# 8-2 Snell’s Laws

In the preceding sections, we examined reflection and transmission of plane waves that are normally incident upon a planar interface between two different media. We now consider the oblique-incidence case depicted in Fig. 8-9, and for simplicity, we assume all media to be lossless. The z=0 plane forms the boundary between media 1 and 2 with constitutive parameters (ε 1,μ 1) and (ε 2,μ 2), respectively. The two lines in Fig. 8-9 with direction k ˆ i represent rays drawn normal to the wavefront of the incident wave, and those along directions k ˆ r and k ˆ t are similarly associated with the reflected and transmitted waves. The angles of incidence, reflection, and transmission (or refraction), which are defined with respect to the normal to the boundary (the z axis), are θ i, θ r, and θ t, respectively. These three angles are interrelated by Snell’s laws, which we derive shortly by considering the propagation of the wavefronts of the three waves. Rays of the incident wave intersect the boundary at O and O′. Here A i O represents a constant-phase wavefront of the incident wave. Likewise, A r O′ and A t O′ are constant-phase wavefronts of the reflected and transmitted waves, respectively (Fig. 8-9). The incident and reflected waves propagate in medium 1 with the same phase velocity u p 1=1/μ 1 ε 1, while the transmitted wave in medium 2 propagates with a velocity u p 2=1/μ 2 ε 2. The time it takes for the incident wave to travel from A i to O′ is the same as the time it takes for the reflected wave to travel from O to A r, and also the time it takes the transmitted wave to travel from O to A t. Since time equals distance divided by velocity, it follows that

A i O′¯u p 1=O A r‾u p 1=O A t¯u p 2. (8.26)

# Figure 8-9 Wave reflection and refraction at a planar boundary between different media.

Figure 8 9 Full Alternative Text

From the geometries of the three right triangles in Fig. 8-9, we deduce that

A i O′¯=O O′¯ sin θ i, (8.27a)
O A r¯=O O′¯ sin θ r, (8.27b)
O A t¯=O O′¯ sin θ t. (8.27c)

Use of these expressions in Eq. (8.26) leads to

The index of refraction of a medium, n, is defined as the ratio of the phase velocity in free space (i.e., the speed of light c) to the phase velocity in the medium. Thus,

In view of Eq. (8.29), Eq. (8.28b) may be rewritten as

sin θ t sin θ i=n 1 n 2=μ r 1 ε r 1 μ r 2 ε r 2. (8.30)

For nonmagnetic materials, μ r 1=μ r 2=1, in which case

Usually, materials with higher densities have higher permittivities. Air, with μ r=ε r=1, has an index of refraction n 0=1. Since for nonmagnetic materials n=ε r, a material is often referred to as more dense than another material if it has a greater index of refraction.

At normal incidence (θ i=0), Eq. (8.31) gives θ t=0, as expected. At oblique incidence θ t<θ i when n 2>n 1 and θ t>θ i when n 2<n 1.

# Figure 8-10 Snell’s laws state that θ r=θ i and sin θ t=(n 1/n 2)sin θ i. Refraction is (a) inward if n 1<n 2 and (b) outward if n 1>n 2; (c) the refraction angle is 90° if n 1>n 2 and θ i is equal to or greater than the critical angle θ c=sin−1(n 2/n 1).

(a) n1 < n2

Figure 8 10-a Full Alternative Text

(b) n1 > n2

Figure 8 10-b Full Alternative Text

(c) n1 > n2 and θi = θc

Figure 8 10-c Full Alternative Text

A case of particular interest is when θ t=π/2, as shown in Fig. 8-10(c); in this case, the refracted wave flows along the surface and no energy is transmitted into medium 2. The value of the angle of incidence θ i corresponding to θ t=π/2 is called the critical angle θ c and is obtained from Eq. (8.30) as

If θ i exceeds θ c, the incident wave is totally reflected, and the refracted wave becomes a nonuniform surface wave that travels along the boundary between the two media. This wave behavior is called total internal reflection.

# Example 8-4: Light Beam Passing through a Slab

A dielectric slab with index of refraction n 2 is surrounded by a medium with index of refraction n 1, as shown in Fig. 8-11. If θ i<θ c, show that the emerging beam is parallel to the incident beam.

# Figure 8-11 The exit angle θ 3 is equal to the incidence angle θ 1 if the dielectric slab has parallel boundaries and is surrounded by media with the same index of refraction on both sides (Example 8-4).

Figure 8 11 Full Alternative Text

Solution: At the slab’s upper surface, Snell’s law gives

sin θ 2=n 1 n 2 sin θ 1. (8.33)

Similarly, at the slab’s lower surface,

sin θ 3=n 2 n 3 sin θ 2=n 2 n 1 sin θ 2. (8.34)

Substituting Eq. (8.33) into Eq. (8.34) gives

sin θ 3=n 2 n 1 n 1 n 2 sin θ 1=sin θ 1.

Hence, θ 3=θ 1. The slab displaces the beam’s position, but the beam’s direction remains unchanged.