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How to explain optical invariant in a simple way ?
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Mikko OH2HVJ
Oct 22, 2011, 2:43:56 AM10/22/11
to
Hi,
We're building a device with a light source that has multiple LEDs
combined to give suitable wavelength distribution. Earlier I asked
here advice for doing this.
We ended up with a structure where the LEDs are mounted on a PCB,
surrounded with high-reflectivity foil. The LEDs are shining on
a semisphere from the same material and the exit fiber is mounted
in the middle of the LEDs. Kind of an integrating sphere, the
wavelenght distribution, spatial form are excellent and the
solution is very stable and easy to manufacture. Also, we've
got about 90% of the theoretical coupling maximum, which is
surprisingly good for a quick prototype. (The theoretical coupling
maximum is nothing to write home about..)
Ok, now we've got a good working prototype which fulfills the
original measurement target.
The customers product manager has some technical background and keeps on
suggesting different solutions which might in his opinion improve the
efficiency. I'm usually very good at explaining complicated technical
issues to people of different leve, but in this case I feel completely
unarmed!
So, the question is: how would you explain the idea of optical invariant
as simply as possible ?
--
Mikko
We're building a device with a light source that has multiple LEDs
combined to give suitable wavelength distribution. Earlier I asked
here advice for doing this.
We ended up with a structure where the LEDs are mounted on a PCB,
surrounded with high-reflectivity foil. The LEDs are shining on
a semisphere from the same material and the exit fiber is mounted
in the middle of the LEDs. Kind of an integrating sphere, the
wavelenght distribution, spatial form are excellent and the
solution is very stable and easy to manufacture. Also, we've
got about 90% of the theoretical coupling maximum, which is
surprisingly good for a quick prototype. (The theoretical coupling
maximum is nothing to write home about..)
Ok, now we've got a good working prototype which fulfills the
original measurement target.
The customers product manager has some technical background and keeps on
suggesting different solutions which might in his opinion improve the
efficiency. I'm usually very good at explaining complicated technical
issues to people of different leve, but in this case I feel completely
unarmed!
So, the question is: how would you explain the idea of optical invariant
as simply as possible ?
--
Mikko
Helmut Wabnig
Oct 22, 2011, 2:59:12 AM10/22/11
to
On Sat, 22 Oct 2011 09:43:56 +0300, Mikko OH2HVJ <oh2...@sral.fi>
wrote:
What is an "optical invariant"?
w.
wrote:
w.
Salmon Egg
Oct 22, 2011, 1:25:04 PM10/22/11
to
In article <m2hb31o...@sral.fi>, Mikko OH2HVJ <oh2...@sral.fi>
wrote:
I think you overestimate your ability to explain technical concepts
simply. You did not even explain what you wanted in the example just
given.
I presume that by "optical invariant" you mean either conservation of
energy or, more subtly, brightness. Try again. Good luck.
--
Sam
Conservatives are against Darwinism but for natural selection.
Liberals are for Darwinism but totally against any selection.
wrote:
simply. You did not even explain what you wanted in the example just
given.
I presume that by "optical invariant" you mean either conservation of
energy or, more subtly, brightness. Try again. Good luck.
--
Sam
Conservatives are against Darwinism but for natural selection.
Liberals are for Darwinism but totally against any selection.
anorton
Oct 22, 2011, 2:30:55 PM10/22/11
to
"Mikko OH2HVJ" <oh2...@sral.fi> wrote in message
news:m2hb31o...@sral.fi...
defined in terms of marginal rays and chief rays and so is not intuitive to
non-raytracers. It is really only useful when applied to imaging optics.
A related quantity is etendue. The overly simple definition is that it is
the product of the area of a light beam and the solid angle diverging from
it. As you trace rays that are not vignetted through the optical system, the
etendue can not decrease. If you find that it is, then you are vignetting
rays somewhere. Conservation of etendue basically says you can not squeeze
light energy into a smaller "volume" in area*angle space. The
conservation is law can be proven by conservation of energy. Etendue can
increase due to scattering for, example, but if it does, then it means your
downstream optical system needs to handle more etendue (e.g. larger
aperture, smaller f/#) if you do not want to lose light.
--
Adam Norton
Norton Engineered Optics
www.nortonoptics.com
(Remove antispam feature before replying)
Salmon Egg
Oct 22, 2011, 3:48:22 PM10/22/11
to
In article <wa2dnaTVrOWOkT7T...@earthlink.com>,
"optical" or "Lagrange" invariant. I I ever knew what it meant, I have
forgotten. I never was much of a lens designer or ray-tracer. I seldom
used the term etendue. I preferred to stick with brightness (or
radiance) and its conservation (or loss). This seemed to be closer to
the physics. It was easy to see that looking at the images formed by a
lens, they were as bright as the object being imaged, if if the lens
distorted. I presume that Lagrange came from the variational formulation
of light propagation.
"anorton" <ano...@removethis.ix.netcom.com> wrote:
> The optical invariant is also known as the Lagrange invariant. It is
> defined in terms of marginal rays and chief rays and so is not intuitive to
> non-raytracers. It is really only useful when applied to imaging optics.
>
> A related quantity is etendue. The overly simple definition is that it is
> the product of the area of a light beam and the solid angle diverging from
> it. As you trace rays that are not vignetted through the optical system, the
> etendue can not decrease. If you find that it is, then you are vignetting
> rays somewhere. Conservation of etendue basically says you can not squeeze
> light energy into a smaller "volume" in area*angle space. The
> conservation is law can be proven by conservation of energy. Etendue can
> increase due to scattering for, example, but if it does, then it means your
> downstream optical system needs to handle more etendue (e.g. larger
> aperture, smaller f/#) if you do not want to lose light.
Although having worked in optics for decades, I never used the term
> The optical invariant is also known as the Lagrange invariant. It is
> defined in terms of marginal rays and chief rays and so is not intuitive to
> non-raytracers. It is really only useful when applied to imaging optics.
>
> A related quantity is etendue. The overly simple definition is that it is
> the product of the area of a light beam and the solid angle diverging from
> it. As you trace rays that are not vignetted through the optical system, the
> etendue can not decrease. If you find that it is, then you are vignetting
> rays somewhere. Conservation of etendue basically says you can not squeeze
> light energy into a smaller "volume" in area*angle space. The
> conservation is law can be proven by conservation of energy. Etendue can
> increase due to scattering for, example, but if it does, then it means your
> downstream optical system needs to handle more etendue (e.g. larger
> aperture, smaller f/#) if you do not want to lose light.
"optical" or "Lagrange" invariant. I I ever knew what it meant, I have
forgotten. I never was much of a lens designer or ray-tracer. I seldom
used the term etendue. I preferred to stick with brightness (or
radiance) and its conservation (or loss). This seemed to be closer to
the physics. It was easy to see that looking at the images formed by a
lens, they were as bright as the object being imaged, if if the lens
distorted. I presume that Lagrange came from the variational formulation
of light propagation.
Mikko OH2HVJ
Oct 22, 2011, 4:01:33 PM10/22/11
to
"anorton" <ano...@removethis.ix.netcom.com> writes:
> A related quantity is etendue. The overly simple definition is that it
> is the product of the area of a light beam and the solid angle
> diverging from it. As you trace rays that are not vignetted through
> the optical system, the etendue can not decrease. If you find that it
> is, then you are vignetting rays somewhere. Conservation of etendue
> basically says you can not squeeze light energy into a smaller
> "volume" in area*angle space. The conservation is law can be proven
> by conservation of energy. Etendue can increase due to scattering
> for, example, but if it does, then it means your downstream optical
> system needs to handle more etendue (e.g. larger aperture, smaller
> f/#) if you do not want to lose light.
Actually conservation of etendue what I was thinking of; how to explain in simple terms
> A related quantity is etendue. The overly simple definition is that it
> is the product of the area of a light beam and the solid angle
> diverging from it. As you trace rays that are not vignetted through
> the optical system, the etendue can not decrease. If you find that it
> is, then you are vignetting rays somewhere. Conservation of etendue
> basically says you can not squeeze light energy into a smaller
> "volume" in area*angle space. The conservation is law can be proven
> by conservation of energy. Etendue can increase due to scattering
> for, example, but if it does, then it means your downstream optical
> system needs to handle more etendue (e.g. larger aperture, smaller
> f/#) if you do not want to lose light.
that light from multiple large-area LEDs cannot be focused through a
small-diameter fiber. I remember having read the proof by conservation of
energy, I just can't get it into my head anymore or find it from anywhere!
I think that one was quite simple and should be understandable for my
client!
--
Mikko
Mikko OH2HVJ
Oct 22, 2011, 4:13:33 PM10/22/11
to
Salmon Egg <Salm...@sbcglobal.net> writes:
> In article <m2hb31o...@sral.fi>, Mikko OH2HVJ <oh2...@sral.fi>
> In article <m2hb31o...@sral.fi>, Mikko OH2HVJ <oh2...@sral.fi>
>> So, the question is: how would you explain the idea of optical invariant
>> as simply as possible ?
>> as simply as possible ?
> I presume that by "optical invariant" you mean either conservation of
> energy or, more subtly, brightness. Try again. Good luck.
Correct.. I'm trying to get my customer understand that with the
> energy or, more subtly, brightness. Try again. Good luck.
current limitations (total of 15mm^2 LED surface radiating into
almost half-space and a 0.8mm NA 0.5 fiber) it is impossible
to get more light into the fiber. Also, I'd like to be able to
explain the same to our EEs.
--
Mikko
Louis Boyd
Oct 22, 2011, 8:27:02 PM10/22/11
to
Mikko OH2HVJ wrote:
>
> Correct.. I'm trying to get my customer understand that with the
> current limitations (total of 15mm^2 LED surface radiating into
> almost half-space and a 0.8mm NA 0.5 fiber) it is impossible
> to get more light into the fiber. Also, I'd like to be able to
> explain the same to our EEs.
Have you tried bonding the end of the fiber directly to the face of a
>
> Correct.. I'm trying to get my customer understand that with the
> current limitations (total of 15mm^2 LED surface radiating into
> almost half-space and a 0.8mm NA 0.5 fiber) it is impossible
> to get more light into the fiber. Also, I'd like to be able to
> explain the same to our EEs.
white LED with about the same area as the aperture of the fiber. That
would be over 100 times more efficient than you can do with 15mm^2 area
using multiple LEDs. I'd expect around 10% total efficiency but I
don't' think you'll do better without a coherent source. The small led
will have a limited power which may be helped some with active cooling.
How much actual white light power does the application need though the
fiber in milliwatts? What is a minimum practical life for the light?
To get more energy through the fiber a brighter (energy per unit angular
area) light source is required. A short arc xenon lamp is the
brightest practical white light source I'm aware of. I'd expect you
could get close to watt though a .8mm fiber with a properly focused 100
watt lamp. That's not near as efficient as the direct coupled LED but
much better than what you're proposing with 225 square millimeters of
LEDs. The arc lamp should have an emission area under 10 square
millimeters. It needs a relay lens to match the NA of the fiber. A
rear spherical reflector should help by a little less than a factor of
two. The power supply for a short arc lamp is not tiny and the wasted
power has to be dissipated. That may not be practical at all
Phil Hobbs
Oct 22, 2011, 8:51:58 PM10/22/11
to
spontaneously flow from cold to hot.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
845-480-2058
hobbs at electrooptical dot net
http://electrooptical.net
Salmon Egg
Oct 22, 2011, 11:22:10 PM10/22/11
to
In article <m2aa8s7...@sral.fi>, Mikko OH2HVJ <oh2...@sral.fi>
wrote:
> Actually conservation of etendue what I was thinking of; how to explain in
> simple terms
> that light from multiple large-area LEDs cannot be focused through a
> small-diameter fiber. I remember having read the proof by conservation of
> energy, I just can't get it into my head anymore or find it from anywhere!
> I think that one was quite simple and should be understandable for my
> client!
The conservation of brightness, is more an indication of the SECOND LAW
of thermodynamics rather than the FIRST LAW which expresses the
conservation of energy. Two black bodies can be kept in thermal
equilibrium by interacting via a lossless narrow band filter. Which way
power flows between the two black bodies depends upon which body has the
higher temperature and consequently the higher spectral radiance within
the interacting band.
wrote:
> Actually conservation of etendue what I was thinking of; how to explain in
> simple terms
> that light from multiple large-area LEDs cannot be focused through a
> small-diameter fiber. I remember having read the proof by conservation of
> energy, I just can't get it into my head anymore or find it from anywhere!
> I think that one was quite simple and should be understandable for my
> client!
of thermodynamics rather than the FIRST LAW which expresses the
conservation of energy. Two black bodies can be kept in thermal
equilibrium by interacting via a lossless narrow band filter. Which way
power flows between the two black bodies depends upon which body has the
higher temperature and consequently the higher spectral radiance within
the interacting band.
Salmon Egg
Oct 22, 2011, 11:27:50 PM10/22/11
to
In article <m2mxcs3...@sral.fi>, Mikko OH2HVJ <oh2...@sral.fi>
wrote:
The best, I can come up with quickly is the case of antenna gain. To get
high antenna gain, you need a large antenna.
wrote:
high antenna gain, you need a large antenna.
boxman
Oct 24, 2011, 10:31:27 AM10/24/11
to
trying to simplify the explanation for the non optics type.
http://e2e.ti.com/blogs_/b/dlp_mems_blog/archive/2009/12/12/who-is-ed-tondue-and-why-due-i-care.aspx
boxman
Oct 24, 2011, 10:33:18 AM10/24/11
to
On 10/22/2011 3:13 PM, Mikko OH2HVJ wrote:
Phil Hobbs
Oct 24, 2011, 11:25:16 AM10/24/11
to
On 10/22/2011 11:27 PM, Salmon Egg wrote:
> In article<m2mxcs3...@sral.fi>, Mikko OH2HVJ<oh2...@sral.fi>
> wrote:
>
>> Salmon Egg<Salm...@sbcglobal.net> writes:
>>
>>> In article<m2hb31o...@sral.fi>, Mikko OH2HVJ<oh2...@sral.fi>
>>>> So, the question is: how would you explain the idea of optical invariant
>>>> as simply as possible ?
>>
>>> I presume that by "optical invariant" you mean either conservation of
>>> energy or, more subtly, brightness. Try again. Good luck.
>>
>>
>> Correct.. I'm trying to get my customer understand that with the
>> current limitations (total of 15mm^2 LED surface radiating into
>> almost half-space and a 0.8mm NA 0.5 fiber) it is impossible
>> to get more light into the fiber. Also, I'd like to be able to
>> explain the same to our EEs.
>
> The best, I can come up with quickly is the case of antenna gain. To get
> high antenna gain, you need a large antenna.
>
You have to be careful with that one. There's a class of electrically
> In article<m2mxcs3...@sral.fi>, Mikko OH2HVJ<oh2...@sral.fi>
> wrote:
>
>> Salmon Egg<Salm...@sbcglobal.net> writes:
>>
>>> In article<m2hb31o...@sral.fi>, Mikko OH2HVJ<oh2...@sral.fi>
>>>> So, the question is: how would you explain the idea of optical invariant
>>>> as simply as possible ?
>>
>>> I presume that by "optical invariant" you mean either conservation of
>>> energy or, more subtly, brightness. Try again. Good luck.
>>
>>
>> Correct.. I'm trying to get my customer understand that with the
>> current limitations (total of 15mm^2 LED surface radiating into
>> almost half-space and a 0.8mm NA 0.5 fiber) it is impossible
>> to get more light into the fiber. Also, I'd like to be able to
>> explain the same to our EEs.
>
> The best, I can come up with quickly is the case of antenna gain. To get
> high antenna gain, you need a large antenna.
>
small antennas, basically highish Q resonators, that can muddy the water
some.
The thermodynamic argument is summarized in an exchange in this very
boutique some years back. If you google for "sufficiently small heat
engine" (in quotes) it comes right up.
Cheers
Phil Hobbs
Helpful person
Oct 24, 2011, 11:48:50 AM10/24/11
to
image cannot be "hotter", "brighter" (choose your own word) than the
original source. The best that can be achieved is to collect a full
sphere of energy and and send it to the image. No tricks can improve
on this.
In addition, the flux in a collected solid angle (the apparent
intensity) cannot be brighter than the original object.
This is the same as saying that the brightness at a camera's image is
dependent on the F/# and nothing else. Even if you cannot get them to
understand the reason they will accept it because they know it's true.
(Now try to explain to an astronomer that for an extended object the
illumination at the image depends only on the F/#, not on the physical
aperture.)
http://www.richardfisher.com
Mikko OH2HVJ
Oct 25, 2011, 9:57:23 AM10/25/11
to
Phil Hobbs <pcdhSpamM...@electrooptical.net> writes:
> The thermodynamic argument is summarized in an exchange in this very
> boutique some years back. If you google for "sufficiently small heat
> engine" (in quotes) it comes right up.
Thanks! I was almost certain I had seen the explanation in your book
> The thermodynamic argument is summarized in an exchange in this very
> boutique some years back. If you google for "sufficiently small heat
> engine" (in quotes) it comes right up.
and even tried quickly to find it before posting my question.
Your book has been invaluable to me, I've had great fun building
instruments based on knowledge learnt from them!
--
Mikko
Phil Hobbs
Oct 25, 2011, 4:16:11 PM10/25/11
to
Cheers
Phil Hobbs
Pat Bigelow
Oct 26, 2011, 10:10:18 PM10/26/11
to
snip snip
> So, the question is: how would you explain the idea of optical invariant
> as simply as possible ?
Oops by invariant you mean "conservation of etendue" (as per Clausius)
or the so-called "Lagrange-Helmholtz invariant", probably the same thing,
more or less I assume.
> So, the question is: how would you explain the idea of optical invariant
> as simply as possible ?
or the so-called "Lagrange-Helmholtz invariant", probably the same thing,
more or less I assume.
Salmon Egg
Oct 29, 2011, 1:23:45 AM10/29/11
to
In article <ofOdnf3wmakyHjjT...@supernews.com>,
not very useful. Oner way of getting super-resolution on an optical
channel is to use a grating to see spatial frequencies higher than what
you would get from the Rayleigh criterion. Aside from knowing something
about it as a theoretical possibility, I try not to waste much time
thinking about it. Maybe someone spending an inordinate amount of effort
on the subject could come up with some useful gadget. As for me, when
that happens, I will just say that I thought about it but someone else
made the money.
Phil Hobbs <pcdhSpamM...@electrooptical.net> wrote:
> You have to be careful with that one. There's a class of electrically
> small antennas, basically highish Q resonators, that can muddy the water
> some.
>
> The thermodynamic argument is summarized in an exchange in this very
> boutique some years back. If you google for "sufficiently small heat
> engine" (in quotes) it comes right up.
>
Yeah! There also are things like super resolution optics, but they are
> You have to be careful with that one. There's a class of electrically
> small antennas, basically highish Q resonators, that can muddy the water
> some.
>
> The thermodynamic argument is summarized in an exchange in this very
> boutique some years back. If you google for "sufficiently small heat
> engine" (in quotes) it comes right up.
>
not very useful. Oner way of getting super-resolution on an optical
channel is to use a grating to see spatial frequencies higher than what
you would get from the Rayleigh criterion. Aside from knowing something
about it as a theoretical possibility, I try not to waste much time
thinking about it. Maybe someone spending an inordinate amount of effort
on the subject could come up with some useful gadget. As for me, when
that happens, I will just say that I thought about it but someone else
made the money.
chris
Nov 11, 2011, 4:58:47 AM11/11/11
to
this optical invariant is something like conservation of brightness.
For engineers start explaining optical magnification:
If you magnify the object-size you de-magify the view angles.
If you have a small light source with large angle you can change it to large with small angle (Laser beam).
If you have large beam with small angle (laser) you can change it to small area with large angle - focusing.
But if you start with large area and large angle you can not gain anything.
When creating light in LEDs it is always with half space angles. Fibers have less then half space. You need to decrease
angles; but then you increase size.
The only solution is to look for LEDs with small emitters. The smallest one will be a laser diode.
BTW there are 440nm 1W laser diodes for less than 100$ available. You could couple hundreds of them into one of that
huge fiber. :-)
Does that help?
Engineering Calculations
Nov 17, 2011, 9:00:17 PM11/17/11
to
Alan Smithee wrote: There are three approaches to dealing with a
customer who brings you a problem, the solution to which requires the
violation of the optical invariant.
1. Try to explain why what they want you to do is impossible, based
upon sound physical principles.
2. Smile and tell them you will do the best that you can possibly do,
given their budget and time constraints.
Approaches 1 and 2 never yielded an optimal result for me. In each
case I came out looking either like a denileistic non-team player who
should never again be consulted or an opportunistic feeder who took
their money and failed to deliver.
3. Explain that the task is beyond your experience level or that your
work load will not allow time to work their job. Then refer them to
another designer (preferably someone you don't like and who is either
too stupid to see that the task requires violation of causality or who
is dishonest enough to take their money and deliver no result) or
refer them to an outside contracter (who you know to be either too
stupid or sufficiently dishonest).
Approach 3. leaves you looking clean, honest and helpful. You don't
get to do the work but you also won't get the eventual blame for a
certain failure. If you are wrong and the person you refered designes
the customer a great system, you get the credit for knowing who to go
to and you get promoted to a management position.
This is a quote from the famous optical designer Alan Smithee
James E. Klein
Engineering Calculations
KDP2 Optical Design Program and Design Work
www.ecalculations.com
1-818-823-4121
1-818-507-5705
1377 E. Windsor Rd., #317
Glendale, CA 91205
customer who brings you a problem, the solution to which requires the
violation of the optical invariant.
1. Try to explain why what they want you to do is impossible, based
upon sound physical principles.
2. Smile and tell them you will do the best that you can possibly do,
given their budget and time constraints.
Approaches 1 and 2 never yielded an optimal result for me. In each
case I came out looking either like a denileistic non-team player who
should never again be consulted or an opportunistic feeder who took
their money and failed to deliver.
3. Explain that the task is beyond your experience level or that your
work load will not allow time to work their job. Then refer them to
another designer (preferably someone you don't like and who is either
too stupid to see that the task requires violation of causality or who
is dishonest enough to take their money and deliver no result) or
refer them to an outside contracter (who you know to be either too
stupid or sufficiently dishonest).
Approach 3. leaves you looking clean, honest and helpful. You don't
get to do the work but you also won't get the eventual blame for a
certain failure. If you are wrong and the person you refered designes
the customer a great system, you get the credit for knowing who to go
to and you get promoted to a management position.
This is a quote from the famous optical designer Alan Smithee
James E. Klein
Engineering Calculations
KDP2 Optical Design Program and Design Work
www.ecalculations.com
1-818-823-4121
1-818-507-5705
1377 E. Windsor Rd., #317
Glendale, CA 91205
Mike S
Jun 19, 2012, 5:27:03 AM6/19/12
to
the focal point of each parabola was at the end of the fiber. Or am I
missing the point?
Mike
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