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<h2 class="hd hd-2 unit-title">Introduction to Reflection and Transmission of EM Waves</h2>
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Now, we are in a position to discuss how electromagnetic waves reflect and transmit at the boundary between two materials. </p><p>
Previously, we discussed how scalar 2D waves transmit and reflect at the boundary between two media (Week 12—Snell's law). We have also discussed the special case of electromagnetic waves that reflect off of perfect conductors at normal incidence (Week 8). Now we will derive the most general case for EM waves at any angle of incidence. </p><p>
Importantly, we will find that there are different boundary conditions for different electric field components of a wave that is incident on a boundary (and similarly for magnetic field components). But, fortunately, we are completely prepared to undertake the analysis! </p>
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<h2 class="hd hd-2 unit-title">L34v1: How to Get Good Contrast in Photos</h2>
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<h3 class="hd hd-2">L34v1: How to get good contrast in photos</h3>
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<h2 class="hd hd-2 unit-title">L34Q1: Polarization of Sunlight and Reflected Light</h2>
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Polarization of Sunlight and Reflected Light
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The observation that motivates this lesson is that "polarizers make pictures look better," and we discussed two examples: </p>
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(a) Viewing the sky through a polarizer will make it appear dark (compared to clouds). </p>
<p>
(b) Viewing surfaces, like glass or water, through a polarizer will reduce glare. </p>
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For each case, identify the correct explanation for this phenomenon. </p>
<p><b class="bfseries">(Part a)</b> Viewing the sky through a polarizer will make it appear dark (compared to clouds). </p>
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<text> a) Sunlight is inherently polarized and a polarizing filter will reduce the intensity depending on its angle, but this does not depend on where you look in the sky.</text>
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<text> b) Sunlight is inherently unpolarized and a polarizing filter will reduce the intensity by the same amount no matter the orientation of the polarizer, and no matter where you look in the sky.</text>
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<text> c) Sunlight is inherently unpolarized but will become partially polarized when it scatters off of molecules in the sky, independent of angle&#8212;a polarizing filter will reduce the intensity depending on its angle, but this does not depend on where you look in the sky.</text>
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<text> d) Sunlight is inherently unpolarized but will become partially polarized when it scatters off of molecules in the sky, depending on the angle&#8212;a polarizing filter will reduce the intensity depending on its angle, and this will be different along different directions in the sky.</text>
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Polarization of sunlight and reflected light
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<p><b class="bfseries">(Part b)</b> Viewing surfaces, like glass or water, through a polarizer will reduce glare. </p>
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<text> a) This only works for polarizers with an anti-glare coating.</text>
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<text> b) For this phenomenon to occur, the light that hits a surface MUST be polarized&#8212;then a polarizer will reduce the intensity of reflected light depending on its angle.</text>
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<text> c) A surface will partially polarize the light that it reflects&#8212;a polarizer will reduce the intensity of reflected light depending on its angle.</text>
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<text> d) A polarizing filter will reduce the intensity of reflected light by the same amount, no matter the orientation of the polarizer.</text>
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<h2 class="hd hd-2 unit-title">L34Q2: Boundary Conditions from the Wave Solution</h2>
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Boundary Conditions from the Wave Solution
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Consider the following requirements for reflection of waves at a surface, where the indices [mathjaxinline]I[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], and [mathjaxinline]T[/mathjaxinline] refer to incident, reflected, and transmitted waves, respectively. Select the correct explanation/reasoning for each requirement listed below. </p>
<p><b class="bfseries">(Part a)</b> [mathjaxinline]\vec{k}_{I} \cdot \vec{r} = \vec{k}_{R} \cdot \vec{r} = \vec{k}_{T} \cdot \vec{r}[/mathjaxinline] where [mathjaxinline]\vec{k}[/mathjaxinline] is the wave vector for each wave. </p>
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<text> a) The incident, reflected, and transmitted waves all have the same velocity.</text>
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<text> b) The magnitudes of the incident, reflected, and transmitted wave vectors are equal.</text>
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<text> c) The wave vectors of the incident and reflected waves have the same magnitude, but their components perpendicular to the surface have opposite sign.</text>
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<text> d) Since the wave is continuous at the boundary, the arguments of all sinusoidal functions must be equal at all times and locations along the boundary.</text>
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<text> e) This requirement is not actually necessary, it just makes the calculation easier.</text>
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<p><b class="bfseries">(Part b)</b> [mathjaxinline]\omega _{I}=\omega _{R}=\omega _{T}[/mathjaxinline] where [mathjaxinline]\omega[/mathjaxinline] is the angular frequency of each wave. </p>
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<text> a) The incident, reflected, and transmitted waves all have the same velocity.</text>
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<text> b) The magnitudes of the incident, reflected, and transmitted wave vectors are equal.</text>
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<text> c) The wave vectors of the incident and reflected waves have the same magnitude, but their components perpendicular to the surface have opposite sign.</text>
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<text> d) Since the wave is continuous at the boundary, the arguments of all sinusoidal functions must be equal at all times and locations along the boundary.</text>
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<text> e) This requirement is not actually necessary, it just makes the calculation easier.</text>
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<p><b class="bfseries">(Part c)</b> [mathjaxinline]\theta _{I}=\theta _{R}[/mathjaxinline] where [mathjaxinline]\theta[/mathjaxinline] is the angle between the direction the wave is moving and a line perpendicular to the surface. </p>
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<text> a) The incident, reflected, and transmitted waves all have the same velocity.</text>
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<text> b) The magnitudes of the incident, reflected, and transmitted wave vectors are equal.</text>
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<text> c) The wave vectors of the incident and reflected waves have the same magnitude, but their components perpendicular to the surface have opposite sign.</text>
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<text> d) Since the wave is continuous at the boundary, the arguments of all sinusoidal functions must be equal at all times and locations along the boundary.</text>
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<text> e) This requirement is not actually necessary, it just makes the calculation easier.</text>
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Boundary Conditions from Maxwell&#39;s Equations - part a
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<p><b class="bfseries">(Part a)</b> Note which of the following Maxwell's equations yields each of the boundary conditions listed below (what additional assumptions are made?). </p>
<table id="a0000000002" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
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<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \mathrm{(i)}\,[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \nabla \cdot \vec{D} = \rho _{f}[/mathjaxinline]
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</tr>
<tr id="a0000000004">
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[mathjaxinline]\displaystyle \mathrm{(ii)}\,[/mathjaxinline]
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<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \nabla \times \vec{E} = -\dfrac {\partial \vec{B}}{\partial t}[/mathjaxinline]
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<td style="width:40%; border:none">&#160;</td>
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<tr id="a0000000005">
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[mathjaxinline]\displaystyle \mathrm{(iii)}\,[/mathjaxinline]
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[mathjaxinline]\displaystyle \nabla \cdot \vec{B} = 0[/mathjaxinline]
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[mathjaxinline]\displaystyle \mathrm{(iv)}\,[/mathjaxinline]
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[mathjaxinline]\displaystyle \nabla \times \vec{H} = \vec{J}_{f} + \dfrac {\partial \vec{D}}{\partial t}[/mathjaxinline]
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<p><b class="bfseries">(1)</b> [mathjaxinline]\varepsilon _{1}E_{\perp }^{(1)} = \varepsilon _{2}E_{\perp }^{(2)}[/mathjaxinline]: </p>
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<p><b class="bfseries">(2)</b> [mathjaxinline]E_{||}^{(1)} = E_{||}^{(2)}[/mathjaxinline]: </p>
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Boundary Conditions from Maxwell&#39;s Equations - part b
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<p><b class="bfseries">(Part b)</b> Media 1 and 2 have indices of refraction [mathjaxinline]n_{1}[/mathjaxinline] and [mathjaxinline]n_{2}[/mathjaxinline]. What is the ratio [mathjaxinline]\varepsilon _{2}/\varepsilon _{1}[/mathjaxinline] in terms of these indices? You may assume that [mathjaxinline]\mu _{1}=\mu _{2}[/mathjaxinline]. </p>
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Express your answer using <code>n_1</code> for [mathjaxinline]n_{1}[/mathjaxinline] and <code>n_2</code> for [mathjaxinline]n_{2}[/mathjaxinline] (assume that the relative permeabilities of both mediums are equal). </p>
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<p style="display:inline">[mathjaxinline]\varepsilon _{2}/\varepsilon _{1}=[/mathjaxinline] </p>
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<td class="formulainput">enter <code> x^(n+1) </code> for [mathjaxinline] x^{n+1} [/mathjaxinline]</td>
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<td class="formulainput">enter <code> v_0 </code> for [mathjaxinline] v_0 [/mathjaxinline] </td>
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<td class="formulainput">use <code>( )</code> to clarify order of operations</td>
<td class="formulainput"> enter <code>(2+3)*2 </code> for 10 <br/>
enter <code> 2+3*2 </code> for 8 </td>
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<td class="formulainput">enter (english) name of letter</td>
<td class="formulainput">enter <code>alpha </code> for [mathjaxinline] \alpha [/mathjaxinline]<br/>
enter <code>lambda </code> for [mathjaxinline]\lambda [/mathjaxinline]
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<td class="formulainput">enter <code>arctan(x^2/3) </code> for [mathjaxinline]\tan^{-1}\left(\frac{x^2}{3}\right) [/mathjaxinline]</td>
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<td class="formulainput">enter <code>cosh(4*x+y) </code> for [mathjaxinline]\cosh(4x+y) [/mathjaxinline]</td>
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<td class="formulainput">matrix</td>
<td class="formulainput">enter <code>[[1,0],[0,-1]]</code> for [mathjaxinline]\begin{pmatrix} 1 &amp; &amp; 0 \\ 0 &amp; &amp; -1 \end{pmatrix}[/mathjaxinline]</td>
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<td class="formulainput">enter <code>[[1],[2],[3]]</code> for [mathjaxinline]\begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix}[/mathjaxinline]</td>
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<td class="formulainput">enter <code>[[1,2,3]]</code> for [mathjaxinline]\begin{pmatrix} 1 &amp; &amp; 2 &amp; &amp; 3 \end{pmatrix}[/mathjaxinline]</td>
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<h2 class="hd hd-2 unit-title">L34Q4: Review of Terminology</h2>
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In the last video segment, we discussed a particular case of polarization. Before continuing, let's note the different ways that light can be polarized with respect to a surface and go over the corresponding terminology. </p><p><h2>TERMINOLOGY</h2></p><p>
The <i class="itshape">interface</i> is the surface on which the wave is incident. The <i class="itshape">incident angle</i> is the angle that the incident wave vector makes with respect to the normal of the interface. </p><center><img src="/assets/courseware/v1/16e0e9dfac1a27c76e8fa4f93f202585/asset-v1:MITx+8.03x+1T2020+type@asset+block/images_lect_22_05a.png" width="495"/></center><p style="margin-bottom: 0px; margin-top: 0px; display: block; padding-bottom: 20px;" class="gap"/><p>
The <i class="itshape">plane of incidence</i> is the plane in which the incident, reflected, and transmitted wave vectors lie. This is also called the <i class="itshape">scattering plane</i>. It is orthogonal to the interface. </p><center><img src="/assets/courseware/v1/bb1692d0a81fa45cd89185be0bd88a5e/asset-v1:MITx+8.03x+1T2020+type@asset+block/images_lect_22_05b.png" width="495"/></center><p style="margin-bottom: 0px; margin-top: 0px; display: block; padding-bottom: 20px;" class="gap"/><p><h2>DESCRIBING THE POLARIZATION</h2></p><p>
Note, the electric field points orthogonally to the direction of propagation and can be polarized in any direction that is orthogonal to the wave vector. We will describe two cases, and any additional cases can be characterized by a superposition of these two. </p><p><b class="bfseries">Case 1:</b> The electric field is polarized parallel to the plane of incidence (shown in lecture). In this case, the electric field will alway have a component that is perpendicular to the interface (unless propagating at normal incidence). This is sometimes referred to as <i class="itshape">p-polarization</i>, and is shown below for incident light only. </p><center><img src="/assets/courseware/v1/11e4e6817da55e623967defcec8a2ee0/asset-v1:MITx+8.03x+1T2020+type@asset+block/images_lect_22_05c.png" width="495"/></center><p style="margin-bottom: 0px; margin-top: 0px; display: block; padding-bottom: 20px;" class="gap"/><p><b class="bfseries">Case 2:</b> The electric field is polarized perpendicular to the plane of incidence. In this case, the electric field will always be parallel to the interface. This is sometimes referred to as <i class="itshape">s-polarization</i>, and is shown below for incident light only. </p><center><img src="/assets/courseware/v1/88b3346872b3eefd9f9ea92b2cb6b306/asset-v1:MITx+8.03x+1T2020+type@asset+block/images_lect_22_05d.png" width="495"/></center><p style="margin-bottom: 0px; margin-top: 0px; display: block; padding-bottom: 20px;" class="gap"/>
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Other Polarization
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In the lesson, we went over <i class="itshape">Case 1</i>. To complete a similar analysis for <i class="itshape">Case 2</i> (i.e., derive the reflection and transmission coefficients), what aspects of the derivation would be different? </p>
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<text> a) No aspects would be different.</text>
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<text> b) The electric field has no perpendicular component, so another constraint on the electric field must be found using something other than Maxwell's equations.</text>
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<text> c) We must use the other Maxwell's equations to derive constraints on [mathjaxinline]\vec{B}[/mathjaxinline].</text>
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<h2 class="hd hd-2 unit-title">L34Q5: Different Incident Angles I</h2>
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Different Incident Angles - part Ia
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Consider the case where [mathjaxinline]\vec{E}[/mathjaxinline] is polarized parallel to the plane of incidence (as done in lecture). In the next problem, we will compute the fresnel reflection and transmission coefficients for various incident angles. First, we must simplify the expressions for [mathjaxinline]\alpha[/mathjaxinline] and [mathjaxinline]\beta[/mathjaxinline]. </p>
<p><b class="bfseries">(Part a)</b> First, express the coefficient [mathjaxinline]\beta[/mathjaxinline] in terms of [mathjaxinline]n_{1}[/mathjaxinline] and [mathjaxinline]n_{2}[/mathjaxinline] ONLY. Assume that the relative permeabilities of the two mediums are identical. Use <code>n_1</code> for [mathjaxinline]n_{1}[/mathjaxinline] and <code>n_2</code> for [mathjaxinline]n_{2}[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]\beta =[/mathjaxinline] </p>
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<code>2520</code>
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<td class="formulainput">fractions</td>
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<code>2/3</code>
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<td class="formulainput"><code>3.14</code>, <code>.98</code></td>
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<td class="formulainput"><code>+ - * /</code> (add, subtract, multiply, divide)</td>
<td class="formulainput">enter <code> (x+2*y)/(x-1)</code> for [mathjaxinline] \displaystyle \frac{x+2y}{x-1} [/mathjaxinline] </td>
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<td class="formulainput"><code>^</code> (raise to a power)</td>
<td class="formulainput">enter <code> x^(n+1) </code> for [mathjaxinline] x^{n+1} [/mathjaxinline]</td>
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<td class="formulainput"><code>_</code> (add a subscript)</td>
<td class="formulainput">enter <code> v_0 </code> for [mathjaxinline] v_0 [/mathjaxinline] </td>
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enter <code>lambda </code> for [mathjaxinline]\lambda [/mathjaxinline]
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<td class="formulainput">enter <code>arctan(x^2/3) </code> for [mathjaxinline]\tan^{-1}\left(\frac{x^2}{3}\right) [/mathjaxinline]</td>
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<td class="formulainput">enter <code>cosh(4*x+y) </code> for [mathjaxinline]\cosh(4x+y) [/mathjaxinline]</td>
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<td class="formulainput">matrix</td>
<td class="formulainput">enter <code>[[1,0],[0,-1]]</code> for [mathjaxinline]\begin{pmatrix} 1 &amp; &amp; 0 \\ 0 &amp; &amp; -1 \end{pmatrix}[/mathjaxinline]</td>
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<td class="formulainput">enter <code>[[1],[2],[3]]</code> for [mathjaxinline]\begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix}[/mathjaxinline]</td>
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<td class="formulainput">enter <code>[[1,2,3]]</code> for [mathjaxinline]\begin{pmatrix} 1 &amp; &amp; 2 &amp; &amp; 3 \end{pmatrix}[/mathjaxinline]</td>
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Different Incident Angles - part Ib
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<p><b class="bfseries">(Part b)</b> Next, we want to express [mathjaxinline]\alpha[/mathjaxinline] solely in terms of the incident angle [mathjaxinline]\theta _{I}[/mathjaxinline]. Write this expression for [mathjaxinline]\alpha[/mathjaxinline], using <code>n_1</code> for [mathjaxinline]n_{1}[/mathjaxinline], <code>n_2</code> for [mathjaxinline]n_{2}[/mathjaxinline], and <code>theta_I</code> for [mathjaxinline]\theta _{I}[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]\alpha =[/mathjaxinline] </p>
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<code>2520</code>
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<code>2/3</code>
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<td class="formulainput">enter <code> (x+2*y)/(x-1)</code> for [mathjaxinline] \displaystyle \frac{x+2y}{x-1} [/mathjaxinline] </td>
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<td class="formulainput"><code>^</code> (raise to a power)</td>
<td class="formulainput">enter <code> x^(n+1) </code> for [mathjaxinline] x^{n+1} [/mathjaxinline]</td>
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<td class="formulainput"><code>_</code> (add a subscript)</td>
<td class="formulainput">enter <code> v_0 </code> for [mathjaxinline] v_0 [/mathjaxinline] </td>
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<td class="formulainput">use <code>( )</code> to clarify order of operations</td>
<td class="formulainput"> enter <code>(2+3)*2 </code> for 10 <br/>
enter <code> 2+3*2 </code> for 8 </td>
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<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">enter (english) name of letter</td>
<td class="formulainput">enter <code>alpha </code> for [mathjaxinline] \alpha [/mathjaxinline]<br/>
enter <code>lambda </code> for [mathjaxinline]\lambda [/mathjaxinline]
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<code>e, pi</code>
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<code>abs, ln, sqrt</code>
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<td class="formulainput">enter <code>abs(x+y) </code> for [mathjaxinline] \left|x+y \right| [/mathjaxinline]<br/>
enter <code>sqrt(x^2-y) </code> for [mathjaxinline] \sqrt{x^2-y} [/mathjaxinline]
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<th class="formulainput" scope="row" rowspan="3">Trigonometric <br/> functions</th>
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<code>sin, cos, tan, sec, csc, cot</code>
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<td class="formulainput">enter <code>sin(4*x+y)^2 </code> for [mathjaxinline]\sin^2(4x+y) [/mathjaxinline]</td>
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<td class="formulainput"><code>arcsin, arccos, arctan</code>, etc.</td>
<td class="formulainput">enter <code>arctan(x^2/3) </code> for [mathjaxinline]\tan^{-1}\left(\frac{x^2}{3}\right) [/mathjaxinline]</td>
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<td class="formulainput"><code>sinh, cosh, arcsinh</code>, etc.</td>
<td class="formulainput">enter <code>cosh(4*x+y) </code> for [mathjaxinline]\cosh(4x+y) [/mathjaxinline]</td>
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<th class="formulainput" scope="row" rowspan="3">Matrices<br/>&amp; Vectors</th>
<td class="formulainput">matrix</td>
<td class="formulainput">enter <code>[[1,0],[0,-1]]</code> for [mathjaxinline]\begin{pmatrix} 1 &amp; &amp; 0 \\ 0 &amp; &amp; -1 \end{pmatrix}[/mathjaxinline]</td>
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<td class="formulainput">column vector</td>
<td class="formulainput">enter <code>[[1],[2],[3]]</code> for [mathjaxinline]\begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix}[/mathjaxinline]</td>
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<td class="formulainput">row vector</td>
<td class="formulainput">enter <code>[[1,2,3]]</code> for [mathjaxinline]\begin{pmatrix} 1 &amp; &amp; 2 &amp; &amp; 3 \end{pmatrix}[/mathjaxinline]</td>
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<h2 class="hd hd-2 unit-title">L34Q6: Different Incident Angles II</h2>
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Different Incident Angles II
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Finally, use your results from the previous problem to determine [mathjaxinline]r=\dfrac {E_{0R}}{E_{0I}}[/mathjaxinline] and [mathjaxinline]t=\dfrac {E_{0T}}{E_{0I}}[/mathjaxinline] for the following incident angles, using [mathjaxinline]n_{1}=1[/mathjaxinline] and [mathjaxinline]n_{2}=2[/mathjaxinline]. </p>
<p>
Note, we diverge from the lesson videos by using the notation [mathjaxinline]r[/mathjaxinline] and [mathjaxinline]t[/mathjaxinline] for the reflection and transmission coefficients, instead of [mathjaxinline]R[/mathjaxinline] and [mathjaxinline]\tau[/mathjaxinline]. This is because [mathjaxinline]R[/mathjaxinline] is typically reserved for a property called "reflectance" and [mathjaxinline]T[/mathjaxinline] for "transmittance," which are related to (but different from) the coefficients defined above. </p>
<p><b class="bfseries">(Part i)</b> [mathjaxinline]\theta _{I} = 0[/mathjaxinline] </p>
<p>
<p style="display:inline">[mathjaxinline]r =[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]t =[/mathjaxinline] </p>
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<p><b class="bfseries">(Part ii)</b> [mathjaxinline]\theta _{I} = \pi /4[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]r =[/mathjaxinline] </p>
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<th class="formulainput" scope="row" rowspan="3">Numbers</th>
<td class="formulainput">integers</td>
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<code>2520</code>
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<td class="formulainput">fractions</td>
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<code>2/3</code>
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<td class="formulainput"><code>3.14</code>, <code>.98</code></td>
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<th class="formulainput" scope="row" rowspan="4">Operators</th>
<td class="formulainput"><code>+ - * /</code> (add, subtract, multiply, divide)</td>
<td class="formulainput">enter <code> (x+2*y)/(x-1)</code> for [mathjaxinline] \displaystyle \frac{x+2y}{x-1} [/mathjaxinline] </td>
</tr>
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<td class="formulainput"><code>^</code> (raise to a power)</td>
<td class="formulainput">enter <code> x^(n+1) </code> for [mathjaxinline] x^{n+1} [/mathjaxinline]</td>
</tr>
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<td class="formulainput"><code>_</code> (add a subscript)</td>
<td class="formulainput">enter <code> v_0 </code> for [mathjaxinline] v_0 [/mathjaxinline] </td>
</tr>
<tr class="formulainput">
<td class="formulainput">use <code>( )</code> to clarify order of operations</td>
<td class="formulainput"> enter <code>(2+3)*2 </code> for 10 <br/>
enter <code> 2+3*2 </code> for 8 </td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">enter (english) name of letter</td>
<td class="formulainput">enter <code>alpha </code> for [mathjaxinline] \alpha [/mathjaxinline]<br/>
enter <code>lambda </code> for [mathjaxinline]\lambda [/mathjaxinline]
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<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">
<code>e, pi</code>
</td>
<td class="formulainput">enter <code>e^x </code> for [mathjaxinline] e^x [/mathjaxinline]<br/>
enter <code>2*pi </code> for [mathjaxinline] 2\pi [/mathjaxinline]
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<th class="formulainput" scope="row">Basic functions</th>
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<code>abs, ln, sqrt</code>
</td>
<td class="formulainput">enter <code>abs(x+y) </code> for [mathjaxinline] \left|x+y \right| [/mathjaxinline]<br/>
enter <code>sqrt(x^2-y) </code> for [mathjaxinline] \sqrt{x^2-y} [/mathjaxinline]
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<th class="formulainput" scope="row" rowspan="3">Trigonometric <br/> functions</th>
<td class="formulainput">
<code>sin, cos, tan, sec, csc, cot</code>
</td>
<td class="formulainput">enter <code>sin(4*x+y)^2 </code> for [mathjaxinline]\sin^2(4x+y) [/mathjaxinline]</td>
</tr>
<tr class="formulainput">
<td class="formulainput"><code>arcsin, arccos, arctan</code>, etc.</td>
<td class="formulainput">enter <code>arctan(x^2/3) </code> for [mathjaxinline]\tan^{-1}\left(\frac{x^2}{3}\right) [/mathjaxinline]</td>
</tr>
<tr class="formulainput">
<td class="formulainput"><code>sinh, cosh, arcsinh</code>, etc.</td>
<td class="formulainput">enter <code>cosh(4*x+y) </code> for [mathjaxinline]\cosh(4x+y) [/mathjaxinline]</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row" rowspan="3">Matrices<br/>&amp; Vectors</th>
<td class="formulainput">matrix</td>
<td class="formulainput">enter <code>[[1,0],[0,-1]]</code> for [mathjaxinline]\begin{pmatrix} 1 &amp; &amp; 0 \\ 0 &amp; &amp; -1 \end{pmatrix}[/mathjaxinline]</td>
</tr>
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<h2 class="hd hd-2 unit-title">L34Q7: Derivation of Brewster's Angle</h2>
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Derivation of Brewster&#39;s angle
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Consider light that propagates from medium 1, with [mathjaxinline]n_{1}=1[/mathjaxinline], to medium 2, with [mathjaxinline]n_{2}=2[/mathjaxinline]. For the case of p-polarized light, which we have been discussing during this lesson, determine the critical angle at which no light is reflected. This is known as Brewster's angle, [mathjaxinline]\theta _{B}[/mathjaxinline]. Express your answer in units of degrees. </p>
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<p style="display:inline">[mathjaxinline]\theta _{B} =[/mathjaxinline] </p>
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<td class="formulainput">integers</td>
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<code>2520</code>
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<td class="formulainput">fractions</td>
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<code>2/3</code>
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<td class="formulainput"><code>3.14</code>, <code>.98</code></td>
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<td class="formulainput"><code>+ - * /</code> (add, subtract, multiply, divide)</td>
<td class="formulainput">enter <code> (x+2*y)/(x-1)</code> for [mathjaxinline] \displaystyle \frac{x+2y}{x-1} [/mathjaxinline] </td>
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<td class="formulainput"><code>^</code> (raise to a power)</td>
<td class="formulainput">enter <code> x^(n+1) </code> for [mathjaxinline] x^{n+1} [/mathjaxinline]</td>
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<td class="formulainput">use <code>( )</code> to clarify order of operations</td>
<td class="formulainput"> enter <code>(2+3)*2 </code> for 10 <br/>
enter <code> 2+3*2 </code> for 8 </td>
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<td class="formulainput">enter (english) name of letter</td>
<td class="formulainput">enter <code>alpha </code> for [mathjaxinline] \alpha [/mathjaxinline]<br/>
enter <code>lambda </code> for [mathjaxinline]\lambda [/mathjaxinline]
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<td class="formulainput">enter <code>e^x </code> for [mathjaxinline] e^x [/mathjaxinline]<br/>
enter <code>2*pi </code> for [mathjaxinline] 2\pi [/mathjaxinline]
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<td class="formulainput">enter <code>abs(x+y) </code> for [mathjaxinline] \left|x+y \right| [/mathjaxinline]<br/>
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<td class="formulainput">enter <code>cosh(4*x+y) </code> for [mathjaxinline]\cosh(4x+y) [/mathjaxinline]</td>
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<td class="formulainput">matrix</td>
<td class="formulainput">enter <code>[[1,0],[0,-1]]</code> for [mathjaxinline]\begin{pmatrix} 1 &amp; &amp; 0 \\ 0 &amp; &amp; -1 \end{pmatrix}[/mathjaxinline]</td>
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<td class="formulainput">column vector</td>
<td class="formulainput">enter <code>[[1],[2],[3]]</code> for [mathjaxinline]\begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix}[/mathjaxinline]</td>
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<td class="formulainput">row vector</td>
<td class="formulainput">enter <code>[[1,2,3]]</code> for [mathjaxinline]\begin{pmatrix} 1 &amp; &amp; 2 &amp; &amp; 3 \end{pmatrix}[/mathjaxinline]</td>
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