CS647 Assignment 2: Radiometry

1. Because an angular wedge of the hemisphere sky is not visible at the differential area, we can subtract the projected solid angle from to get the new projected solid angle. It’s equivalent to directly compute the projected solid angle of the whole hemisphere without the wedge part (as the picture shows, a unit hemisphere). So the new projected solid angle will be (the projected solid angle of the left part) plus (the projected solid angle of the right part). So, the irradiance at the center point is:

2. We can divide the square light source by the two diagonals into 4 right-angled triangles and since it’s symmetric we only need to compute one of them and multiply the result by four. To compute the projected solid angle of one such triangle, connect the two triangle vertices with the hemisphere center, so it’s clear that the projected area of that sector (formed by the two dotted lines and the arc, which is the projection of the third triangle edge on the hemisphere) is what we want. Using triangle equations, the angle of this sector is, so its area is. The sector has a 45 degree angle with the x-y plane, so the projected area is. So the irradiance at the center is:

3. The irradiance where, and since here the source is from a differential area, the radiance goes down with. So the irradiance

it goes with