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| | | ### 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. |
| | + | $12.1.19$. For a sufficiently large number of conduction electrons per unit volume of metal, the component of the 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 electromagnetic waves are two waves that fall perpendicularly 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 incident wave reaches the $AB$ plane. |
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| | + |  |
| | + | |
| | | ### Solution |
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| | − | 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. |
| | + | Reflection from 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. |
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| | − | Choice of geometry and waves |
| | + | <b>Choice of geometry and waves:</b> |
| | + | Metal surface: plane $x = 0$. The metal occupies $x > 0$; vacuum occupies $x < 0$. |
| | + | Incident wave (traveling toward $+x$): |
| | + | $$\vec{E}_i = E_0 \cos(\omega t - kx)\hat{y}, \quad \vec{B}_i = \frac{E_0}{c} \cos(\omega t - kx)\hat{z}$$ |
| | + | Reflected wave (real, traveling toward $-x$): |
| | + | $$\vec{E}_r = -E_0 \cos(\omega t + kx)\hat{y}, \quad \vec{B}_r = \frac{E_0}{c} \cos(\omega t + kx)\hat{z}$$ |
| | + | The sign reversal of $\vec{E}_r$ ensures that the total tangential component vanishes at $x = 0$. |
| | | |
| | − | Metal surface: plane x = 0. The metal occupies x > 0; vacuum occupies x < 0. |
| | − | Incident wave (traveling toward +x) |
| | + | <b>Electromagnetic field outside the metal ($x < 0$):</b> |
| | + | Superposing the incident and reflected waves creates a standing wave: |
| | + | $$\vec{E}(x, t) = E_0 [\cos(\omega t - kx) - \cos(\omega t + kx)]\hat{y} = 2E_0 \sin(kx) \sin(\omega t)\hat{y}$$ |
| | + | $$\vec{B}(x, t) = \frac{E_0}{c} [\cos(\omega t - kx) + \cos(\omega t + kx)]\hat{z} = \frac{2E_0}{c} \cos(kx) \cos(\omega t)\hat{z}$$ |
| | | |
| | − | $ \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}}$ |
| | + | <b>Instant when the crest of the incident wave reaches the surface:</b> |
| | + | The crest (positive maximum of $\vec{E}_i$) arrives at $x = 0$ when $\cos(\omega t) = 1$, i.e., $\omega t = 0, 2\pi, \dots$ |
| | + | Taking $t = 0$: |
| | + | $$\vec{E}(x, 0) = 0 \quad \text{(zero everywhere outside the metal)}$$ |
| | + | $$\vec{B}(x, 0) = \frac{2E_0}{c} \cos(kx)\hat{z}$$ |
| | + | Near the metal plane itself ($x \to 0^-$), the magnetic field induction is maximal and equals $\vec{B} = \frac{2E_0}{c}\hat{z}$. |
| | | |
| | − | 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 |
| | − | |
| | − | |
| | − | $\boxed{\mathbf{E}(x,0) = 0 \quad\text{(zero everywhere outside the metal)}}$. |
| | − | |
| | − | |
| | − | $\boxed{\mathbf{B}= \frac{2E_0}{c}\,\hat{\mathbf{z}}}$. |
| | + | $E = 0$; $B = \frac{2E_0}{c}$. |
