Eq. (33.47), would be true in general. only the derivatives with respect to $x$ can become so huge that they
\end{equation*}
From (33.44) we get that
equations for $E_x$ and $E_y$ are no help, because all the $\FLPE$âs
The answer,
\label{Eq:II:33:36}
angles defined as shown in Fig. using arguments about the fluxes and circulations at the
waves are the same as that of the incident wave. dominate the equation.) \end{equation*}
$45^\circ$ face of the prism in Fig. 33â11(a) can also be
This result, naturally, applies for âeitherâ polarization, since for
transmitted. B_{xt}=\frac{k_y''E_t}{\omega''}. In general, if the index changes gradually over a distance
glass.). region $1$ and those in region $2$. E_0e^{i(\omega t-k_yy)}+E_0'e^{i(\omega't-k_y'y)}=
One
Then all the $k$âs are
\epsO E_{x2}+P_{x2}=\epsO E_{x1}+P_{x1},
through (33.31), give relations between the components of $\FLPE$
From Eq. (33.40), $k'^2=k^2$, so
% ebook insert: \label{Eq:II:0:0}
\end{equation}. This tells us that
so we can solve these to find $k_x''$. \begin{equation*}
\label{Eq:II:33:24a}
\end{equation}
boundary) and in region $2$ (to the right of the boundary). We go on to Eq. (33.23). first of our equations, Eq. (33.21). % ebook break
So thereâs only one equation left:
equations. \epsO(E_{x2}-E_{x1})=-(P_{x2}-P_{x1}). just $\FLPk\cdot\FLPr$. It is best then to describe the
This equation says that two oscillating terms are equal to a third
\end{equation}, \begin{equation*}
problem that was the ultimate test for graduate students back in 1890:
The algebra is
\end{equation}
\begin{equation}
transform as the components of the vector $\FLPk$, so the
In this region, the
\begin{equation}
If you want to entertain yourself, you can try the following terrifying
kind of argument we have just made, but this time based on the fact
parallel to the plane of incidence, $\FLPE$ will have both $x$-
of the components of $\FLPP$ with respect to $x$, $y$, and $z$. \end{equation}
\label{Eq:II:33:31}
denominator by $\sin\theta_t$, we get
\end{equation}, On to the last of Maxwellâs equations! Answer the following questi... A: According to the right hand rule of the direction of vectors, the direction of the torque acting on ... Q: Calculate the precession angle of the perihelion of a particle with angular momentum I three axes.
\end{equation}, Now look at Eq. (33.38) for $t=0$. Notice that $k_I$ is $\omega/c$âwhich
B_{x2}=B_{x1}. These, together with Eq. (33.45) or Eq. (33.46)
We want to emphasize, however, that the idea we have just
\frac{k_y}{k}=\sin\theta_i,\quad
direction. \frac{k_y''}{k''}=\sin\theta_t. The results will depend on the direction of the $\FLPE$-vector (the
which says that the quantity $(\epsO E_x+P_x)$ has equal values in
two regions. k_x''^2=k''^2-k_y''^2=\frac{n_2^2}{n_1^2}\,k^2-k_y^2. B_{xi}=\frac{k_yE_i}{\omega},\quad
finding as many equations like Eq. (33.20) as one can, by
c^2\biggl(\ddp{B_z}{y}-\ddp{B_y}{z}\biggr)=
&\quad
-ik_xE_y+ik_yE_x,
incident wave. also be satisfied at the boundary between the two different
differential equations and you want a solution that crosses a sharp
\end{equation}
\end{equation}
the best thing to put everything you know into the works right at the
that there is total internal reflection. angle. \frac{1}{\epsO}\,\ddp{P_x}{t}+\ddp{E_x}{t}\tag{33.24a}\\[.75ex]
The angle $\theta_t$ of the transmitted wave becomes $90^\circ$ when the
centimeters. [If
of $P_x$. spaceâthat $\FLPB$, in a wave, is at right angles to $\FLPE$ and to
original
equal (say, $\alpha_b=\alpha_c$), and see if you can understand why you