# 8.5 Reflectivity and Transmissivity

The reflection and transmission coefficients derived earlier are ratios of the reflected and transmitted electric field amplitudes to the amplitude of the incident electric field. We now examine power ratios, starting with the perpendicular polarization case. Figure 8-18 shows a circular beam of electromagnetic energy incident upon the boundary between two contiguous, lossless media. The area of the spot illuminated by the beam is A, and the incident, reflected, and transmitted beams have electric-field amplitudes E⊥0 i, E⊥0 r, and E⊥0 t, respectively. The average power densities carried by the incident, reflected, and transmitted beams are

S⊥i=|E⊥0 i|2 2 η 1, (8.73a)
S⊥r=|E⊥0 r|2 2 η 1, (8.73b)
S⊥t=|E⊥0 t|2 2 η 2, (8.73c)

where η 1 and η 2 are the intrinsic impedances of media 1 and 2, respectively. The cross-sectional areas of the incident, reflected, and transmitted beams are

A i=A cos θ i, (8.74a)
A r=A cos θ r, (8.74b)
A t=A cos θ t, (8.74c)

and the corresponding average powers carried by the beams are

P⊥i=S⊥i A i=|E⊥0 i|2 2 η 1 A cos θ i, (8.75a)
P⊥r=S⊥r A r=|E⊥0 r|2 2 η 1 A cos θ r, (8.75b)
P⊥t=S⊥t A t=|E⊥0 t|2 2 η 2 A cos θ t. (8.75c)

# Figure 8-18 Reflection and transmission of an incident circular beam illuminating a spot of size A on the interface.

Figure 8 18 Full Alternative Text

The reflectivity R (also called reflectance in optics) is defined as the ratio of the reflected to the incident power. The reflectivity for perpendicular polarization is then

R⊥=P⊥r P⊥i=|E⊥0 r|2 cos θ r|E⊥0 i|2 cos θ i=E⊥0 r E⊥0 i 2, (8.76)

where we used the fact that θ r=θ i in accordance with Snell’s law of reflection. The ratio of the reflected to incident electric field amplitudes, |E⊥0 r/E⊥0 i|, is equal to the magnitude of the reflection coefficient Γ⊥. Hence,

and similarly, for parallel polarization

The transmissivity T (or transmittance in optics) is defined as the ratio of the transmitted power to incident power:

In fact, in many cases, the transmitted electric field is larger than the incident electric field. Conservation of power requires that the incident power equals the sum of the reflected and transmitted powers. That is, for perpendicular polarization,

P⊥i=P⊥r+P⊥t, (8.80)

or

|E⊥0 i|2 2 η 1 A cos θ i=|E⊥0 r|2 2 η 1 A cos θ r+|E⊥0 t|2 2 η 2 A cos θ t. (8.81)

Use of Eqs. (8.76), (8.79a), and (8.79b) leads to

R⊥+T⊥=1, (8.82a)
R∥+T∥=1, (8.82b)

or

Figure 8-19 shows plots for (R∥,T∥) as a function of θ i for an air–glass interface. Note that the sum of R∥ and T∥ is always equal to 1, as mandated by Eq. (8.82b). We also note that, at the Brewster angle θ B, R∥=0 and T∥=1.

# Figure 8-19 Angular plots for (R∥,T∥) for an air–glass interface.

Figure 8 19 Full Alternative Text

Table 8-2 provides a summary of the general expressions for Γ, τ, R, and T for both normal and oblique incidence.

# Table 8-2 Expressions for Γ, τ, R, and T for wave incidence from a medium with intrinsic impedance η 1 onto a medium with intrinsic impedance η 2. Angles θ i and θ t are the angles of incidence and transmission, respectively.

Table 8 2 Full Alternative Text

# Example 8-7: Beam of Light

A 5 W beam of light with circular cross section is incident in air upon the plane boundary of a dielectric medium with index of refraction of 5. If the angle of incidence is 60° and the incident wave is parallel polarized, determine the transmission angle and the powers contained in the reflected and transmitted beams.

Solution: From Eq. (8.56),

sin θ t=n 1 n 2 sin θ i=1 5 sin 60°=0.17

or

θ t=10°.

With ε 2/ε 1=n 2 2/n 1 2=(5)2=25, the reflection coefficient for parallel polarization follows from Eq. (8.68) as

Γ∥=−(ε 2/ε 1) cos θ i+(ε 2/ε 1)−sin 2 θ i(ε 2/ε 1) cos θ i+(ε 2/ε 1)−sin 2 θ i=−25 cos 60°+25−sin 2 60°25 cos 60°+25−sin 2 60°=−0.435.

The reflected and transmitted powers therefore are

P∥r=P∥i|Γ∥|2=5(0.435)2=0.95 W,P∥t=P∥i−P∥r=5−0.95=4.05 W.