33-6 Total Internal Reflection

33-6 Total Internal Reflection#

Prompts

  • When does total internal reflection occur? Which medium must have the higher index of refraction?

  • Define the critical angle \(\theta_c\). What is the angle of refraction when \(\theta_1 = \theta_c\)?

  • Derive the formula \(\theta_c = \arcsin(n_2/n_1)\). Why must \(n_2 < n_1\)?

  • For light in glass (\(n \approx 1.5\)) incident on a glass–air interface, what is the critical angle?

  • How do optical fibers use total internal reflection? Give an application.

Lecture Notes#

Overview#

  • Total internal reflection (TIR): When light in a medium of higher index \(n_1\) strikes an interface with a medium of lower index \(n_2\), and the angle of incidence exceeds a critical angle \(\theta_c\), no refracted ray exists—all light is reflected.

  • At \(\theta_c\): the refracted ray would travel along the interface (\(\theta_2 = 90°\)).

  • Applications: optical fibers, endoscopes, prisms in binoculars.

The critical angle#

Consider light in medium 1 (index \(n_1\)) incident on an interface with medium 2 (index \(n_2\)), where \(n_1 > n_2\) (e.g., glass to air).

As the angle of incidence \(\theta_1\) increases, the angle of refraction \(\theta_2\) (from Snell’s law \(n_1\sin\theta_1 = n_2\sin\theta_2\)) also increases. When \(\theta_2\) reaches 90°, the refracted ray lies along the interface. The corresponding angle of incidence is the critical angle \(\theta_c\):

\[ n_1 \sin\theta_c = n_2 \sin 90° = n_2 \]
(245)#\[ \theta_c = \arcsin\left(\frac{n_2}{n_1}\right) \]

  • \(\theta_1 < \theta_c\): both reflection and refraction; refracted ray enters medium 2.

  • \(\theta_1 = \theta_c\): refracted ray at 90° (along interface).

  • \(\theta_1 > \theta_c\): total internal reflection—no refracted ray; all light stays in medium 1.

Important

Total internal reflection occurs only when light is in the medium with the higher index (\(n_1 > n_2\)). If the light is in the lower-index medium, Snell’s law always gives a real \(\theta_2\); there is no critical angle and no TIR.

Example: glass to air

For glass (\(n_1 \approx 1.5\)) to air (\(n_2 \approx 1\)):

\[ \theta_c = \arcsin\left(\frac{1}{1.5}\right) \approx 42° \]

For water (\(n \approx 1.33\)) to air: \(\theta_c \approx 49°\).

Poll: Optical fiber—which angles keep light in?

An optical fiber has index \(n\) and is surrounded by air (\(n \approx 1\)). Light travels inside the fiber. Which angles of incidence \(\theta\) (at the fiber–air interface) will keep the light inside the fiber?

  • (A) \(\theta > \arcsin(1/n)\)

  • (B) \(\theta < \arcsin(1/n)\)

  • (C) \(\theta > \arcsin(\sqrt{n^2 - 1})\)

  • (D) \(\theta < \arcsin(\sqrt{n^2 - 1})\)

  • (E) Depends on how \(\theta\) is measured

Optical fibers#

Optical fibers guide light by repeated total internal reflection. A core of higher-index glass is surrounded by a cladding of lower-index material. Light entering the core at angles that keep the incidence on the core–cladding interface above \(\theta_c\) is trapped and travels along the fiber.

Applications: telecommunications (fiber-optic cables), endoscopes (viewing inside the body), sensors.

Summary#

  • Total internal reflection: \(n_1 > n_2\), \(\theta_1 > \theta_c\) → no refraction, all light reflected.

  • Critical angle: \(\theta_c = \arcsin(n_2/n_1)\); at \(\theta_c\), refracted ray at 90°.

  • Condition: light must be in the higher-index medium.

  • Applications: optical fibers, endoscopes, prisms.