The angle measure between vectors remains unchanged if the systems of equations which define them are translated, scaled or rotated. Might as well check.
In the image below, N is no longer aligned with the z-axis. The surface has been rotated by an arbitrary, positive angle, φ (phi).
Phi, fie, fo fum.
Ummmmmmmmm....
Oh. Right. The math.
We have, again in two parts which may be done in any order:
\\ (1)\;\;\;\;\;\;\;&\boldsymbol{I} = (-\sin(\theta_1 \!+ \phi),\, \; \cos(\theta_1 \!+ \phi))\\ &\boldsymbol{T} = (\sin(\theta_2 + \phi), \;\,-\cos(\theta_2 + \phi)) \end{align*})
The dot products:
\;\;\;\;\;\;\begin{align*} \boldsymbol{I \!\cdot \!N} &= \, \sin(\theta_1 + \phi)\sin\phi \;+\; \cos(\theta_1 +\phi)\cos\phi\\ &= \cos([\theta_1 + \phi]-\phi) \\ &= \cos\theta_1\\ \boldsymbol{T\!\cdot \!(-N)} &= \sin(\theta_2 + \phi)\sin \phi \;+\; \cos(\theta_2 + \phi)\cos\phi\\ &= \cos([\theta_2 + \phi]-\phi) \\ &= \cos\theta_2\\ \end{align*})
And so, as before:
![sin θ1 = √(1 − (I•N)^2), sin θ2 = (η1/η2)√(1−(I•N)^2), cos θ2 = √(1 − (η1/η2)^2 (1 − [I•N]^2))](http://latex.codecogs.com/gif.latex?(3) \begin{align*} \;\;\;\;\;\;\; \sin\theta_1\;\; &= \;\;\;\; \sqrt{1-\cos^2\theta_1}\;\;\;\; =\;\;\; \sqrt{1-(I\!\cdot\! N)^2}\\ \sin\theta_2 \;\; &= (\eta_1/\eta_2)\sqrt{1-\cos^2\theta_1}\;\; =\; (\eta_1/\eta_2)\sqrt{1-(I\!\cdot\! N)^2}\\ \cos\theta_2 \;\; &=\;\;\;\; \sqrt{1-\sin^2 \! \theta_2} \;\; =\;\;\sqrt{1-(\eta_1/\eta_2)^2(1-[\boldsymbol{I\!\cdot \!N}]^2)} \end{align*})
The sign of the square root is taken positive, as defined in the last section. I makes a positive angle with N less than 90° (π/2 radians).
I. Calculate T for any η1, η2, N and I
The dot products:
And so, as before:
The sign of the square root is taken positive, as defined in the last section. I makes a positive angle with N less than 90° (π/2 radians).
II. Give T as the sum of two vectors
αI + βN = T, α, β constant:
\;\;\;\;\;\;\; \begin{align*} \alpha\! \begin{bmatrix} -\sin(\theta_1 + \phi)\\ \cos(\theta_1+\phi)\end{bmatrix} +\beta\! \begin{bmatrix} -\sin\phi \\ \cos\phi \end{bmatrix}= \begin{bmatrix} \;\sin(\theta_2 + \phi)\\ -\cos(\theta_2 + \phi) \end{bmatrix} \\ \\ \begin{bmatrix} -\sin(\theta_1 + \phi) &-\sin\phi \\ \cos(\theta_1+\phi) &\cos\phi \end{bmatrix} \begin{bmatrix} \alpha\\ \beta \end{bmatrix}= \begin{bmatrix} \;\sin(\theta_2 + \phi)\\ -\cos(\theta_2 + \phi) \end{bmatrix} \end{align*})
Using Cramer's Rule:
 &-\sin\phi \\ \cos(\theta_1 + \phi) &\cos\phi \end{vmatrix} \\ &=-[\sin(\theta_1 \!+ \phi)\cos\phi \;-\; \cos(\theta_1 \!+\phi)\sin \phi]\\ &=-\sin([\theta_1 \!+ \phi] - \phi)\\ &=-\sin\theta_1 \end{align*})
Solve for alpha:
 &-\sin\phi \\ -\cos(\theta_2 \!+ \phi) &\cos\phi \end{vmatrix}} {\Delta} \;=\; \frac {\sin(\theta_2 \!+ \phi)\cos\phi\,-\,\cos(\theta_2 \!+ \phi)\sin\phi} {-\sin\theta_1} \\&=\; \frac {\sin([\theta_2\!+\phi] - \phi)}{-\sin\theta_1} \;\;=\;\; -\frac {\sin\theta_2}{\sin\theta_1} \\&= -\frac{\eta_1}{\eta_2} \end{align*})
Solve for beta:
![β = (η1/η2)[I•N] − √(1 − (η1/η2)^2(1 − [I•N]^2) )](http://latex.codecogs.com/gif.latex?\begin{align*} \boldsymbol{\beta} \; &=\, \frac{\begin{vmatrix} -\sin(\theta_1\!+ \phi) &\sin(\theta_2\!+\phi)\\ \cos(\theta_1 \!+ \phi) &-\cos(\theta_2\!+\phi) \end{vmatrix}} {\Delta}\\ &=\; \frac {\sin(\theta_1 \!+ \phi)\cos(\theta_2\!+\phi)\,-\,\cos(\theta_1 \!+ \phi)\sin(\theta_2\!+\phi)} {-\sin\theta_1}\\ &=\; \frac {\sin([\theta_1\!+\phi] - [\theta_2 \!+ \phi])}{-\sin\theta_1}\;\;\;\;\; =\; \frac {\sin(\theta_1-\theta_2)}{-\sin\theta_1}\\ &= \frac {\sin\theta_1 \cos\theta_2 - \cos\theta_1\sin\theta_2}{-\sin\theta_1} \;=\; \cos\theta_2 \,+\,\cos\theta_1 \left(\frac {\sin\theta_2}{\sin\theta_1}\right )\\ &=\frac{\eta_1}{\eta_2}\cos\theta_1 \,-\,\cos\theta_2 \\ &= (\eta_1/\eta_2)[I \!\cdot \! N] - \sqrt{1 -(\eta_1/\eta_2)^2(1-[I \! \cdot\! N]^2)} \end{align*})
Radness. And at last:
![T = αI + βN = −(η1/η2)I + ( (η1/η2)[I•N] − √(1 − (η1/η2)^2(1 − [I•N]^2) ) ) N](http://latex.codecogs.com/gif.latex?\begin{align*} \boldsymbol{T} &= \alpha\boldsymbol{I}\; +\; \beta\boldsymbol{N}\\ &= -(\eta_1/\eta_2)\boldsymbol{I} + \left ( (\eta_1/\eta_2)[I\! \cdot\! N] -\sqrt{1-(\eta_1/\eta_2)^2(1-[I\! \cdot \! N]^2)} \right )\!\boldsymbol{N} \end{align*})
...being the same equation as Parts I and II. So don't do it this way, because it's harder.
(It's over now.)
Using Cramer's Rule:
Solve for alpha:
Solve for beta:
Radness. And at last:
THE RESULT
...being the same equation as Parts I and II. So don't do it this way, because it's harder.
(It's over now.)
{CodeCogs you are the warp and weft of my Latex heart}
{<−− Part III} {Part V upcoming}
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