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+### Statement
+
+$12.1.19.$ [Insert the problem statement]
+
+### 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
+
+[Insert a concise answer or boxed result]