Problem 12.1.25 Solution | Properties of emission and reflection of electromagnetic waves

Statement

12.1.25.

Using the method described in the problem 12.1.19, prove that the angle of
incidence of an electromagnetic wave is equal to the angle of reflection. Con-
sider the following cases:: a) the vector E of the electromagnetic wave incident
on the metal is parallel to the metal surface; b) the vector B of the electromag-
netic wave is parallel to the metal surface.

Solution

(Look the solution for the problem 12.1.19 firts )

The incident wave has a wave vector $\mathbf{k}_i$ that forms an angle $\theta_i$ with the normal. The reflected (real) wave will have a wave vector $\mathbf{k}_r$ that forms an angle $\theta_r$ For the boundary condition to be satisfied at every instant and at every point on the surface, the phases of the incident and reflected waves must be equal at z = 0. This implies:

$k \sin\theta_i = k \sin\theta_r \quad\Longrightarrow\quad \boxed{\theta_i = \theta_r}$

This is the law of reflection, valid for any polarization.

a) Electric field parallel to the surface

The incident electric field is purely tangential $\mathbf{E}_i = E_0\,\hat{\mathbf{y}}$
For the total tangential field at z=0 to be zero, the reflected wave must satisfy:

$\mathbf{E}_r = -E_0\,\hat{\mathbf{y}} \quad\Longrightarrow\quad \boxed{E' = -E}$

The corresponding magnetic field, being in phase with the incident one at the surface, satisfies:

$\boxed{B' = B}$

b) Magnetic field parallel to the surface

Here it is the incident magnetic field that is tangential: $\mathbf{B}_i = B_0\,\hat{\mathbf{y}}$
The tangential component of the incident electric field must cancel with the reflected one. For this, the reflected electric field maintains the same phase as the incident one at the surface:

$\boxed{E' = E}$

The normal component of the magnetic field reverses to conserve the direction of propagation.

Answer

$\boxed{B' = B}$

$\boxed{E' = E}$

Formulas in this solution the whole sheet

\frac{q_1}{x^2} = \frac{q_2}{(l - x)^2}, \qquad x = \frac{\sqrt{q_1}}{\sqrt{q_1} + \sqrt{q_2}}\,l

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отрпад$\alpha_{\text{отр}} = \alpha_{\text{пад}}, \qquad \frac{\sin\alpha_1}{\sin\alpha_2} = \frac{v_1}{v_2} = \frac{n_2}{n_1}, \qquad \lambda' = \frac{\lambda}{n}$

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Statement

Solution

Answer

Formulas in this solution the whole sheet

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