# Consider a diffuse (Lambertian) emitter

1. Jul 11, 2010

### mnb96

Hello,

I consider a diffuse (Lambertian) emitter which has the shape of a square planar surface, with area A, parallel to the horizontal xy-plane, and centered at the point (0,0,h).

I know the radiant flux at each point of the emitter $\Phi(x,y,z,\Theta)=\mathrm{const}$

What should I do to compute the irradiance at the location (0,0,0) of the surface represented by the xy-plane?

Thanks!

2. Jul 16, 2010

### mnb96

...I thought this question to be really basic...I was probably wrong.

3. Jul 16, 2010

### Andy Resnick

Thanks for the bump- I missed it the first time.

Let me see if I understand your geometry: the source is a Lambertian emitter, which means the the radiance is independent of angle.

The fundamental expression is E = L * Z / R^2, where E is the irradiance at the receiver, L the radiance of the source, R^2 the distance to the receiver, and Z the 'throughput'/entendue/lichtleitwet...

$$dZ = \frac{dA_{1} cos(\theta_{1})dA_{2}cos( \theta_{2})}{\rho^{2}}$$

is the "fundamental equation of radiative transfer" involving two differential areas (source and reciever) separated by a distance 'rho', and their relative orientations with respect to the line of sight.

So in order to calculate the irradiance, you have to integrate over the area of the source which will be cumbersome for a square, but easier for a disk. I'll step through the disk of radius R centered at {0,0,h}: dA_2 = r dr d$\phi$, $\rho = h/ tan(\theta)$, r = h tan($\theta$)

$$dZ = \frac{dA_{1} cos(\theta_{1})dA_{2}cos( \theta_{2})}{\rho^{2}}$$
$$dZ = \frac{dA cos^{2}(\theta) r dr d\theta d\phi}{h^{2} sec^{2}(\theta)}$$
$$dZ = dA sin(\theta) cos(\theta)d\theta d\phi$$
$$E = \frac{LA sin^{2}(\theta)}{h^{2}} = \frac{I}{h^{2}}(1-(R/h)^{2}+(R/h^{4}-...)$$

It's all geometry- once you write down your expressions for dA_2 and rho, it's simply a matter of integrating over dA_2.

Hope this helps. A great source is "Introduction to Radiometry" by Wolfe.

4. Jul 16, 2010

### mnb96

Uhm...I am trying to follow your steps, but there is something I don't understand.
Let's try to spot my mistake:

1) $dA_1 = dA$
2) $cos(\theta_1)=cos(\theta_2)=cos(\theta)$
3) $dA_2=r\cdot dr d\phi$

Let's substitute 1)2)3) into dZ:

$$dZ = \frac{cos^2(\theta) \cdot dA \cdot r dr d\phi}{\rho^2}$$

where does that $$\mathbf{d\theta}$$ in your equations come from???
I noticed it appears also in the book you mentioned, but I can't follow this step.

-- Another question is: why you didn't consider the following:

$$\rho^2 = h^2 + r^2$$

$$\cos^2(\theta) = \frac{h^2}{h^2+r^2}$$

5. Jul 16, 2010

### Andy Resnick

heh- you caught me, I cribbed the derivation directly from the book... :)

Ok- I went from my Eq 1 through Eq. 3 and got the correct answer:

$$Z = \frac{dA_{1}dA_{2}cos(\theta_{1})cos(\theta_{2})}{\rho^{2}}$$

$$dA_{1} = dA$$
$$dA_{2} = r dr d\phi$$
$$cos(\theta_{1}) = cos(\theta_{2}) = cos(\theta)$$
$$r = h tan(\theta)$$

I erred: $$\rho = h/cos(\theta)$$

We substitute:
$$Z = \frac{dA cos^{2}(\theta) r dr d\phi}{h^{2} sec^2(\theta)}$$

now substitute for r and dr:
$$dr = h sec^{2}(\theta) d\theta$$

$$Z = dA cos(\theta) sin(\theta) d\theta d\phi}$$

6. Jul 17, 2010

### mnb96

Thanks!
Now I was finally able to follow all the steps ;)

To conclude, I have a last and more general question related to the study of radiometry.
When you deal with radiometric quantities, how do you usually picture them in your mind? For example when you think of radiant flux do you imagine a certain quantity of photons flowing in many directions through a differential area (in the unit time)?

Also, all the radiometry books I opened surprisingly avoid introducing rigor in the definitions. This is giving me a lot of frustration. As a result, it is not yet clear to me whether:

- radiometric flux is a function of position and direction ?
- is radiometric flux a function at all? If so, what are the input variables?
- does it make sense to talk about radiometric flux through a point or it can be only through a differential surface patch?

Last edited: Jul 17, 2010
7. Jul 17, 2010