What would have happened in Example 34.4 if the prism angle had been $45\text{°}$ rather than $30\text{°}$? The light rays would approach the rear surface of the prism at an angle of incidence ${\theta }_1=45\text{°}$. When we try to calculate the angle of refraction at which the ray emerges into the air, we find

Angle ${\theta }_2$ does not exist because the sine of an angle can’t be larger than 1. The ray is unable to refract through the boundary. Instead, 100% of the light reflects from the boundary back into the prism. This process is called total internal reflection, often abbreviated TIR. That it really happens is illustrated in FIGURE 34.20. Here three laser beams enter a prism from the left. The bottom two refract out through the right side of the prism. The blue beam, which is incident on the prism’s top face, undergoes total internal reflection and then emerges through the right surface.

FIGURE 34.20 The blue laser beam undergoes total internal reflection inside the prism.

FIGURE 34.21 shows several rays leaving a point source in a medium with index of refraction $n_1$. The medium on the other side of the boundary has $n_2<n_1$. As we’ve seen, crossing a boundary into a material with a lower index of refraction causes the ray to bend away from the normal. Two things happen as angle ${\theta }_1$ increases. First, the refraction angle ${\theta }_2$ approaches $90\text{°}$. Second, the fraction of the light energy transmitted decreases while the reflected fraction increases.

FIGURE 34.21 Refraction and reflection of rays as the angle of incidence increases.

A critical angle is reached when ${\theta }_2=90\text{°}$. Because $\text{sin}90\text{°}=1$, Snell’s law $n_1\text{sin}{\theta }_c=n_2\text{sin}90\text{°}$ gives the critical angle of incidence as

The refracted light vanishes at the critical angle and the reflection becomes 100% for any angle ${\theta }_1\ge {\theta }_{\text{c}}$. The critical angle is well defined because of our assumption that $n_2<n_1$. There is no critical angle and no total internal reflection if ${\mathit{n}}_2>{\mathit{n}}_1$.

As a quick example, the critical angle in a typical piece of glass at the glass-air boundary is

The fact that the critical angle is less than $45\text{°}$ has important applications. For example, FIGURE 34.22 shows a pair of binoculars. The lenses are much farther apart than your eyes, so the light rays need to be brought together before exiting the eyepieces. Rather than using mirrors, which get dirty and require alignment, binoculars use a pair of prisms on each side. Thus the light undergoes two total internal reflections and emerges from the eyepiece. (The actual arrangement is a little more complex than in Figure 34.22, to avoid left-right reversals, but this illustrates the basic idea.)

FIGURE 34.22 Binoculars make use of total internal reflection.

EXAMPLE 34.5 Total internal reflection

A small lightbulb is set in the bottom of a 3.0-m-deep swimming pool. What is the diameter of the circle of light seen on the water’s surface from above?

MODEL Use the ray model of light.

VISUALIZE FIGURE 34.23 is a pictorial representation. The lightbulb emits rays at all angles, but only some of the rays refract into the air and are seen from above. Rays striking the surface at greater than the critical angle undergo TIR and remain within the water. The diameter of the circle of light is the distance between the two points at which rays strike the surface at the critical angle.

FIGURE 34.23 Pictorial representation of the rays leaving a lightbulb at the bottom of a swimming pool.

FIGURE 34.23 Full Alternative Text

SOLVE From trigonometry, the circle diameter is $D=2h\text{tan}{\theta }_c$, where $h$ is the depth of the water. The critical angle for a water-air boundary is ${\theta }_{\text{c}}={\text{sin}}^{-1}(1.00/1.33)=48.7\text{°}$. Thus

$$D=2(3.0\text{m})\text{tan}48.7\text{°}=6.8\text{m}$$