the lens area, but if you tilt the light a bit, you won't see the die any more, thus the emission angle (solid angle W) is much smaller than without the lens.
However for throw we want luminous intensity I, which in turn depends on two factors: according to L=I/A => I=L*A the forward intensity is simply the emitter luminance L multiplied with the lens area A.
It's basically the same with a reflector: If you hold a light (off) at arm's length and look right into it, you'll see that almost the complete reflector looks yellow (except for the unused center, the 'dead hole'), that's actually the reflected die (with a big reflector you might need a bigger distance to make the whole reflector look yellow), and all that yellow area is apparent emitter area A for the equation I=L*A, and it all has the same luminance L as the die itself (well, if we still neglect losses, that is).
Taking losses into account, even a good aluminium reflector absorbs about 15% of the light, so the luminance L is reduced at least by 15%. An XM-L's luminance L at 3A should be around 30 cd/mm²; most reflector based lights have an effective luminance of around 21-24 cd/mm² though (which is simply the measured intensity I divided by reflector area A=pi/4*D²*0.9, where the factor of 0.9 accounts for the dead hole), which means an typical effective luminance drop of 20%-30% compared to the LED.
For throw that means d=sqrt(I/E0)=sqrt(L*A/E0)=sqrt(L*pi*0.9 / 4*E0)*D = ~ 8200 * D, which means that a typical XM-L reflector based flashlight has a NEMA throw of about 8200 times it's reflector diameter D. And actually that's what I typically find with well driven XM-L reflector flashlights.

Why am I telling you all of this? Just to make two points clear:
For throw you want to have a emitter with high luminance!
You can't increase luminance with normal optics (like lenses)!

However there's one trick to increase the luminance anyway, and it is based on the fact that the emitter itself is diffuse.


The LED die consists of a GaN layer which produces blue light. On top of it is a thin Ce:YAG phosphor layer (in the magnitude of 100µm) which converts a part of the blue light into yellow light by fluorescence (actually a very 'broad' yellow including green and red). Mixed together this gives white light. A thicker phosphor layer gives a more yellowish light (i.e. a lower color temperature), while a thinner layer gives cool white. Since the phosphor layer also adds some absorption losses, cool white LEDs can be produced with higher luminous flux [lumens].

What the dome does
The GaN die has a refractive index of about 2.5, which makes getting the blue photons out quite difficult: Light from the die hitting the surface perpendicularly has a good chance to get out, however there's always some fraction of the light reflected. With an increasing angle to the surface normal the reflection gets more, and above the critical angle there's only reflection (total inner reflection, TIR). Tha GaN-to-air surface has a critical angle of only 24°, so most photons get reflected. (The YAG layer doesn't help to increase the effective critical angle.) The reflected photons may bounce up and down a while until they get absorbed or they are lucky to hit a surface patch under a better angle (the surface is often roughened to offer such patches) and finally escape.
A silicone dome has a refractive index of about 1.5, and the GaN-to-dome critical angle is 37°, that means photons have a much better chance to escape the GaN crystal. However there's a second surface then: the dome-to-air surface of course gives some reflection, too. But since the die usually in in the center of the dome hemisphere, all photons hit that surface more or less perpendicularly so there's no TIR and only about 4% reflection there. But even those reflected photons aren't lost, they go back to the die and may be scattered or reflected there again to get another chance to get out, or they get absorbed and maybe re-emitted (called photon recycling) into a better angle.
Result: The total amount of light emitted (luminous flux) is higher. That's why the dome is there.

What dedoming does
As written above, without the dome the photon extraction is reduced, most photons don't get out - at the first try. But those photons aren't lost, TIR bounces them back and gives them more chances to be diffused into a better angle and get out. That TIR back to a the diffuse LED and photon reuse is the reason why the luminance can be increased even though normal optics can't do that. The dome actually is just some normal lens and would have no influence on luminance, but without it we get TIR and photon reuse, and those photons add to luminance.

The same trick is used with the Wavien collar: With an SST90 in an aspheric lens thrower, the light emitted to the side is lost, as it's not caught by the lens. An additional lens as 'pre-collimator' may catch more flux from the LED, but that won't give more throw, since it neither increases luminance nor the front lens area (instead the bigger caught flux is packed into a wider beam with same intensity and throw). However some clever people put a reflective collar around the lower part of the dome, so that the light that wouldn't hit the lens anyway was back-reflected onto the die, where it additionally lights up the diffuse surface (and those photons have another chance to escape into the right direction). This effectively increases the luminance.
As a side effect, there's a tint shift towards lower color temperatures, because the back-reflected blue photons have another chance to be converted into yellow light by the phosphor layer.

The TIR does exactly the same for the dedomed XM-L; it increases the luminance and shifts the tint towards warmer side.

Another observation can be explained by this: XM-Ls have quite some angular tint shift, i.e. the light emitted to the side is substantially more yellowish, while with the dedomed LED this effect is greatly reduced. The YAG phosphor has a refractive index of about 1.8. the critical angle agains the dome (~1.5) is about 55°, which means that such a ray to the very side has traveled a 74% longer distance through the phosphor compared to a perpendicular ray and more blue photons get converted to yellow. Without the dome, the critical angle against air is 33°, resulting in only 19% longer distance in the phosphor and thus much less angular tint shift. See these nice images by Tecmo: https://budgetlightforum.com/t/-/11595?page=8#comment-254514 (post #431)

Wrong/incomplete explanations

The die is smaller so the light density is higher - well, if you get about 30% less flux out of about 50% less (apparent) area with similar beam profile (lambertian), that should result in about 50% more luminance. However that's only half of the truth: If you try to get a smaller image of the die by using a diverging lens, or by filling something around the dome to get a flat surface to cancel the magnifying effect of the dome, that would not succeed. To increase luminance the TIR light needs to be reflected back to the die, thus the reflecting surface must be very near to the die. Either you remove the dome completely or you only leave a small layer (<0.5mm).

Due to the collimating dome missing, the light from the LED has a bigger beam angle and thus more light hits the reflector - actually no. First, the beam is about lambertian (i.e. 120° FWHM angle) with or without dome (see Tecmo's images). Second, that wouldn't explain why aspherics get increased throw, too. And third, even if you use a more collimating dome (like XR-E), that won't change luminance, because normal optics can't. (The XR-E doesn't throw so well because of the smaller beam angle, but because the EZ900 has a quite small area, thus more flux per area and thus a high luminance).
Actually the A*W rule does make the beam profile broader, broader than the hemisphere by back reflection that is. That itself doesn't change luminance. Only if that back-reflected light hits the diffuse die again, luminance increases.

4 Thanks

Thanks for the detailed explanation , DrJones .

Thank you. This addresses a question I had about dedoming.

To increase luminance the TIR light needs to be reflected back to the die, thus the reflecting surface must be very near to the die. Either you remove the dome completely or you only leave a small layer (<0.5mm).
I had Wondered if there was a range of apparent focus/tint shift. Edit relative to the amount of dome removed.

One of the biggest problems I see here is that the surface of the blue chip is not flat, not even close to flat, so while the critical angle in a GaN:Air interface is still 24 degrees, the way the surface is photons emitted from the active layer at many angles still have a chance to escape.

EDIT: Missed the part where he mentioned this.

So that's the short version .....

Thanks Dr., very illuminating (pun not intended) , I have been thinking a lot about led optics (without being a physicist -I am a cell biologist actually-), but there is not that much written down in a way I can understand it. This read helps a lot, some things I knew, some things I suspected, some things I knew wrong, some things are new to me .

I wouldn't mind the long version sometime (illustrated and well?)

Thanks for taking the time to do this excellent article, I’m about 1/2 way through, but determined to finish it.

whoa, i missed this before, thanks for the btt, tabetha, and for the science, Dr. Jones.

I’ll be studying this before I decide whether or not to de-dome a green sst-90 J)

I want to leave this god awful country I live in!!
Sorry rant over!!


Which high powered LED now has the highest luminance? Is it still XR-E, or is there anything that surpasses it? The XP-G2, XP-E2, XP-C or anything else?

Also, is there any breakdown anywhere that lists the luminance of all the different LEDs, like the Cree, Nichia, Luminus, etc, so that we could look at that data and decide which LEDs we want to play with and which ones we would not?

Thanks and thank you for the article,


this doesn’t have all the data you’re looking for, and you’ve probably seen it, but anyway, a good reference


What he needs to find is the flux per mm2. I may go add that to my LED sheet.

I just did a bit of math and I think a XP-E2 would actually beat a XR-E.

XR-E=114 (max brightness at 350 ma)/.81 (area of EZ900 die)=140.7407

XP-E2=142 (max brightness at 350 ma)/1 (area of die)=142

So this means that a XP-E2 has a higher surface brightness than a XR-E.

Would it also beat an xre r2? If so does that mean it would out throw the xre r2?

Well, this shows that the XP-E2 produces more light per mm2. Of course the XP-E2 has a wider viewing angle at 110 degrees v. the 90 of the XR-E. Then you have to take into account the XP-E2's efficiency at higher currents.

You can beat the h*ll out of a XRE (heat&current-wise that is ) without destroying it, this remains to be seen with the XPE2. In other words: what does it do at 2 Amps instead of 350mA.

Excellent post, thank you very much Dr.Jones!

Thats an effective write up! Thanks, you made leds and even etendue much more clear to me.

Thanks much for this. Great write-up!

Very Good Info.