Whos knows about optics ?

Thread Starter

drjohsmith

Joined Dec 13, 2021
852
not a home work question,
radiance....
https://en.wikipedia.org/wiki/Radiance

Bene reading an optics book, for "fun"
did not know there were so many ways to define how "bright" a light source is

Am I half there saying
that if I have alight source puts out an amount of light that hits a flat target of say 2mm square,
if I point the same light at the same target , but light is now 1 mm square due to optics,
the radiance is the same ?
 

MrAl

Joined Jun 17, 2014
11,715
not a home work question,
radiance....
https://en.wikipedia.org/wiki/Radiance

Bene reading an optics book, for "fun"
did not know there were so many ways to define how "bright" a light source is

Am I half there saying
that if I have alight source puts out an amount of light that hits a flat target of say 2mm square,
if I point the same light at the same target , but light is now 1 mm square due to optics,
the radiance is the same ?
Hi,

You mean using a lens?
For that example I think the radiance would increase by 4 times because the solid angle decreased to 1/2 of what it was before. This is not including the efficiency of the lens however, which will almost always reduce the total power by some amount and may be significant.

We can think of it like this. If you want to burn a hole in paper with the suns light you use a magnifying lens. If you don't use a lens the paper just gets warm, but if you use a (convex) lens then the point of focus (which means the area has decreased) becomes hotter. That would mean the radiance increased. Higher radiance means more surface damage and is a measure of optical safety.

I happen to know a PhD who majored in optics I'll ask him about this too just to be sure.
 

BobTPH

Joined Jun 5, 2013
9,269
What is irradiance vs radiance?


Irradiance: The amount of energy incident on a given area of a surface in a given amount of time (W/m2). Radiance: The amount of energy scattered in a particular direction (W/m2/sr).
Using this definition, no. You have increased the irradiance at the target, but the radiance is a property of the emitter and has not changed.
 

Ya’akov

Joined Jan 27, 2019
9,264
not a home work question,
radiance....
https://en.wikipedia.org/wiki/Radiance

Bene reading an optics book, for "fun"
did not know there were so many ways to define how "bright" a light source is

Am I half there saying
that if I have alight source puts out an amount of light that hits a flat target of say 2mm square,
if I point the same light at the same target , but light is now 1 mm square due to optics,
the radiance is the same ?
I think the answer is "no", but the uncertainty has to do with how you've framed the question. Let me reverse it to clarify.

If you have a certain power in W from an emitter, the radiance will increase or decrease proportional to the area over which that power is spread.

It is easier to understand this from the point of view of "seeing" the radiance than transmitting it.
 

Thread Starter

drjohsmith

Joined Dec 13, 2021
852
I think the answer is "no", but the uncertainty has to do with how you've framed the question. Let me reverse it to clarify.

If you have a certain power in W from an emitter, the radiance will increase or decrease proportional to the area over which that power is spread.

It is easier to understand this from the point of view of "seeing" the radiance than transmitting it.
This is my "confusion" / "question"
I think radiance is over a square area,
if you put the same amount of light onto the same target at half the area,
the radience I think stays same,
but to me the "brightness" will increase.

BTW: to others, yep Im assuming perfect optics , zero loss in system this is a mind experiment only ...
 
Last edited:

BobTPH

Joined Jun 5, 2013
9,269
By the definitions I posted, you are talking about irradiance, not radiance.

I can see how it is confusing, but look at the units, they are are different.
 

Ya’akov

Joined Jan 27, 2019
9,264
This is my "confusion" "question"
as radiance is over a square area,
if you put the same amount of light onto the same target at half the area,
h eradience I think stays same,
but to me the "brightness" will increase.

BTW: to tother,s yep Im assuming perfect optics , zero loss in system this is a mind experiment only ...
Does this help? [source]

The radiance of a source is increased by increasing its emitted power, by making the emitting area of the source smaller or by emitting the radiation into a smaller solid angle. Strictly speaking, radiance is defined at every point on the emitting surface, as a function of position, and as a function of the angle of observation. Often, as in the example above we use radiance of a source to mean the radiance averaged over a finite sized aperture and over some solid angle of interest.

Radiance is a conserved quantity in an optical system so that radiance measured as watts per unit area per unit solid angle incident on a detector will not exceed the radiance at the emitter. In practice, for any bundle of rays mapping an emitter to a detector, the radiance seen at the detector will be diminished by the light which is absorbed along the way or scattered out of the solid angle of the bundle of rays reaching the detector.
 

BobTPH

Joined Jun 5, 2013
9,269
I would say you have increased the radiance of the lens / emitter as a system in that particular direction, at the expense of other directions.
 

MrAl

Joined Jun 17, 2014
11,715
I just asked the optics expert a little while ago it may take a few hours to get a reply.
I could be reading this wrong too so I wanted to double check with someone who knows this completely.

It is interesting that the solid angle decreases but then again so does the area.
The expression is in units W/(m^2*sr).
The solid angle decreased by 1/2 and the area decreased by 1/2 so i think that would be a factor of 8 not 4 because of the m^2:
W/(1/2)^2/(1/2)=W/(1/4)/(1/2)=W/(1/8)=8*W
I must have mistaken m^2 to be just 'm' the first time around that's how i got 4 instead of 8: W/(1/2)/(1/2)=4*W.
On the flip side, looking through a telescope does not increase the radiance and that would imply that ideal lensing does not change anything.

Yes i think irradiance is in W/(m^2).
 
Last edited:
Top