Question

Consider the normal incidence of a plane electromagnetic wave with electric field given by \[ \vec{E} = E_0 \exp{[i(k_1 z - \omega t)]} \hat{x} \] over an interface at \( z = 0 \) separating two media [wave velocities \( v_1 \) and \( v_2 \) (with \( v_2>v_1 \)) and wave vectors \( k_1 \) and \( k_2 \), respectively], as shown in the figure. The magnetic field vector of the reflected wave is 

• A. \( \frac{E_0}{v_1} \exp{[i(k_1 z - \omega t)]} \hat{y} \)
• B. \( \frac{E_0}{v_1} \exp{[i(-k_1 z - \omega t)]} \hat{y} \)
• C. \( -\frac{E_0}{v_1} \exp{[i(-k_1 z - \omega t)]} \hat{y} \)
• D. \( -\frac{E_0}{v_1} \exp{[i(k_1 z - \omega t)]} \hat{y} \)

Hint:

For reflection, the electric field vector changes direction and its wave vector also reverses.

Answer 1

The Correct Option is C

Step 1: Understanding the situation.
In the given setup, a plane electromagnetic wave is incident normally on an interface at \( z = 0 \) separating two media with different wave velocities \( v_1 \) and \( v_2 \). For the reflected wave, the electric field vector will reverse direction, and its wave vector will change direction as well. The general form of the electric field in the second medium (after reflection) will be the same as the incident wave but with a reversed sign in the exponential factor.
Step 2: Identifying the correct reflected field.
The electric field in the reflected wave should have the same magnitude but an opposite direction. Thus, the reflected electric field vector will be: \[ \vec{E}_{\text{reflected}} = -\frac{E_0}{v_1} \exp{[i(-k_1 z - \omega t)]} \hat{y} \]
Step 3: Conclusion.
Thus, the correct answer is (C), where the reflected wave has the electric field vector in the opposite direction with the same magnitude but the opposite sign in the exponential term.