• When light falls on the surface of an object it can be:
  • Absorbed (and converted to heat)
  • Reflected
  • Transmitted (i.e., moved through the object)
• Visible objects:
  • Reflect some light, in some wavelengths, to the eye

• Properties of light that impact reflection:
  • Wavelengths and intensity of the light
  • Direction and geometry of the source
• Properties of objects that impact reflection:
  • Surface orientation
  • Surface properties (e.g., roughness, conductivity)

• Perfectly Transparent Materials:
  • Transmit all of the light
• Opaque Materials:
  • Materials that reflect some of the light and perhaps absorb some of the light
• Translucent Materials:
  • Transmit (and perhaps refract/bend) some of the light and reflect some of the light (including Dielectric materials, which are insulators that can be polarized by an electric field)

• Diffuse Reflection:
  • A perfect diffuser reflects light equally in all directions making the object appear matte/flat and appear the same to all viewers
• Specular Reflection:
  • The light isn't scattered (i.e., all of the light is reflected at one angle) and its properties are (essentially) unchanged making the object appear shiny

• Assume:
  • A point source of light
  • A perfect diffuser (sometimes called a Lambertian surface)
• Lambert's Law:
  • The intensity of light reflected is proportional to the cosine of the angle between the light source and the surface normal

• Simplifying the Notation:
  • We will develop the model for the point $\bs{0}$
  • $\bs{l}$ denotes the (normalized) vector for the light source (which is also the position of the light source since the point of interest is $\bs{0}$)
  • $\bs{n}$ denotes the surface normal vector (which is perpendicular to the surface at the point)
• Visualization:
  • 
• Lambert's Law:
  • For each color (or wavelength):
  • $D = L \kappa_{D} \cos \theta$
  • where $D$ is the intensity of the diffuse reflection, $L$ is the intensity of the light source, and $\kappa_{D}$ is an "absorption parameter"

• From the Law of Cosines:
  • $\cos \theta = \frac{\bs{n} \cdot \bs{l}} {||\bs{n}|| \enspace ||\bs{l}||}$
  • Since $\bs{n}$ and $\bs{l}$ have been normalized (i.e., have unit length) we know $||\bs{n}|| \enspace ||\bs{l}|| = 1$
  • Hence, $\cos \theta = \bs{n} \cdot \bs{l}$
• Substituting into the Model:
  • For each color (or wavelength):
  • $D = L \kappa_{D} \bs{n} \cdot \bs{l}$

• For Perfectly Reflecting Surfaces:
  • The angle of reflection equals the angle of incidence
• We Will Assume:
  • A point source of light
  • An imperfect reflecting surface

• Notation:
  • $\bs{r}$ denotes the reflected ray
  • $\bs{v}$ denotes the location of the viewer
• Visualization:
  • 
• An Empirical Model for Imperfect Reflectors (Romney):
  • For each color (or wavelength):
  • $S = L \kappa_{S} \cos^{\delta} \phi$
  • where $S$ is the intensity of the specular reflection, $\delta$ is a "shininess" parameter (that is often specific to a material), and $\kappa_{S}$ is an "absorption parameter"

• From the Law of Cosines:
  • Assuming that $\bs{r}$ and $\bs{v}$ have been normalized (i.e., have unit length):
  • $\cos^{\delta} \phi = (\bs{r} \cdot \bs{v})^{\delta}$
• Substituting into the Model:
  • For each color (or wavelength):
  • $S = L \kappa_{S} (\bs{r} \cdot \bs{v})^{\delta}$

• Ambient Light:
  • We will assume the ambient intensity is given by $A \kappa_{A}$
• An Attenuation Effect:
  • To model attenuation, we will divide the non-ambient intensities by $\beta(d)$ where $\beta$ is some function and $d$ is the distance.
• The Complete Model:
  • $I = A \kappa_{A} + \frac{L}{\beta(d)} [\kappa_{D} (\bs{n} \cdot \bs{l}) + \kappa_{S} (\bs{r} \cdot \bs{v})^{\delta}]$
  • where $I$ denotes the total intensity

• Incorporating Multiple Light Sources:
  • Have non-ambient terms for each light source and add them up
• Incorporating Multiple Wavelengths/Colors Components:
  • People often have a separate model for each color component (e.g., red, green, and blue)

• Parameters:
  • $\beta(d) = 1$ for all $d$
  • $A = 1$, $L = 10$
  • $\kappa_{S} = 0.65$, $\kappa_{D} = \kappa_{A} = 0.15$ (Note: $\kappa_{S} + \kappa_{D} + \kappa_{A} = 0.95$ implies 5 percent of the light is absorbed)
  • $\delta=5$
• The Point of Interest:
  • $\bs{0}$
• Normal at $\bs{0}$:
  • $\bs{n} = [0.0 \quad 1.0 \quad 0.0]^T$
• The Location of the Light Source and the Viewer:
  • Light Source: $\bs{s} = [-1.0 \quad 2.0 \quad -1.0]^T$
  • Viewer: $\bs{e} = [1.0 \quad 1.5 \quad 0.5]^T$

• The Normalized Vectors at $\bs{0}$:
  • $\bs{n} = [0 \quad 1 \quad 0]^T$
  • $\bs{l} = \frac{[-1.0 \quad 2.0 \quad -1.0]^T} {|| [-1.0 \quad 2.0 \quad -1.0]^T ||} = [-0.408 \quad 0.816 \quad -0.408]^T$
  • $\bs{v} = \frac{[1.0 \quad 1.5 \quad 0.5]^T} {||[1.0 \quad 1.5 \quad 0.5]^T||} = [0.534 \quad 0.801 \quad 0.267]^T$
  • $\bs{r} = [0.408 \quad 0.816 \quad 0.408]^T$
• The Angles:
  • $\theta = \cos^{-1}(\bs{n} \cdot \bs{l}) = \cos^{-1}(0.816) = 35.26$
  • $\phi = \cos^{-1}(\bs{r} \cdot \bs{v}) = \cos^{-1}(0.982) = 10.89$

• The Intensity:
  • $I = (1)(0.15) + (\frac{10}{1}) [ (0.15)(0.816) + (0.65)(0.982^{5}) ] = 7.31$
• Interpretation:
  • The sight vector is almost coincident with the reflection vector so this point is bright

• For Triangles:
  • Use the cross product
• For Planar Polygons:
  • Use the equation of the plane

• Getting Started:
  • Normalize all of the vectors
• The Conditions that must be Satisfied:
  • The angle of incidence equals the angle of reflection
  • All of the rays lie in the same plane

• The First Condition:
  • $\theta_{i} = \theta_{r}$
  • $\Rightarrow \cos \theta_{i} = \cos \theta_{r}$
  • $\Rightarrow \bs{l} \cdot \bs{n} = \bs{n} \cdot \bs{r}$
• The Second Condition:
  • $\gamma \bs{n} = \bs{l} + \bs{r}$ for some real number $\gamma$
  • $\Rightarrow \bs{r} = \gamma \bs{n} - \bs{l}$
• Combining:
  • $\bs{l} \cdot \bs{n} = (\gamma \bs{n} - \bs{l}) \cdot \bs{n}$
  • $\Rightarrow \bs{l} \cdot \bs{n} = \gamma \bs{n} \cdot \bs{n} - \bs{l} \cdot \bs{n}$
  • $\Rightarrow \bs{l} \cdot \bs{n} = \gamma - \bs{l} \cdot \bs{n}$ (since $\bs{n}$ has a norm of 1)
  • $\Rightarrow \gamma = 2 \bs{l} \cdot \bs{n}$
• Substituting:
  • $\bs{r} = 2(\bs{l} \cdot \bs{n}) \bs{n} - \bs{l}$

• An Aside (for those who care):
  • $\bs{r} = 2(\bs{l} \cdot \bs{n}) \bs{n} - \bs{l}$
  • $\Rightarrow \bs{r} = 2[\bs{l}^{T} \bs{n}] \bs{n} - \bs{l}$
  • $\Rightarrow \bs{r} = 2[\bs{n} \bs{n}^{T}] \bs{l} - \bs{l}$
  • $\Rightarrow \bs{r} = (2[\bs{n} \bs{n}^{T}] - \bs{I}) \bs{l}$
• Some Observations:
  • $[\bs{n} \bs{n}^{T}]$ is a symmetric matrix with rank 1 (i.e., a dyadic matrix)
  • $\bs{H} = (2[\bs{n} \bs{n}^{T}] - \bs{I})$ is a Householder matrix (which is a type of matrix that is often used in the solution of linear systems)

• Be Careful:
  • There is often confusion involving the distinction between points and direction vectors
• Visualization:
  • 
• In practice:
  • We are often given the position of the light source, $\bs{s}$, (which usually isn't normalized) and we must work with (and normalize) the direction vector, $\bs{s}-\bs{p}$
  • Artists usually provide the normal vector, $\bs{n}$, at the point, $\bs{p}$ (and it is usually normalized)

• Parameters, Light Source and Viewer (As Before):
  • Light Source: $\bs{s} = [-1.0 \quad 2.0 \quad -1.0]^T$
  • Viewer: $\bs{e} = [1.0 \quad 1.5 \quad 0.5]^T$
• The Point of Interest (and the Normal):
  • $\bs{p} = [1.0 \quad -1.0 \quad 0.0]^T$
  • $\bs{n} = [1.0 \quad 0.0 \quad 0.0]^T$
• The Vectors of Interest:
  • $\bs{l} = \frac{\bs{s}-\bs{p}}{||\bs{s}-\bs{p}||} = \frac{[-2.0 \quad 3.0 \quad -1.0]^T} {3.742} = [-0.534 \quad 0.801 \quad -0.267]^T$
  • $\bs{v} = \frac{\bs{e}-\bs{p}}{||\bs{e}-\bs{p}||} = \frac{[0.0 \quad 2.5 \quad 0.5]^T} {2.549} = [0.000 \quad 0.981 \quad 0.196]^T$
  • $\bs{r} = 2(\bs{l} \cdot \bs{n})\bs{n} - \bs{l} = 2(-0.534)\left[\begin{array}{c}1.0\\0.0\\0.0\end{array}\right] - \left[\begin{array}{c}-0.534\\0.801\\-0.267\end{array}\right] = \left[\begin{array}{c}-0.534\\-0.801\\0.267\end{array}\right]$
• The Intensity:
  • $I = (1)(0.15) + (\frac{10}{1}) [ (0.15)(-0.534) + (0.65)(-0.734^{5}) ] = -2.036$

• Comparison of the Points:
  • The light source is to the left of $\bs{0}$; the normal at $\bs{0}$ is directed up
  • This point is to the right of and below $\bs{0}$; the normal at this point is directed away from the light source
• Comparison of the Intensities:
  • This point will be darker than the origin

• The Issues:
  • Some or all of the light is transmitted through the medium
  • The transmitted light may be refracted (i.e., bent)
• The Main Result:
  • Snell's Law describes the relationship between the angle of incidence and the angle of transmission when light passes through a boundary between two isotropic (i.e., uniform) media

Understanding the Transmission/Refraction Angle

Where $\eta_i$ and $\eta_t$ denote the index of refraction of the initial medium and transmitting medium, respectively, $\bs{n}$ denotes the normal vector, and $\bs{b}$ is chosen so as to form a basis with $\bs{n}$ (i.e., $\bs{b} = \frac{\bs{i} + \cos(\theta_i)\bs{n}}{\sin(\theta_i)}$).

• Snell's Law:
  • $\frac{\sin(\theta_i)}{\sin(\theta_t)} = \frac{\eta_t}{\eta_i}$
• The Implication:
  • $\bs{t} = \frac{\eta_i}{\eta_t} \left(\bs{i} - (\bs{i} \cdot \bs{n}) \bs{n}\right) - \bs{n} \sqrt{1 - \frac{\eta_i^2}{\eta_t^2} \left(1 - (\bs{i} \cdot \bs{n})^2\right)}$
• A Special Case:
  • When the square root is imaginary (which can only happen when $\eta_t \lt \eta_i$) the light is reflected internally (i.e., total internal reflection)

1. From Snell's Law: $\sin(\theta_t) = \frac{\eta_i}{\eta_t} \sin(\theta_i)$
2. $\sin^2(\alpha) + \cos^2(\alpha) = 1$ for any angle, $\alpha$, so:
   $\sin^2(\alpha) = \left(1 - \cos^2(\alpha) \right) \Rightarrow \sin(\alpha) = \sqrt{\left(1 - \cos^2(\alpha) \right)} \\ \cos^2(\alpha) = \left(1 - \sin^2(\alpha) \right) \Rightarrow \cos(\alpha) = \sqrt{\left(1 - \sin^2(\alpha) \right)}$
3. Substituting the first equation in (2) into (1): $\sin(\theta_t) = \frac{\eta_i}{\eta_t} \sqrt{\left(1 - \cos^2(\alpha) \right)}$
4. From our study of the standard basis and and the unit circle:
   $\bs{t} = \sin(\theta_t)\bs{b} - \cos(\theta_t)\bs{n} \\ \bs{i} = \sin(\theta_i)\bs{b} - \cos(\theta_i)\bs{n}$
5. From the second equation in (4): $\sin(\theta_i)\bs{b} = \bs{i} + \cos(\theta_i)\bs{n}$
6. Substituting Snell's Law into the first equation in (4): $\bs{t} = \frac{\eta_i}{\eta_t} \sin(\theta_i)\bs{b} - \cos(\theta_t)\bs{n}$
7. Substituting (5) into (6): $\bs{t} = \frac{\eta_i}{\eta_t} \left( \bs{i} + \cos(\theta_i)\bs{n} \right) - \cos(\theta_t)\bs{n}$
8. Since $\cos(\theta_i) = -\bs{i} \cdot \bs{n}$, it follows from (7) that:
   $\bs{t} = \frac{\eta_i}{\eta_t} \left(\bs{i} - (\bs{i} \cdot \bs{n}) \bs{n} \right) - \cos(\theta_t) \bs{n}$
9. Substituting (3) into the second equation in (2): $\cos^2(\theta_t) = 1 - \frac{\eta_i^2}{\eta_t^2} \left( 1 - \cos^2(\theta_i) \right)$
10. Substituting (9) into (8): $\bs{t} = \frac{\eta_i}{\eta_t} (\bs{i} - (\bs{i} \cdot \bs{n}) \bs{n}) - \bs{n} \sqrt{1 - \frac{\eta_i}{\eta_t^2} (1 - \cos^2(\theta_i)}$
11. Since $\cos(\theta_i) = -\bs{i} \cdot \bs{n}$ and $(-\cos(\theta_i))^2 = \cos^2(\theta_i)$:
    $\bs{t} = \frac{\eta_i}{\eta_t} (\bs{i} - (\bs{i} \cdot \bs{n}) \bs{n}) - \bs{n} \sqrt{1 - \frac{\eta_i^2}{\eta_t^2} \left( 1 - (\bs{i} \cdot \bs{n})^2 \right)}$

• A Simplifying Assumption:
  • Assume all objects are in the air with an index of refraction of 1 and surface normals that point towards the air
• The Implication:
  • You can determine whether a vector is headed into or out of an object using the sign of the inner product of the vector and the surface normal
  • 

• Fresnel's Equations:
  • Describe the ratio of the intensity of the reflected and transmitted light
• Schlick's Approximation of the Reflection Coefficient:
  • $R(\theta) = R_0 + (1 - R_0) \left( 1 - \cos(\theta)\right)^5$
  • where:
  • $R_0 = \left(\frac{\eta_i - \eta_t} {\eta_i + \eta_t}\right)^2$