Image Synthesis Assignment 2: Radiometry

Due: February 15

Description

The goal of this written assignment is to first do some exercises in radiometry, and then to work on camera models. The mathematics that you derive for camera models will be used in your next programming assignment.

Problem 1

Assume the sky has constant radiance L over the entire upper hemisphere (it is a perfectly cloudy day). In class, we showed that a constant radiance hemispherical source would contribute irradiance E = π L to a point on the ground.

Suppose at some point on the terrain the ground plane is tilted by an angle θ with the respect to the horizon. These means a angular wedge of the hemispherical sky will not be visible at that point because parts of the sky are below the local horizon of the surface.

What is the irradiance E at this point?

Answer 1

When a sphere is tilted by a given θ, the projected area onto the base of the unit sphere is reduced only on the side opposite the rotation. Thus in order to calculate the irradiance we need only to determine how this area decreases as a function of theta. The following picture details this:

The area of the wedge in question is given by projecting the total area of a half circle onto this plane. This is accomplished by multiplying by a cosθ term because the radius is 1. Plugging this in gives:

Giving As a check notice that when θ = 0, we get the normal solution E = π L and when we rotate to π/2 we get half the original irradiance because half of the sphere is exposed.

Problem 2

Consider a square light source with constant radiance L and vertices (-1,-1,1), (-1,1,1), (1,1,1), and (1,-1,1).  What is the irradiance E at the point (0,0,0), assuming that the surface at (0,0,0) has a normal vector of (0,0,1) [i.e., it points straight towards the center of the square]?

Answer 2

First we note that if we split the square into 4 triangles from the center we can reduce the problem to finding the area projected on to a unit sphere's base centered at (0,0,0). Thus we need only to calculate the area of one of the triangles and multiply by 4. The projection of one of these triangles is given by the following equation which governs the area of a wedge of a circle:

Where θ is π/2 and R is calulated by sin(arctan(√(2))). By substituting in the proper values we get E = 4L[(-1+π/2)/3] = 4/3(-1+π/2)L

Problem 3

A real optical system exhibits vignetting. Vignetting causes the exposure to vary from a maximum at the center of the image to a lesser value at the edges.

Assume you have a camera with a infinitesimal disk aperture da (e.g., not a pinhole camera).  Behind the aperture at distance equal to the focal length f is the film plane. Show that the irradiance on the film plane falls off as cos^4 θ.

Answer 3

First of all, we need to write down the form of irradiance at a plane from a circular source. This form goes as follows:

From the following picture it's easy to see that θi = θo = θ and r = f/cos(θ):

Where

Plugging these facts into the equation above gives:

Which clearly shows that the irradiance falls of as cos^4 θ.