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Imagine a surface which radiates light equally in all directions from all points

if we measure the radiance of the surface from two positions

one opposite the surface

and one at a 45 degree angle

would the measured radiance be the same?

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I will confess this one left me confused for a while. Yes, the radiance is the same, assuming the emitter is at the same distance (it's not clear from the diagram). The reason for this is because the emitter is always at normal incidence to the detector's view angle.

There's this neat theorem called conservation of radiance which is really helpful in puzzling situations like these, which states that radiance measured at the emitter is the same as radiance measured at the detector, assuming no radiance is lost between emitter and detector. In this case, it becomes obvious the radiance should be the same, as in both cases the emitters are identical up to rotation (which conserves areas, distances, and basically everything you could possibly care about when working with radiance).

There are also some helpful notes under the Wikipedia Talk page for radiance, see the "cosine term" paragraph. The key point is that the cosine term in the definition of radiance does not use the detector's surface normal but the emitter's surface normal. Notably, the sentence "The cosine factor in the denominator reflects the fact that the apparent size of the source goes to zero as your angle of view approaches 90°." highlights this. In your case, the apparent size of the source is quite clearly constant due to the nature of your setup. This can get horribly confusing especially when you see computer graphics papers such as the rendering equation which apparently use the cosine term with the detector's surface normal, but they make entirely different assumptions.

Another, perhaps more intuitive way of deducing this result is by looking at the units of radiance: watts per meter squared per steradian. Also known as power per area per solid angle. Just to be clear, "power" is the power of the emitter, "area" is the surface area of the emitter, and "solid angle" is the solid angle subtended by the emitter from the detector's point of view. Now, for both cases considered:

- power is constant, since the emitter is obviously still emitting the same amount of light in both cases

- area is constant, as the emitter's surface area hasn't changed

- solid angle is constant, because the emitter is held facing the detector at normal incidence in both cases

Therefore, measured radiance is the same in both cases.

That said, I could be wrong on this. Please feel free to correct me if I made a mistake.

EDIT: actually, I don't even know anymore. It's all foggy

EDIT 2: no, I think, radiance should really be interpreted as emitted intensity (power / area) per solid angle (direction). There, that's better  still mighty confused though,

Edited by Bacterius

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Radiometry is a puzzle, but using this special non-Lambertian surface makes things easier

I believe the _flux_ incident to the detector is equal in both cases

but the measured _radiance_ is 1.0 / 0.707 = 1.414 times higher for the angled detector

because from the angled detector's point of view the emitter looks smaller (it has smaller projected area)

which means you have the same amount of flux in a smaller area which increases flux density .... and therefore "brightness"

this is due to the confusing projected area term in the radiance equation:

radiance = flux / solid_angle / projected_area

and THAT is why Lambertian reflectance needs a cosine law term ... to balance out the increase in radiance due to viewing angle

that is why Lambertian surface gives constant radiance when viewing angle changes

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(edit) rubbish

Edited by skytiger

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I think it's best to always talk about radiance with respect to a imaginary surface perpendicular to the chosen direction. In all settings I ever encountered you were free to choose the surface normal, so why not chose the easiest configuration, in which the confusing cosine term simply disappears.

Assuming the receiver is left and the emitter right in your images:
Incident radiance (say) at the center of the receiver (x) coming from the direction pointing to the center of the emitter (w) is the same in both configurations. It's also the same as the exitant radiance from the center of the emitter toward the center of the receiver.
Irradiance at every point of the receiver is smaller for the angled configuration. The solid angle covered by the emitter is the same, but in the case of irradiance you are not free to choose the normal of the surface (since you asking specifically about irradiance of a concrete surface). So in this case the cosine factor is not 1 and it'll give you a smaller value.
Power collected in total by the receiver is smaller, too (obviously, if irradiance at every point is).

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Irradiance at every point of the receiver is smaller for the angled configuration. The solid angle covered by the emitter is the same, but in the case of irradiance you are not free to choose the normal

I do not understand exactly what you mean by this. Could you elaborate a bit?

The way I thought about it: imagine a hemisphere around the receiver center point's normal. In the first configuration, the emitting surface would be close to the Zenith of this hemisphere. In the second configuration, the surface would be closer to the Azimuth and thus the cosinus factor would be different.

Is this what you mean?

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I do not understand exactly what you mean by this. Could you elaborate a bit?

The way I thought about it: imagine a hemisphere around the receiver center point's normal. In the first configuration, the emitting surface would be close to the Zenith of this hemisphere. In the second configuration, the surface would be closer to the Azimuth and thus the cosinus factor would be different.
Is this what you mean?

Yes, that is correct. The incident beam would be smaller due to the cosine factor, hence irradiance decreases by a corresponding amount. That is not radiance, though, but irradiance.

Assuming the receiver is left and the emitter right in your images:

That was my assumption as well. Perhaps this was not the expected interpretation?

Edited by Bacterius

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The emitter is on the left

and the receivers are on the right

the blue represents the solid angle subtended by the receivers

the magenta shows the area and the projected area of the emitter

they are meant to be infinitesimally small differential areas - so the direction is from the center of emitter to the center of receiver

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The emitter is on the left

and the receivers are on the right

the blue represents the solid angle subtended by the receivers

the magenta shows the area and the projected area of the emitter

they are meant to be infinitesimally small differential areas - so the direction is from the center of emitter to the center of receiver

Well in that case, yes, radiance is lower. Think about what happens when you aim the emitter perpendicular to the receiver, the projected area goes to zero and no light moves towards the receiver so radiance is zero, as expected. So a lower emitter projected area for the emitter -> lower radiance, by the factor you stated in your first post.

The blue illustration in the diagram was confusing, though. And it's important to not confuse the cosine term in the emitter's projected area with the cosine term for the irradiance's incidence area, they are not the same!

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Irradiance at every point of the receiver is smaller for the angled configuration. The solid angle covered by the emitter is the same, but in the case of irradiance you are not free to choose the normal

I do not understand exactly what you mean by this. Could you elaborate a bit?

The way I thought about it: imagine a hemisphere around the receiver center point's normal. In the first configuration, the emitting surface would be close to the Zenith of this hemisphere. In the second configuration, the surface would be closer to the Azimuth and thus the cosinus factor would be different.
Is this what you mean?

That's what I meant. Although it wasn't quite correct of me to speak of 'the' cosine factor when a finite solid angle is involved, because there are many cosine factors; one for each direction in the direction bundle that makes up the solid angle. But each factor is smaller in (b) than the corresponding factor in (a), so it's save to say that the irradiance will be lower due to 'the' cosine factor.

The emitter is on the left
and the receivers are on the right

the blue represents the solid angle subtended by the receivers
the magenta shows the area and the projected area of the emitter

they are meant to be infinitesimally small differential areas - so the direction is from the center of emitter to the center of receiver

Well in that case, yes, radiance is lower. Think about what happens when you aim the emitter perpendicular to the receiver, the projected area goes to zero and no light moves towards the receiver so radiance is zero, as expected. So a lower emitter projected area for the emitter -> lower radiance, by the factor you stated in your first post.

The blue illustration in the diagram was confusing, though. And it's important to not confuse the cosine term in the emitter's projected area with the cosine term for the irradiance's incidence area, they are not the same!

I have to disagree, in my opinion you'll approximately get the same results.
- Radiance is the same. You look up from the receiver with in a single direction. Either you see the emitter, then you get the radiance it emits, or you don't see the emitter, then you get zero. The orientation of the emitter doesn't matter since it's defined to emit equally in all directions.
- Irradiance is lower in b than in a. This time not because of the cosine factor(s), but because of the smaller solid angle.
- Power also lower.

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