Question

A plane wave is normally incident on a dielectric slab of permittivity elementof_r and thickness d, where d = lambda_0/(4 Squareroot elementof_r) and lambda_0 is the free-space wavelength of the incident wave, as shown in the accompanying figure. If free-space exists on both sides of the slab, find the reflection coefficient of the wave reflected from the front of the slab.

Answer 1

The reflection coefficient for a plane wave normally incident on a dielectric slab depends on the permittivity of the slab and can be calculated using the Fresnel equations for normal incidence.

Step-by-step explanation:

The reflection coefficient for a wave normally incident on a dielectric slab can be found by considering the impedance mismatch at the interface between free space and the dielectric material. The dielectric slab has a given permittivity \( \epsilon_r \) and a specific thickness d. The value of d is chosen to be \( \lambda_0 / (4\sqrt{\epsilon_r}) \), where \( \lambda_0 \) is the free-space wavelength of the incident wave. To find the reflection coefficient at the front of the slab, we use the Fresnel equations for normal incidence, which tell us that the reflection coefficient r is

\[ r = \frac{\sqrt{\epsilon_r} - 1}{\sqrt{\epsilon_r} + 1} \]

However, because of the specific thickness of the slab (d), which is a quarter-wavelength optical thickness, the reflection from the second interface can interfere destructively with the reflection from the first interface, potentially leading to zero reflection for perfect matching. The specifics of this interference depend on \( \epsilon_r \) and would affect the overall reflection coefficient.