Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 6.4.2

Total internal reflection in a prism

A right-angle glass prism ( $n = 1.5$ , nonmagnetic) turns a beam of light of free-space wavelength $λ_{0} = 600 nm$ through $9 0^{\circ}$ : inside the glass the beam meets the hypotenuse face at $θ_{i} = 4 5^{\circ}$ , with air outside.

Show that the beam is totally reflected.
Find $Γ_{⊥}$ and $Γ_{∥}$ (magnitude and phase).
Find the attenuation constant of the field in the air just outside the hypotenuse, and the distance in which it falls to $1/ e$ .
Show that no time-average power flows into the air.

Try it yourself first.

Strategy

Compare $θ_{i}$ with $θ_{c}$ . Beyond $θ_{c}$ Snell's law gives $sin θ_{t} > 1$ , and $cos θ_{t}$ is the imaginary root $- j sin^{2} θ_{t} - 1$ that makes the field decay into the air, Eq. (6.4.18). With it the Fresnel coefficients still hold. Work with impedances divided by $η_{0}$ : medium 1 is the glass, $η_{1} \to \frac{2}{3}$ , medium 2 the air, $η_{2} \to 1$ .

$θ_{c} = arcsin (1/1.5) = 41. 8^{\circ} < 4 5^{\circ}$ : the beam is totally reflected.

$sin θ_{t} = 1.5 sin 4 5^{\circ} = 1.0607$ , $cos θ_{t} = - j 1.125 - 1 = - j 0.3536$ , $cos θ_{i} = 0.7071$ :

Γ_{⊥} Γ_{∥} = \frac{0.7071 + j \frac{2}{3} ( 0.3536 )}{0.7071 - j \frac{2}{3} ( 0.3536 )} = \frac{0.7071 + j 0.2357}{0.7071 - j 0.2357} = 0.8 + j 0.6 = 1∠36. 9^{\circ}, = \frac{- j 0.3536 - \frac{2}{3} ( 0.7071 )}{- j 0.3536 + \frac{2}{3} ( 0.7071 )} = \frac{- 0.4714 - j 0.3536}{0.4714 - j 0.3536} = - 0.28 - j 0.96 = 1∠ - 106. 3^{\circ} .

$β_{0} = 2 π / λ_{0} = 1.047 \times 1 0^{7} rad/m$ , $β_{1} = 1.5 β_{0}$ , $β_{2} = β_{0}$ , and from Eq. (6.4.19)

α_{2} = β_{1}^{2} sin^{2} θ_{i} - β_{2}^{2} = β_{0} 1. 5^{2} sin^{2} 4 5^{\circ} - 1 = 1.047 \times 1 0^{7} \times 0.3536 = 3.70 \times 1 0^{6} Np/m,

so the field falls to $1/ e$ in $1/ α_{2} = 270 nm$ $= 0.45 λ_{0}$ . Along the face it varies as $e^{- j β_{1} y s i n θ_{i}}$ with $β_{1} sin θ_{i} = 1.11 \times 1 0^{7} rad/m$ .

S case: $E_{z}^{t} = E_{0}^{t} e^{- α_{2} x} e^{- j β_{1} y s i n θ_{i}}$ and $H_{y}^{t} = - (cos θ_{t} / η_{2}) E_{z}^{t}$ . The normal component of the time-average Poynting vector is

P_{x} = \frac{1}{2} Re {E_{y} H_{z}^{*} - E_{z} H_{y}^{*}} = \frac{∣ E _{0}^{t} ∣ ^{2} e ^{- 2 α_{2} x}}{2 η _{2}} Re {cos θ_{t}^{*}} = \frac{∣ E _{0}^{t} ∣ ^{2} e ^{- 2 α_{2} x}}{2 η _{2}} Re {j 0.3536} = 0.

In the P case $E_{y}^{t} \propto cos θ_{t}$ is in quadrature with $H_{z}^{t}$ , and again $P_{x} = 0$ .

Answer

θ_{c} = 41. 8^{\circ}, Γ_{⊥} = 1∠36. 9^{\circ}, Γ_{∥} = 1∠ - 106. 3^{\circ}, α_{2} = 3.70 \times 1 0^{6} Np/m, 1/ α_{2} = 270 nm

Check

Numerator and denominator of each $Γ$ are complex conjugates (apart from the sign in $Γ_{∥}$ ), so $∣ Γ_{⊥} ∣ = ∣ Γ_{∥} ∣ = 1$ . This agrees with $P_{x} = 0$ : the reflected wave must carry all the incident power. The phases follow from Eq. (6.4.20): $2 arctan \frac{( 2/3 ) \times 0.3536}{0.7071} = 2 arctan \frac{1}{3} = 36.8 7^{\circ}$ .

Problem 6.4.2. The phases of \Gamma_\perp and \Gamma_\parallel for glass (n=1.5) to air beyond the critical angle, where |\Gamma_\perp|=|\Gamma_\parallel|=1. The dots are the answers of (b). The envelope of the evanescent field at 45^\circ is the middle curve of Figure 6.4.7b. — **Figure 6.4.17.** Problem 6.4.2. The phases of $Γ_{⊥}$ and $Γ_{∥}$ for glass ( $n = 1.5$ ) to air beyond the critical angle, where $∣ Γ_{⊥} ∣ = ∣ Γ_{∥} ∣ = 1$ . The dots are the answers of (b). The envelope of the evanescent field at $4 5^{\circ}$ is the middle curve of Figure 6.4.7b.

What the graphs say

The field in the air is not zero: $τ_{⊥} = 1 + Γ_{⊥} = 1.8 + j 0.6$ , so $∣ E_{0}^{t} ∣ = 1.90 E_{0}^{i}$ right at the surface, and $P_{y} = \frac{1}{2} Re {E_{z} H_{x}^{*}} \neq = 0$ : power flows along the face inside the evanescent layer but not away from it. Just above $θ_{c}$ the field reaches far into the air ( $1.1 μ m$ at $4 2^{\circ}$ ): the reflection is total but not sharp. The phases of $Γ_{⊥}$ and $Γ_{∥}$ differ, so total reflection turns linear into elliptical polarisation. The Fresnel rhomb uses two such reflections to make circular polarisation, and Exercise 6.4.5 follows the same effect through a Porro prism.

Problem 6.4.2: Total internal reflection in a prism