TEM Wave Oblique Incidence Total Reflection

The problem asks for the angle of incidence ($\theta_i$) at which total reflection occurs when a TEM wave moves from a dielectric medium 1 ($\epsilon_{r1} = 2$) to a dielectric medium 2 ($\epsilon_{r2} = 1$) at an oblique angle.

Calculating Refractive Indices

For non-magnetic dielectric materials ($\mu_r = 1$), the refractive index ($n$) is related to the relative permittivity ($\epsilon_r$) by the formula:

$n = \sqrt{\mu_r \epsilon_r} = \sqrt{\epsilon_r}$

Calculate the refractive indices for both media:

• Medium 1: $n_1 = \sqrt{\epsilon_{r1}} = \sqrt{2}$
• Medium 2: $n_2 = \sqrt{\epsilon_{r2}} = \sqrt{1} = 1$

Determining the Critical Angle

Total internal reflection occurs when the wave travels from a medium with a higher refractive index to one with a lower refractive index ($n_1 > n_2$), and the angle of incidence exceeds the critical angle ($\theta_c$). In this case, $n_1 = \sqrt{2}$ and $n_2 = 1$, so $n_1 > n_2$, and total reflection is possible.

The critical angle $\theta_c$ is found using the relationship derived from Snell's Law, where the angle of refraction is $90^\circ$:

$\sin(\theta_c) = \frac{n_2}{n_1}$

Substitute the calculated refractive indices:

$\sin(\theta_c) = \frac{1}{\sqrt{2}}$

Finding the Angle of Incidence

To find the angle $\theta_c$, we take the arcsine:

$\theta_c = \arcsin\left(\frac{1}{\sqrt{2}}\right)$

This value corresponds to:

$\theta_c = 45^\circ$

Therefore, the angle of incidence for total reflection is $45^\circ$. This matches Option C.