Example of calculation of radiance

[mnb96]

Hello,

I found http://omlc.ogi.edu/classroom/ece532/class1/radiance_flashlight.html" an example of calculation of radiance.
I found this example quite confusing, and now I am not sure whether it's me who didn't understand some concepts, or the author who silently made a bunch of simplifying assumptions or even mistakes.

The most important point which bothers me is the following:
first the author assumes the flashlight behaves as a point-source having constant http://en.wikipedia.org/wiki/Radiant_intensity" received by the wall.
According to the author, it is sufficient to divide the radiant intensity by the area of the lit surface of the wall.

At this point, either I misunderstood something (very possible), or his reasoning is plain wrong, as he omitted the weighting factor $$cos(\theta)$$ to account for the projected differential area of wall. If I am correct, the surface of wall directly in front of the flashlight should receive more radiance than the one received at the periphery of the projected beam.

 

[Andy Resnick]

Radiometry can be very confusing, that is true. What the author wrote is correct; let's see if we can untangle your confusion.

First- intensity is power per solid angle: where the power goes. The radiance is the power per area per steriadian: not only where the power goes, but what the power density is in a particular direction.

First- the assumption that the source is a point located 10.7 cm behind the face- assuming the source is a point *greatly* simplifies the radiance calculations. A point source radiates equally in all directions, so the intensity is constant.

Also, because the wall and flashlight face are parallel, and oriented normal to the optical axis, the cos(theta) term is not needed- the projected area is the actual area.

Now, to calculate the radiance, the author says things like "L = 0.453 W/cm^2*sr in one direction, 0 elsewhere", which can be hard to parse out. Sometimes it's helpful to 'work backwards'- if you were on the wall, a little tiny eyeball on the wall, you would only see the flashlight if you looked in a particular direction (since the flashlight is a point source).

Does this help?

 

[mnb96]

Hi Andy!

it's nice that you are always able to provide some help whenever one comes up with a problem in radiometry :) Unfortunately I think I am still in trouble with this problem...
I understood your explanations about the point-source and the definition of radiant intensity. All my troubles probably lie in your following statement:

...Also, because the wall and flashlight face are parallel, and oriented normal to the optical axis, the cos(theta) term is not needed- the projected area is the actual area...

To me, the above assertion would sound correct, if the beam of light was collimated by a parabolic mirror, which theoretically would be our case; however it cannot be our case because the author explicitly chooses to model the problem by resorting to a point-source radiating with equal intensity in every direction. The point-source gives rise to this conic-shaped beam.

--- --- ---

So let's stick with the author's assumption.
If we place a point-source in front of a wall (=infinite plane), I guess your above assertion is not true anymore, and we do have to take into account the cosine term. Am I right?

I just pictured the situation for those differential elements of the wall that are infinitely far away from the optical axis: the projected area decrease with the squared distance from the point source (that tends to infinity), and with the cosine of the angle (that tends to pi/2) ==> radiance at those points should be definitely zero, while radiance is maximum in correspondence with the optical axis. Analogous difference between a surface lit by the Sun at noon, and at sunset.

Imho, the author is not allowed to arbitrarily mix two (very) different models and expect to get a reasonable conclusion.

What do you think about all this?

 

[Andy Resnick]

I have to check my reference (Wolfe's Introduction to Radiometry) which is in my office, but I think the calculation for the radiance is correct. If you want to instead calculate the irradiance on the wall (the flux of light incident on the wall [W/m^2]), you will indeed have a cos(theta) dependence on the projected area.

Again, I sometimes fall back on considering what I would see if I was on the receiving surface (and could only see along a single direction). So, if the surface was oriented parallel to the 'line of sight to the source', I would still see the source when I looked at 90 degrees, but the irradiance I receive is zero.

 

[mnb96]

Hi Andy,

I will patiently wait that you check Wolfe's book (I am probably going to purchase it), and hopefully help me clarifying this.

I could make few considerations, though,
If I recall correctly the definition of radiance does take into account the cos factor for the projected area, while the cos factor does not appear in the definition of irradiance, which should be given by $$E(\mathbf{x})=\int_\Omega L(x\leftarrow \Theta_\omega)cos(\theta)d\omega$$I also believe that if we have a point source placed in front of an infinite wall, and we assume that each point on the wall receives the same radiance, as the author claims, we have to admit the point source is emitting an infinite amount of energy.
I am still led to think the author is wrong, but at the same time I feel a bit puzzled.

Thanks again.

 

[Andy Resnick]

Ok- William Wolfe to the rescue. The radiant incidance/excitance are defined as

$$E = \frac{\partial\Phi}{\partial A}$$

Where $\Phi$ is the flux of power [W] and A the area. Similarly, the radiant intensity is

$$I = \frac{\partial\Phi}{\partial \Omega}$$

Where $\Omega$ is the solid angle. Sort-of combining the two is the radiance:$$L = \frac{\partial\Phi}{\partial (A cos\theta) \partial \Omega}$$

Where A_p = A cos$\theta$ is the projected area.

The radiance is used in the basic equation of radiative transfer:

$$d\Phi = L\frac{dA^{1}_{p}dA^{2}_{p}}{r^{2}}$$

Where r is the line-of-sight distance from dA1 to dA2, and the projected area is used because the surface may not be oriented along the line of sight.

So, it appears the website contains an error- the projected area should be used, introducing a cos(8.86) factor for light emitted at the edge of the flashlight beam.