diff --git a/exam-2/exam2.md b/exam-2/exam2.md index 7aa58f2..6ae2d36 100644 --- a/exam-2/exam2.md +++ b/exam-2/exam2.md @@ -15,8 +15,6 @@ author: | \newcommand{\now}[1]{\textcolor{blue}{#1}} \newcommand{\todo}[0]{\textcolor{red}{\textbf{TODO}}} -[ 3 8 9 ] - ## Reflection and Refraction 1. \c{Consider a sphere $S$ made of solid glass ($\eta$ = 1.5) that has radius @@ -186,7 +184,7 @@ author: | $z$-axis, the rotation matrix would look like: $$ - M = + M_1 = \begin{bmatrix} \cos(23.5^\circ) & \sin(23.5^\circ) & 0 & 0 \\ -\sin(23.5^\circ) & \cos(23.5^\circ) & 0 & 0 \\ @@ -205,15 +203,15 @@ author: | $y$-axis, so the matrix looks like: $$ - M' = - M^{-1} + M_2 = + M_1^{-1} \begin{bmatrix} \cos(\theta(t)) & 0 & \sin(\theta(t)) & 0 \\ 0 & 1 & 0 & 0 \\ -\sin(\theta(t)) & 0 & \cos(\theta(t)) & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} - M + M_1 $$ c. \c{[5 points extra credit] What series of rotation matrices could you use @@ -222,7 +220,39 @@ author: | In the image, the globe itself does not rotate, but I'm going to assume it revolves around the sun at a different angle $\phi(t)$. The solution here - would be to + would be to rotate the globe _after_ the translation to whatever its position + is. + + - First, rotate the globe to the $23.5^\circ$ tilt, using $M_2$ as described + above. + + - Then, translate the globe to its position, which I'm going to assume is + $(r, 0, 0)$. + + $$ + M_3 = + \begin{bmatrix} + 1 & 0 & 0 & r \\ + 0 & 1 & 0 & 0 \\ + 0 & 0 & 1 & 0 \\ + 0 & 0 & 0 & 1 \\ + \end{bmatrix} + $$ + + - Finally, rotate by $\phi(t)$ around the $y$-axis _after_ the translation, + using: + + $$ + M_4 = + \begin{bmatrix} + \cos(\phi(t)) & 0 & \sin(\phi(t)) & r \\ + 0 & 1 & 0 & 0 \\ + -\sin(\phi(t)) & 0 & \cos(\phi(t)) & 0 \\ + 0 & 0 & 0 & 1 \\ + \end{bmatrix} + $$ + + The resulting transformation matrix is $M = \boxed{M_4 M_3 M_2}$. ## The Camera/Viewing Transformation