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\section*{Chapter 6 Solutions}
6.1 Point sources produce a very harsh lighting. Such images are
characterized by abrupt transitions between light and dark.
The ambient light in a real scene is dependent on {\em both} the
lights on the scene and the reflectivity properties of the objects
in the scene, something that cannot be computed correctly with OpenGL.


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6.3 If we were to take into account a light source being obscured
by an object, we would have to have all polygons available so as
to test for this condition. Such a global calculation is incompatible
with the pipeline model that assumes we can shade each polygon independently
of all other polygons as it flows through the pipeline.

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6.5 Materials absorb light from sources. Thus, a surface that appears
red under white light appears so beacuse the surface absorbs all wavelengths
of light except in the red range---a subtractive process. To be compatible
with such a model, we should use  surface absorbtion constants
that define the materials for cyan, magenta and yellow, rather than
red, green and blue.

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6.7 Let $\psi$ be the angle between the normal and the halfway vector,
$\phi$ be the angle between the viewer and the reflection angle, and $\theta$
be the angle between the normal and the light source. If all the vectors 
lie in the same plane, the angle between the light source and the viewer
can be computer either as $\phi+2\theta$ or as $2(\theta+\psi)$. Setting
the two equal, we find $\phi=2\psi$. If the vectors are not coplanar then
$\phi<2\psi$.

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6.13 Without loss of generality, we can consider the problem in two dimensions. 
Suppose that the first material
has a velocity of light of $v_1$ and the second material has a light
velocity of $v_2$. Furthermore, assume the axis $y=0$ separates
the two materials.

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Place a point light source at $(0,h)$ where $h>0$ and a viewer at
$(x,y)$ where $y<0$. Light will travel in a straight line from the
source to a point $(t,0)$ where it will leave the first material and
enter the second. It will then travel from this point in a straight
line to $(x,y)$. We must find the $t$ that minimizes the time traveled.

Using some simple trigonometry, we find the line from the source to
$(t,0)$ has length $l_1=\sqrt{h^2+t^2}$ and the line from there to the viewer 
has length  $l_2=\sqrt{y^2+(x-t)^2}$.
The total time light travels is thus $\frac{l_1}{v_1}+\frac{l_2}{v_2}$.
Minimizing over $t$ gives desired result when we note the two desired
sines are $\sin \theta_1=\frac{h}{\sqrt h^2+t^2}$ and
$\sin \theta_2=\frac{-y}{\sqrt(y^2+(x-t)^2}$.

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6.15 The transmiitted light obeys the equation
$${\bf T}=\frac{1}{\eta}({\bf L}+(\cos \theta_2-\frac{1}{\eta}\cos\theta_1){\bf N}$$
If the most light is to be transmitted in the direction of a viewer behind
the surface, we want to ${\bf T}$ by a vector proportional to ${\bf V}$.
Hence, 
$${\bf N}=-\alpha ({\bf V}+\frac{1}{\eta}{\bf L})$$.
Normalizing, we find
$${\bf N}=-\frac{1}{\eta-1} ({\bf V}+\frac{1}{\eta}{\bf L})$$.

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