I was kind of deterred by the wording of the last sentence of the slide, and was still slightly uncertain after lecture - If we have half the distance (not the square of the distance, but the distance itself), the number of rays should quadruple. My heuristic was that this is because the solid angle subtended by either dA element would quadruple as a result, due to the formula that the differential solid angle depends on the differential area element divided by radius squared. Therefore, assuming rays distributed uniformly in all directions from every point, 4x the differential solid angle corresponds to 4x the count of rays.

I was kind of deterred by the wording of the last sentence of the slide, and was still slightly uncertain after lecture - If we have half the distance (not the square of the distance, but the distance itself), the number of rays should quadruple. My heuristic was that this is because the solid angle subtended by either dA element would quadruple as a result, due to the formula that the differential solid angle depends on the differential area element divided by radius squared. Therefore, assuming rays distributed uniformly in all directions from every point, 4x the differential solid angle corresponds to 4x the count of rays.