Nits to lumens

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simon....@gmail.com

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Mar 14, 2022, 5:50:11 PMMar 14
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Hi, well it's been a long time since I contributed to sci.optics - about 20 years I guess. It's great to see that it's still active.

A quick question on photometric units - I'm trying to compare the light output from Oleds with that from LEDs. I'd like to compare total flux emitted by the two sources. Here's my calculation:

An LED die has 30 candela output
A 1 candela lambertian emits pi lumens into a hemisphere.
Therefore the LED emits 30pi lumens

An OLED emits about 400 nits which is 400 candela/m^2
The OLED is 18mm x 7mm , so the OLED emits 400 * 0.018 * 0.007 candela = 0.05 candela
As above, assuming the OLED is a lambertian emitter
OLED emits 0.05*pi lumens.

So my conclusion here would be that my LED has a flux 600 times that of the OLED.

My question is about the nit to candela calculation - I've not found any references to this. It seems sensible, but photometrics are a very strange beast!

Thanks,

Simon

Phil Hobbs

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Mar 14, 2022, 8:12:35 PMMar 14
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simon....@gmail.com wrote:
> Hi, well it's been a long time since I contributed to sci.optics -
> about 20 years I guess. It's great to see that it's still active.

Well, there are some diehards in the world. You should join up--there
are still a few openings. ;)

> A quick question on photometric units - I'm trying to compare the
> light output from Oleds with that from LEDs. I'd like to compare
> total flux emitted by the two sources. Here's my calculation:
>
> An LED die has 30 candela output A 1 candela lambertian emits pi
> lumens into a hemisphere. Therefore the LED emits 30pi lumens.

Assuming that it's really Lambertian. Some are reasonably close, mostly
the ones with textured surfaces. LEDs with flat facets aren't
Lambertian because of Snell's law and the Fresnel reflection at the
surface, both of which change the angular distribution of the emission.

> An OLED emits about 400 nits which is 400 candela/m^2 The OLED is
> 18mm x 7mm , so the OLED emits 400 * 0.018 * 0.007 candela = 0.05
> candela As above, assuming the OLED is a lambertian emitter OLED
> emits 0.05*pi lumens.
>
> So my conclusion here would be that my LED has a flux 600 times that
> of the OLED.

It depends on the surface roughness. The two main problems with LEDs
are: (1) while the emission is almost 100% efficient, it occurs deep
within a highly-absorbing substrate; and (2) the substrate has a
refractive index of 3.3 to 3.5, so that almost all the light gets
internally reflected at the surface. The great strides in LED
efficiency that we've seen in the last few decades mostly come from
fixing those two problems.

A good chunk of the confusion IMO comes from the nomenclature. I mean,
how scientific-sounding are quantities such as "luminous intensity" and
"spectral radiance"? They sound like something out of Edgar Allan Poe,
or maybe a romance novel. (Give me "flux density" and so forth any day.)

Much of the remainder comes from People Who Know Best deciding to
redefine perfectly well-defined pre-existing terms to mean something
completely different. For instance, "intensity" means watts per square
metre to normal people, but watts per steradian to a radiometrist.

> My question is about the nit to candela calculation - I've not found
> any references to this. It seems sensible, but photometrics are a
> very strange beast!

Yup. There's a lot of job security stuff in radiometry, and even more
in photometry. Making the candela a SI base unit is weird as hell, for
a start. The lumen is derivable from the joule, the metre, and the
second, times exactly 683 lumens/W peak and the Anointed International
Photometric Eyeball curve. (As I tend to point out to any vaguely
interested person, 683 appears to be the largest prime number ever used
for unit conversion.)

However, they standardized the _candela_, which imports the idea of
solid angle. Which solid angle--projected or spherical? There's a
factor of cos(theta) difference between the two, and the choice depends
on whether you're illuminating a surface or a volume.

How do you make a traceable measurement of solid angle, when the
transmittance of an optical system depends on both angle and position?
Confused yet? (Me too.) And then there's the nearly impenetrable
thicket of redundant units.

But yeah, OLEDs are pretty dim in general.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC / Hobbs ElectroOptics
Optics, Electro-optics, Photonics, Analog Electronics
Briarcliff Manor NY 10510

http://electrooptical.net
http://hobbs-eo.com

simon....@gmail.com

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Jun 2, 2022, 12:33:07 PMJun 2
to
Thanks for this Phil. I did come across a wikipedia article which discusses my exact question. The specific question was: 'for a lambertian emitter of Lv nits, the flux in lumens is F=pi*Lv*A'. The link for the wikipedia is:
https://en.wikipedia.org/wiki/Lambert%27s_cosine_law

Also, I was pondering about placing a lambertian source in front of a window and calculating the integral of the fresnel coefficients across angle. There's a very clear paper on this by Mathieu Hébert (who also references a 1942 paper by Judd):
http://paristech.institutoptique.fr/site.php?id=797&fileid=11468.

The conclusion of the calculation is that for an uncoated n-bk7 window, a lambertian emitter has 9.4% reflected across all angles at the entrance to the window. About 4.1% is reflected from the second surface in the window. So, 86.8% of lambertian light incident on a window passes through it. (My zemax simulation gives 84.7% transmission. The difference is that the closed form solution assumes that there is a lambertian profile within the glass medium. However, this isn't exactly true - high angle rays in the input lambertian are attenuated more by the Fresnel loss than axial ones).

My children will be very proud that I've now understood how a window works....

Simon

Phil Hobbs

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Jun 2, 2022, 7:44:28 PMJun 2
to
simon....@gmail.com wrote:
> Thanks for this Phil. I did come across a wikipedia article which
> discusses my exact question. The specific question was: 'for a
> lambertian emitter of Lv nits, the flux in lumens is F=pi*Lv*A'.
> The link for the wikipedia is:
> https://en.wikipedia.org/wiki/Lambert%27s_cosine_law

Well, yes, but a good rant is far more satisfying for everyone. ;)

>
> Also, I was pondering about placing a lambertian source in front of
> a window and calculating the integral of the fresnel coefficients
> across angle. There's a very clear paper on this by Mathieu Hébert
> (who also references a 1942 paper by Judd):
> http://paristech.institutoptique.fr/site.php?id=797&fileid=11468.
>
> The conclusion of the calculation is that for an uncoated n-bk7
> window, a lambertian emitter has 9.4% reflected across all angles at
> the entrance to the window. About 4.1% is reflected from the second
> surface in the window. So, 86.8% of lambertian light incident on a
> window passes through it. (My zemax simulation gives 84.7%
> transmission. The difference is that the closed form solution
> assumes that there is a lambertian profile within the glass medium.
> However, this isn't exactly true - high angle rays in the input
> lambertian are attenuated more by the Fresnel loss than axial ones).

Plus they hit the edges sooner.

>
> My children will be very proud that I've now understood how a window
> works....
>
> Simon
>

Now if you'll just explain to me exactly what a lens does, we'll all be
further ahead. ;)

Cheers

Phil Hobbs

simon....@gmail.com

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Jun 2, 2022, 11:06:27 PMJun 2
to
Too funny. On several occasions during my career I've concluded that I don't understand how lenses work due to the aberration, mirrors are a complete nightmare since I can't explain left-right inversion to an elementary school kid without using the word 'magic' but planar optics was ok. Until I worked with prisms and discovered how hard those are. I've always concluded that my mastery of windows is unsurpassed, but clearly I'm still on a journey to understand the basics of a window.

s

Dieter Michel

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Jun 3, 2022, 9:30:34 AMJun 3
to
Hi Simon,

> [...] mirrors are a complete nightmare since
> I can't explain left-right inversion to an
> elementary school kid without using the word 'magic'

from the little I know about mirrors I seem to recall
that left-right inversion is hardly explainable because
it doesn't exist.

Mirrors do not swap left and right, but front and back.
The impression that left and right would be swapped
by the mirror only comes about when one (involuntarily)
puts oneself in the place of the mirror image and thus
changes the coordinate system without being aware of it.
Then one's own left hand magically seems to be the
right hand of the mirror image and vice versa.

If one does not change the coordinate system, left remains
left and right remains right.

Just my 2 cents,

Dieter

Phil Hobbs

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Jun 3, 2022, 11:09:03 AMJun 3
to
Yup. The left-right inversion is an illusion, caused by the approximate
bilateral symmetry of common objects such as trees and humans. The
front-back inversion is much more obvious in chirally asymmetric things
such as hands and screw threads.

*Re: What lenses do:*

It's very complicated, so typically we pick out whatever set of
approximate properties is most useful at the moment. A lens can be
described as:

1. A ray bender

2. A parabolic phase function

3. (most useful for what I do) A Fourier transformer between pupil and
focus.

Then we may have to patch up the polarization effects, edge diffraction
contributions, and so forth until we get a good enough approximation.

Except at high NA, we can usually stick with scalar optics, which is a help.

There are various bits of heuristic reasoning available that help too.
For instance, a focused beam with undercorrected spherical aberration
shows strong interference rings inside focus (i.e. closer to the lens)
but much weaker ones outside focus. Overcorrected spherical does the
opposite.

Mentally combining the wave and ray pictures, we can get an intuitive
idea of why.

The total optical intensity goes like |E|**2, so if we notionally divide
the beam up into separate center and edge contributions, this becomes

|E_center + E_edge|^2 =
|E_center|**2 + |E_edge|^2 + 2 Re{ E_center + E_edge*}.

The first two terms don't have phase information, and so ideally should
be the same on both sides of focus.

The third (interference) term is phase sensitive. Because the
interference is proportional to the product of the center and edge field
strengths, the edge contributions produce the strongest fringes when
they interfere with the largest E field of the rest of the beam, namely
near where the edge rays are crossing the beam axis.

That happens inside focus for undercorrected and outside focus for
overcorrected. While this is not too much use in computing the actual
fringe pattern, arguments of this sort help intuition considerably.
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