# A Irradiance distribution of projected beam

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1. Sep 10, 2017

### moundman9

Hey folks, I'm having a little bit of trouble shaking the rust off of my radiometry.

I'm projecting a circular beam of some known divergence and power onto a surface angled relative to the center of the beam. Say holding a flashlight, with a cone of half angle θ, at my waist some height h off the ground, projecting it onto the ground at an incidence angle α.

Assuming the beam has constant intensity across solid angle from the source, I want to determine the irradiance as a function of position across the spot. I can easily calculate the area of the projected ellipse to determine the average irradiance across the spot, but I'm interested in the difference between irradiances at the near and far ends of the spot.

Basically I'm trying to find a function to relate differential solid angle of the beam to differential projected area of the spot, and I'm getting stuck figuring out the proper way to set it up.

Any and all help is much appreciated.

2. Sep 10, 2017

If the beam intensity (watts/sr) is uniform across the illuminated solid angle $\Omega$ over which it shines, you can call the intensity $I_o=\frac{P}{\Omega}$. Then irradiance $E=\frac{I_o}{s^2}$(watts/m^2), but this is onto a surface that faces the source. For a surface whose normal is tilted at angle $\theta$ relative to the line from the flashlight to the point on the ground, you will find $E_{ground}=\frac{I_o}{s^2} \, \cos(\theta)$.

3. Sep 11, 2017

### Andy Resnick

Check your units- I believe what you call 'irradiance' is actually 'radiance'. The OP is on the right track, and just simply needs to carefully think about the differential projected area, which is a function of both α and θ.

4. Sep 11, 2017

Radiance (also known as brightness) is $L$, and has units of $watts/(sr \, m^2)$ . You might think that irradiance $E=I_o/s^2$ should have units of $watts/(sr \, m^2)$, but the steradian is actually dimensionless, (solid ange $\Omega=A/s^2$), and no $sr$ appears in the units for the irradiance $E$. $\\$ (Note: This same question actually came up at a conference on radiometry that I attended at the University of Colorado at Boulder in 1982. Many people ask how the unit for steradian "sr" can just come and go in the equations depending on whether it is applicable or not.)

Last edited: Sep 11, 2017
5. Sep 11, 2017

### Andy Resnick

I agree that radiometric units are rather clumsy, I have my reference book in easy reach. Irradiance, like emittance, incidance, and excitance, is a measure of flux density and has units of [power/area]. Intensity, defined radiometrically, has units of [power/solid angle]. Radiance has units of flux per unit projected area (A*cos(θ)) and per unit solid angle, [power/(area*sr)] and is typically associated with an extended source. An extended, diverging, beam incident onto a surface means that the projected area may vary with position on the surface, in addition to any directional variation in intensity.

6. Sep 11, 2017

### moundman9

I get that in general, and that's how I know the average irradiance. I'm not sure if there's a simple step that I'm missing. For differential area and solid angle, the incidence angle across the projected spot (using the notation from my original post) will not be a constant α, but a function of θ. Do I just need to figure out a function α(θ), and then set up an integral over the spot that should evaluate to the beam's total power to check? In that case maybe my problem is just an issue of figuring out the limits of the integral, probably in spherical coordinates.

7. Sep 11, 2017

To try to compute the integral of the incident light per unit area and show that it equates to the total power would likely be a very difficult integral to evaluate in closed form. This one is much more readily solved for individual selected positions on the ground.

8. Sep 11, 2017

It necessarily will by conservation of energy, but the mathematics could get rather complex.

9. Sep 11, 2017

### Andy Resnick

It's a geometry problem- somewhat complicated, but not too terrible.

First, the beam specification: say like Figure II.11 (http://light-measurement.com/calculation-of-radiometric-quantities/). The beam appears to originate from a point source located somewhere behind the bulb, such that a cone of light with divergence angle θ0 has a diameter of 'd' at the 'bright end'. The irradiance E at that exit plane can be uniform (as in 'constant intensity across the solid angle') or be nonuniform, which we can parameterize in terms of the angular coordinate θ, E = E(θ).

That's the easy part.

The conical beam intersects the ground as a conic section (http://www3.ul.ie/~rynnet/swconics/planes_cutting_coneA.htm), with the intersecting plane inclined to the illumination cone axis by an angle α0. Now, you "simply" need to parameterize the ellipse, parabola, or hyperbola in terms of α0 and h, which I can't do off the top of my head. But once you do that, you can then 'map' E (or E(θ)) as the radiant incidance onto the hyperbolic section, and if you did everything correctly, there will be a 1/r^2 dependence of the incidance, where 'r' is the distance from a point on the ground to the apparent source.

Does that help?

10. Sep 12, 2017

Additional comment on the energy conservation: The integral over the surface by treating it as a conic section does not look simple. The quickest way I know to show energy conservation here is to treat the source as one that originates at the origin and obeys an inverse square law. It can be assumed to be of the form $\vec{S}(r,\theta, \phi)=\frac{f(\theta,\phi)}{r^2} \hat{a}_r$. This field $\vec{S}$ could readily be shown to obey $\nabla \cdot \vec{S}=0$ at all points except the origin, so that by Gauss' law $\int \vec{S} \cdot \hat{n}dA$ over any enclosed volume must be zero. Thereby the power that gets transmitted through any plane that intercepts a cone of the light coming from the origin will be the same, regardless of how far away that plane is or how it is oriented, since the contribution to the surface integral from the surface of the cone is zero since $\vec{S}$ and $\hat{n}$ are perpendicular on the surface of the cone.