Electromagnetic waves > Properties of emission and reflection of electromagnetic waves > Problem 12.1.19

Statement

$12.1.19.$
For a sufficiently large number of conduction electrons per unit volume of
metal, the component electric field strength of the wave parallel to the metal
surface is weakened to almost zero. Therefore, the solution of the problem of
the interaction of an electromagnetic wave with a metal is reduced to finding
two such traveling waves near its surface, the superposition of which gives a
zero component of the electric field strength along the surface. Such electro-
magnetic waves are two waves that fall perpendicular to a metal surface: one
actually moves in space outside the metal, and another fictitious ”inverted”
wave moves towards the first one inside the metal (in the figure, this area
along with the fictitious wave is located to the right of the AB plane). The
dummy wave becomes real as soon as it goes beyond the AB boundary, where
it overlaps with the first wave. The superposition of these waves to the left of
the AB plane gives zero electric field strength along AB and, therefore, solves
the problem.
Using the described technique, find the electric field strength and magnetic
field induction near the metal plane at the moment when the top of the inci-
dent wave reaches the AB plane.

Solution

Reflection in a perfect metal is modeled using a fictitious wave that, inside the metal, propagates toward the surface. Outside the metal there exists the real incident wave; inside, the fictitious wave moves in the opposite direction. By superposing both, the tangential electric field vanishes at the surface (plane AB). This fictitious wave, upon crossing the boundary, becomes the real reflected wave.

Choice of geometry and waves

Metal surface: plane x = 0. The metal occupies x > 0; vacuum occupies x < 0.
Incident wave (traveling toward +x)

$ \mathbf{E}_i = E_0 \cos(\omega t - kx),\hat{\mathbf{y}}, \qquad
\mathbf{B}_i = \frac{E_0}{c} \cos(\omega t - kx),\hat{\mathbf{z}}$

Reflected wave (real, traveling toward -x)

$ \mathbf{E}_r = -E_0 \cos(\omega t + kx),\hat{\mathbf{y}}, \qquad
\mathbf{B}_r = \frac{E_0}{c} \cos(\omega t + kx),\hat{\mathbf{z}}$

The sign reversal of$ E_r $ ensures that the total tangential component vanishes at x = 0.

Electromagnetic field outside the metal (x < 0)

Superposing the incident and reflected waves:

$ \mathbf{E}(x,t) = E_0\bigl[\cos(\omega t - kx) - \cos(\omega t + kx)\bigr]\hat{\mathbf{y}}
= 2E_0 \sin(kx)\sin(\omega t),\hat{\mathbf{y}}$

$\mathbf{B}(x,t) = \frac{E_0}{c}\bigl[\cos(\omega t - kx) + \cos(\omega t + kx)\bigr]\hat{\mathbf{z}}
= \frac{2E_0}{c} \cos(kx)\cos(\omega t),\hat{\mathbf{z}}$

Instant when the crest of the incident wave reaches the surface

The crest (positive maximum of $ E_i$) arrives at x = 0 when $\cos(\omega t) = 1, i.e., \omega t = 0,\, 2\pi,\dots Taking t = 0$:

$\boxed{\mathbf{E}(x,0) = 0 \quad\text{(zero everywhere outside the metal)}}$.

$\boxed{\mathbf{B}(x,0) = \frac{2E_0}{c}\cos(kx)\, \hat{\mathbf{z}} \; \xrightarrow{x\to 0^-}\; \frac{2E_0}{c}\, \hat{\mathbf{z}}}$.

Answer

$\boxed{\mathbf{E}(x,0) = 0 \quad\text{(zero everywhere outside the metal)}}$.

$\boxed{\mathbf{B}= \frac{2E_0}{c}\, \hat{\mathbf{z}}}$.