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# System prepended metadata

title: Discussion Errata

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# Discussion Errata

Hi everyone,

I made some mistakes while explaining radiometry concepts today, and here are some corrections. Apologies for sending a notification to everyone, but this is the most reliable way to reach out to everyone who attended my section, and I thought alternative explanations of why radiance is constant over a distance may be helpful to many. So here we go!

Erratum 1. In my discussion today, I talked about why radiance is constant with respect to distance:![Screen_Shot_2022-02-23_at_2.18.52_PM.png](/redirect/s3?bucket=uploads&prefix=paste%2Fjg6lw88k7nv447%2Ff7fbe03543eed747274a44b79c381ee87effa77a3a606ffdb3b4d5fe47a88cce%2FScreen_Shot_2022-02-23_at_2.18.52_PM.png)

When we compared the solid angles at surface 2 vs surface 1, I said we can hold the area of a test surface constant at both distances, in which case the solid angle at surface 2 becomes $$1/r^2$$ of the surface 1. However, this does not match with the image displayed (the image suggests that we should consider the solid angle subtended by the light source instead), which was very confusing and I apologize for the inconsistency. In this post, I will offer two explanations - one that matches with the diagram shown in the discussion, and another one that matches with the explanation that I gave in the discussion (but this time with a separate correct diagram). Both explanations should be correct, but the models of the light sources are slightly different.

### Explanation 1 (Area Light)

As shown in the discussion slide above, let’s assume we are dealing with a uniform area light.

**Irradiance:** As shown in [lecture](https://cs184.eecs.berkeley.edu/sp22/lecture/11-29/radiometry-and-photometry), irradiance falls off proportional to $$\frac{1}{r^2}$$. This is kind of hard to show with an area light, but rather intuitive with a point light (see explanation 2).

If we believe that irradiance falls off proportional to $$\frac{1}{r^2}$$, and if the irradiance at surface 1 is $$E$$, then the irradiance at surface 2 is $$\frac{E}{r^2}$$.

**Solid Angle:**

Earlier in the discussion, we’ve seen that solid angle is defined as $$\Omega = \frac{A}{r^2}$$.

Let the area of the light be $$A$$.

At surface 1, the solid angle subtended by the light is given to us as $$\omega = \frac{A}{1^2} = A$$.

At surface 2, the solid angle subtended by the light is $$\frac{A}{r^2} = \frac{\omega}{r^2}$$

**Radiance:**

Using the equation $$L = \frac{dE}{d\omega \cos{\theta}}$$, we see that the $$\frac{1}{r^2}$$ terms cancel out for surface 2, which leaves us with $$L\_1 = L\_2$$. (For both surfaces, because we are interested in the radiance directed perpendicular to the surfaces, $$\cos{\theta} = \cos{0} = 1$$).

### Explanation 2 (Point Light)

Here’s an explanation of why radiance is constant for a point light (which does not have an area). Again, we assume the irradiance at surface 1 is $$E$$, and the solid angle at surface 1 is $$\omega$$.

![IMG_0A18433F2A2F-1.jpeg](/redirect/s3?bucket=uploads&prefix=paste%2Fjg6lw88k7nv447%2F5743d6f22b2fae2360ded8302543695910e6444face116713bc700ae0c66f001%2FIMG_0A18433F2A2F-1.jpeg)

**Irradiance:** At surface 1, the total area of the sphere enclosing the entire light source is $$4\pi 1^2=4\pi$$. Therefore, the irradiance is

$$E = \frac{d\Phi(p\_1)}{dA\_1}$$

$$= \frac{\Phi}{A\_1}$$ (by symmetry, the irradiance of a point light is uniform at a given radius)

$$= \frac{\Phi}{4\pi}$$ (substitute the area)

At surface 2, the total area of the sphere enclosing the entire light source is $$4\pi r^2$$. Therefore, our irradiance at surface 2 is

$$\frac{d\Phi(p\_2)}{dA\_2} = \frac{\Phi}{A\_2} = \frac{\Phi}{4\pi r^2} = \frac{\Phi}{4\pi} \* \frac{1}{r^2} = E \* \frac{1}{r^2}$$

**Solid Angle:** We don’t have an area occupied by the light readily available anymore. However, we can still use our test area at both surface 1 and surface 2 to calculate solid angles for use in our radiance equation.

Let there be test areas at surfaces 1 & 2, both of which have the same differential (i.e. very small) area $$\epsilon$$. We use a differential area here because we want the quantity that we are measuring (in this case the radiance) to be uniform across the area.

At surface 1, let’s again assume the solid angle is $$\omega = \frac{A\_1}{r^2} = \frac{\epsilon}{1^2} = \epsilon$$.

At surface 2, our solid angle is $$\frac{A\_2}{r^2} = \frac{\epsilon}{r^2} = \frac{\omega}{r^2}$$

**Radiance:**

Again, using the equation $$L = \frac{dE}{d\omega \cos{\theta}}$$, we see that the $$\frac{1}{r^2}$$ terms cancel out for surface 2, which leaves us with $$L\_1 = L\_2$$

Erratum 2. In my discussion today, I made another mistake of implying that radiant intensity may change if we move a differential test surface farther away from a point light source while keeping the test surface’s size constant. This is completely incorrect, because as the test surface moves away, although it receives less power from the light source, the solid angle that it subtends also decreases at the same rate ($$1/r^2$$), so radiant intensity stays constant.

It’s good to remember that out of all the radiometry quantities we’ve learned, the only quantity that decreases with distance is **irradiance**.

Hope that was helpful! If you have any questions, feel free to leave them down below, or email me at markz@berkeley.edu. Have a good evening everyone!
