4
$\begingroup$

A question related to radiometry:

Irradiance $E$ at a point $x$ can be written as:

$$E = \int_\Omega L(x, \omega) cos(\theta) d\omega$$

I understand this formula and where it comes from. The equation for radiance can be written as:

$$L = {d^2\Phi \over {d\omega dA^\perp cos(\theta)}}$$

What I don't understand is if we substitute this last equation in the first one for irradiance, don't we get:

$$E = \int_\Omega {d^2\Phi \over {d\omega dA^\perp cos(\theta)}} cos(\theta) d\omega \rightarrow \int_\Omega {d^2\Phi \over {dA^\perp}} \rightarrow \int_\Omega d({d\Phi \over dA^\perp}) \rightarrow \int_\Omega dE$$

which doesn't make sense to me? What am I missing? Is irradiance the integral of differential irradiance over the hemisphere?

$\endgroup$
2
  • $\begingroup$ If I understand your notation right, the far LHS of your equation $E= \cdots \rightarrow \int_\Omega dE$ is the total optical power through the surface $\Omega$, whereas you use $E$ for irradiance elsewhere. Is this right? $\endgroup$ Jul 24, 2013 at 4:31
  • $\begingroup$ I am not too sure to understand what you mean, sorry. I have used E for irradiance and dE for differential irradiance! I just would like to confirm if it is technically to say that irradiance (w/m^2) can be computed from integrating differential irradiance over the hemisphere which doesn't makes sense to me!? I would really like someone to confirm. Thank you. $\endgroup$
    – user18490
    Jul 26, 2013 at 13:41

2 Answers 2

3
$\begingroup$

Radiance is the correct term for what we loosely describe as "brightness."

Its units are Watts per square meter, per steradian.

Irradiance is simply Watts per square meter, and describes the power per unit of area impinging on a surface.

Radiance is invariant under all optical transformations (reflection, refraction, etc.)

The only way to change the radiance is to introduce statistical processes (scattering by diffuse materials).

This always involves energy losses, due to large angle and back scattering, and absorption in the diffusing medium.

The constancy of radiance is a consequence of the second law of thermodynamics. If it were possible to losslessly change radiance, we could increase the radiance of an emission from a black body source, and impinge it on a second black body, which would drive that body to a higher temperature than the source. Because of bi-directionality, if you can't increase radiance, you can't decrease it either (losslessly).

$\endgroup$
2
  • 1
    $\begingroup$ You might also like to mention that if you're used to thinking about light in terms of Maxwell's equations or quantum optics, you'll likely to use the word "Intensity" for "irradiance". There is a great deal of confusion between words and I am ashamed to admit I have to look all these words up every time I use them. Great answer BTW. $\endgroup$ Jul 24, 2013 at 4:32
  • $\begingroup$ Thank you for your answer George but ultimately it doesn't answer my question is which is to know/confirm that irradiance (w/m^2) can be computed from integrating differential irradiance over the hemisphere. This is what I would like to know. Thank you again. $\endgroup$
    – user18490
    Jul 26, 2013 at 13:42
0
$\begingroup$

In your expansion of the integral equation for irradiance, you should expand like this:

$$E(x) = \int_\Omega {d^2\Phi \over {d\omega dA cos(\theta)}} cos(\theta) d\omega \rightarrow \int_\Omega {dE(x) \over {d\omega}} d\omega \rightarrow \int_\Omega dE(x).$$

This will add up all the differential irradiances over the point, resulting in the total irradiance at that point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.