Reflection Vector (Ray Tracing) - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-09-20T18:26:31Z https://physics.stackexchange.com/feeds/question/454361 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/454361 0 Reflection Vector (Ray Tracing) Doubloon Conroy https://physics.stackexchange.com/users/167081 2019-01-15T12:20:40Z 2019-01-16T08:32:09Z <p>Snell's law of refraction at the interface between 2 isotropic media is given by the equation: <span class="math-container">\begin{align} \tag{1} n_1 \,\text{sin} \,\theta_1 = n_2 \, \text{sin}\,\theta_2 \end{align}</span></p> <p><span class="math-container">$\qquad$</span> where <span class="math-container">$\theta_1$</span> is the angle of incidence and <span class="math-container">$\theta_2$</span> the angle of refraction. <span class="math-container">$n_1$</span> is the refractive index of the optical medium in front of the interface and <span class="math-container">$n_2$</span> is the refractive index of the optical medium behind the interface.</p> <p>Eq.(1) can be expressed in vector form as <span class="math-container">\begin{equation}\tag{2} n_1(\textbf{i} \times \textbf{n}) = n_2 (\textbf{t} \times \textbf{n}) \end{equation}</span> <span class="math-container">$\qquad$</span> where <span class="math-container">$\textbf{i}$</span> and <span class="math-container">$\textbf{t}$</span> are the unit <em>directional</em> vector of the incident and transmitted ray respectively. <span class="math-container">$\textbf{n}$</span> is the unit <em>normal</em> vector to the interface between the two media pointing from medium 1 with refractive index <span class="math-container">$n_1$</span> into medium 2 with refractive index <span class="math-container">$n_2$</span>. Similarly <span class="math-container">$\textbf{r}$</span> is the reflected ray vector.</p> <p>How can the equation <span class="math-container">\begin{align}\tag{3} \textbf{t} = \mu \textbf{i} + n\sqrt{1- \mu^2[1-(\textbf{ni})^2]} - \mu \textbf{n}(\textbf{ni}) \end{align}</span> be used to derive the equation <span class="math-container">\begin{equation}\tag{4} \textbf{n} = \dfrac{\textbf{i}-\textbf{r}}{\sqrt{2[1-(\textbf{i}\textbf{r})]}}? \end{equation}</span></p> <p><span class="math-container">$\qquad$</span>Here <span class="math-container">$\mu = \dfrac{n_1}{n_2}$</span> and <span class="math-container">$\textbf{n}\textbf{i}= n_{\text{x}} i_{\text{x}} + n_{\text{y}}i_{\text{y}} + n_{\text{z}} i_{\text{z}}$</span> denotes the dot (scalar) product of vectors <span class="math-container">$\textbf{n}$</span> and <span class="math-container">$\textbf{i}$</span>.</p> <p>In Ref. it says that from Eq.(3) follows <span class="math-container">\begin{align}\tag{5} \textbf{r} = \textbf{i} - 2\textbf{n}(\textbf{n}\textbf{i}) \end{align}</span> By simple modification <span class="math-container">\begin{align}\tag{6} \textbf{n} = \dfrac{\textbf{i}-\textbf{r}}{2(\textbf{n}\textbf{i})} \end{align}</span> It says that "...by calculating the dot products of vector <span class="math-container">$\textbf{n}$</span> with both sides of Eq.(6), one can express the dot product <span class="math-container">$(\textbf{n}\textbf{i})$</span> in the form as shown in Eq.(4)" which I can't follow )=</p> <p>Could someone explain how equation (4) is derived?</p> <p>References:</p> <ol> <li>Antonín Mikš and Pavel Novák, Determination of unit normal vectors of aspherical surfaces given unit directional vectors of incoming and outgoing rays: comment, 2012 Optical Society of America, page 1356</li> </ol> https://physics.stackexchange.com/questions/454361/-/454493#454493 2 Answer by Frobenius for Reflection Vector (Ray Tracing) Frobenius https://physics.stackexchange.com/users/110781 2019-01-15T20:18:40Z 2019-01-16T08:32:09Z <p><a href="https://i.stack.imgur.com/zQ8ne.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zQ8ne.png" alt="enter image description here"></a></p> <p>I wonder why all this Physics stuff (refraction, reflection, Snell's Law etc) in order to ask a pure simple mathematical question in Vector Calculus : that of the normalization of a vector.</p> <p><span class="math-container">\begin{equation} \mathbf{n}\boldsymbol{=}\dfrac{\mathbf{i}\boldsymbol{\!-\!}\mathbf{r}}{\Vert\mathbf{i}\boldsymbol{\!-\!}\mathbf{r}\Vert} \tag{01}\label{01} \end{equation}</span> </p> <p><span class="math-container">\begin{equation} \Vert\mathbf{i}\boldsymbol{\!-\!}\mathbf{r}\Vert^{\bf 2}\boldsymbol{=}\Vert\mathbf{i}\Vert^{\bf 2}\boldsymbol{+}\Vert\mathbf{r}\Vert^{\bf 2}\boldsymbol{-}2(\mathbf{i}\boldsymbol{\cdot}\mathbf{r})=1\boldsymbol{+}1\boldsymbol{-}2(\mathbf{i}\boldsymbol{\cdot}\mathbf{r})\boldsymbol{=}2\left[1\boldsymbol{-}(\mathbf{i}\boldsymbol{\cdot}\mathbf{r})\right] \tag{02}\label{02} \end{equation}</span> that is <span class="math-container">\begin{equation} \Vert\mathbf{i}\boldsymbol{\!-\!}\mathbf{r}\Vert\boldsymbol{=}\sqrt{2\left[1\boldsymbol{-}(\mathbf{i}\boldsymbol{\cdot}\mathbf{r})\right]\vphantom{\frac12}} \tag{03}\label{03} \end{equation}</span> so <span class="math-container">\begin{equation} \mathbf{n}\boldsymbol{=}\dfrac{\mathbf{i}\boldsymbol{\!-\!}\mathbf{r}}{\sqrt{2\left[1\boldsymbol{-}(\mathbf{i}\boldsymbol{\cdot}\mathbf{r})\right]\vphantom{\frac12}}} \tag{04}\label{04} \end{equation}</span></p> <p><span class="math-container">$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$</span></p> <p><strong>EDIT</strong></p> <p><span class="math-container">\begin{equation} \left. \begin{cases} \mathbf{n}\boldsymbol{\cdot}\mathbf{i}\boldsymbol{=}\cos\theta\\ \:\mathbf{i}\boldsymbol{\cdot}\mathbf{r}\boldsymbol{=}\cos(\pi\!\boldsymbol{-}\!2\theta)\!\boldsymbol{=}\!\boldsymbol{-}\!\cos2\theta \end{cases}\!\! \right\} \stackrel{\cos2\theta\boldsymbol{=}2\cos^{2}\theta\boldsymbol{-}1}{\boldsymbol{=\!=\!=\!=\!=\!=\!=\!=\!\Longrightarrow}}\boldsymbol{-}(\mathbf{i}\boldsymbol{\cdot}\mathbf{r})\boldsymbol{=}2(\mathbf{n}\boldsymbol{\cdot}\mathbf{i})^{2}\!\boldsymbol{-}\!1 \tag{05}\label{05} \end{equation}</span> that's why your equation (6) <span class="math-container">\begin{equation} \mathbf{n}\boldsymbol{=}\dfrac{\mathbf{i}\!\boldsymbol{-}\!\mathbf{r}}{2(\mathbf{n}\!\boldsymbol{\cdot}\!\mathbf{i})} \tag{06}\label{06} \end{equation}</span></p> <p><span class="math-container">$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$</span></p> <p>Related : <a href="https://physics.stackexchange.com/questions/435512/snells-law-in-vector-form/436252#436252">Snell's law in vector form</a> </p>