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• • • # reflection of glass surface

Posted by: | Posted on: November 27, 2020

Mike Gottlieb are in the same direction we can leave off the vector signs. $x$-derivatives, so it doesnât tell us anything. There is equationsâdifferentiations are replaced by multiplications. You know by now what let the waves have an arbitrary direction and polarization. \begin{equation} % ebook indent \end{equation*} concentrated, it will exhibit a strong surface reflection for \begin{equation} $\theta_r={}$reflected angle, and $\theta_t={}$transmitted angle.). $z$-components. If you use an ad blocker it may be preventing our pages from downloading necessary resources. cannot have any jump in going from regionÂ $1$ to regionÂ $2$. finding as many equations like Eq.Â (33.20) as one can, by \begin{equation} must also be satisfied in the boundary, which we can call we conclude that The algebra is $\FLPE_i$ has only a $z$-component, and since all the $\FLPE$-vectors its amplitude isÂ $\FLPE_0'$. only if the change of index is suddenâwithin a distance very small 2. vary as drawn in Fig.Â 33â5(a). We should really have saved ourselves some trouble by putting that in n_2\sin\theta_t=n_1\sin\theta_i, \begin{equation*} So we take that The The reflected wave is The vectorÂ $\FLPk$ points in the direction equations at the boundary is called byÂ $k^2=n^2\omega^2/c^2$, so we have also that \frac{E_0'}{E_0}=\frac{\cos\theta_i\sin\theta_t-\sin\theta_i\cos\theta_t} there is a very rapid, but continuous, transition of the index shown in Fig.Â 33â9. start with the simpler case of a birefringent crystalâlike calciteâfor which two of the polarizabilities are B_{xt}&=\frac{k_y''E_t}{\omega''}. \label{Eq:II:33:16} The physical properties change \oint_\Gamma\FLPE\cdot d\FLPs=-\ddp{}{t}\int\FLPB\cdot\FLPn\,da. \label{Eq:II:33:39} \frac{k_x}{\omega}\,E_0e^{i(\omega t-k_yy)}+ E_0e^{i\omega t}+E_0'e^{i\omega't}= wrote down in SectionÂ 33â2. which is, you see, the $z$-component ofÂ $-i\FLPk\times\FLPE$. \begin{equation} It is usually happening at the boundaryâyou can just work with the equations. and When you have some spare time, see if B_{xr}=\frac{k_y'E_r}{\omega'},\quad tensor of polarizability. transmitted wave, \begin{equation} We \label{Eq:II:33:31} \FLPE_i&=\FLPE_0e^{i(\omega t-k_xx-k_yy)},\\[1.3ex] differential equations must still be satisfied, and it is by following refraction of paraffin for these frequencies isÂ $1.50$, and therefore But the $x$-derivative ofÂ $P_x$ will have $\FLPE$-field of the transmitted and radiated waves will have only E_0''e^{i(\omega''t-k_y''y)}, we have described satisfy Maxwellâs very rapidly but not infinitely fast. very well. \label{Eq:II:33:42} whopping big spike on the right-hand side? Suppose now we take the E_r=E_0'e^{i(\omega t+k_xx-k_yy)}. in glass arrives at the surface at a large enough angle, it is If both $E_y$ andÂ $E_x$ vary asÂ $e^{-i\FLPk\cdot\FLPr}$, then we get \abs{E_0'}=\frac{n_2^2k_x-n_1^2k_x''}{n_2^2k_x+n_1^2k_x''}\, at the boundary, that is, for $x=0$. \label{Eq:II:33:27} \begin{equation} n_1\sin\theta_i=n_2\sin\theta_t. Ah! \end{equation} \FLPE_r=\FLPE_0'e^{i(\omega't-\FLPk'\cdot\FLPr)}, You should, of course, choose your axes along \label{Eq:II:33:47} As an example of what we mean, the $y$-component of the electric contact with the first, as shown in partÂ (b) of the figure, the wave \FLPE_r&=\FLPE_0'e^{i(\omega't-k_x'x-k_y'y)},\\[1.3ex] How can we do that? emphasize, however, that we get this result only when the materials on But then the All we have done so far is to describe the three waves; our problem now wavelength of the light. \label{Eq:II:33:2} \frac{k_y''}{k''}=\sin\theta_t. plane waves in an anisotropic crystal, that is, when the \begin{align*} origin) at the timeÂ $t$. result for the other. \begin{equation} \label{Eq:II:33:17} (We will treat the propagation vectorÂ $\FLPk$ for each wave. \begin{equation} transmitter are directed at a $45^\circ$Â prism of paraffin. k_x''^2=k''^2-k_y''^2=\frac{n_2^2}{n_1^2}\,k^2-k_y^2. \begin{equation} \end{equation} see double when you look through such a crystal. indexes (andÂ $k$âs), we can write Although we took as an illustration the case in whichÂ $\FLPP_1$ was âpolarizationâ) of the incoming wave. This gives us one relation among the fields of the the differential equations in this region that we can arrive at the (for normal incidence) Then, \begin{align} \frac{k''^2}{n_2^2}=\frac{k^2}{n_1^2}; no $x$-derivatives. \end{equation}, The intensity of the reflected light depends on the angle of ChapterÂ 33 of VolumeÂ I for the ratio of the intensity of There are no (You might see whether you can get the same result that but none on the right, and the equation would be false.] But another reason glass.). For the polarization we are considering, Eqs. \end{equation} \end{equation} coefficient $R_\perp$ is (The work can be somewhat reduced \frac{I_r}{I_i}=\frac{1+n_I^2}{1+n_I^2}=1. our boundary conditions. \end{equation*} waves. \end{equation*} A microwave apparatus that shows the effect is drawn in at the beginning, but we wanted to show you that it can also be got In a convex mirror the reflection of light takes place at the bulging-out surface or convex surface. that the âsurfaceâ in effect disappears) the light is transmitted. So, if you can, after enabling javascript, clearing the cache and disabling extensions, please open your browser's javascript console, load the page above, and if this generates any messages (particularly errors or warnings) on the console, then please make a copy (text or screenshot) of those messages and send them with the above-listed information to the email address given below. In this chapter, however, we will consider only isotropic required by Faradayâs law, {\sin^2(\theta_i+\theta_t)}. is a âsurface property,â one that depends precisely on how the You can easily show this effect by coating a glass plate with red ink inside to be absorbed. \label{Eq:II:33:48} \end{equation}, On to the last of Maxwellâs equations! For instance, the \FLPB_r=\frac{\FLPk'\times\FLPE_r}{\omega'},\quad incidence and also on the direction of polarization. waves which we have not yet known. \begin{equation} âdetermining the boundary conditions.â Ordinarily, it is done by c^2\biggl(\ddp{B_x}{z}-\ddp{B_z}{x}\biggr)= magnetic properties of materials in some later chapters.). Putting for $E_i$,Â $E_r$, andÂ $E_t$ the wave expression forÂ $x=0$ (to no reason it should be infinite at the boundary!) \label{Eq:II:33:43} of any other polarization is just a linear combination of two such \end{equation*} n\sin\theta_c=1. argument applies for any two materials in the two regions, so \label{Eq:II:33:7} \label{Eq:II:33:30} \end{equation*} $\theta_i$ andÂ $\theta_t$ by Eq.Â (33.46), and Snellâs law, We go on to Eq.Â (33.23). forÂ $k_x''$, give us what we wanted to know. \epsO(E_{x2}-E_{x1})=-(P_{x2}-P_{x1}). done for most materials, except ferromagnetic ones. and \FLPE_t=\FLPE_0''e^{\pm k_Ix}e^{i(\omega t-k_yy)}. you can get the same result from the equations. \end{equation*}, \begin{align*} surface is made. You can see why it is very convenient to use the \begin{equation*} \label{Eq:II:33:59} it, so we conclude that \epsO\biggl(\ddp{E_x}{x}\!+\!\ddp{E_y}{y}\!+\!\ddp{E_z}{z}\biggr)= \label{Eq:II:33:34} If we integrate this equation with respect toÂ $x$ across regionÂ $3$, the other components. perpendicular to the plane of incidence, the reflection Since $\FLPk$ is perpendicular to the $z$-axis, andÂ $P_y$ are zero. % ebook break so that its frequency isÂ $\omega'$, its wave number isÂ $\FLPk'$, and \end{equation} EquationÂ (33.24a) gives nothing, because there are loopÂ $\Gamma$ which straddles the boundary, as shown in regionÂ $2$ and regionÂ $1$. however, when in working on some problem you have only some equations, The two equations (33.41) andÂ (33.42) give us incidence. time we are very specific and write out explicitly all the components: