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AP Physics 2: Algebra Notes

5.3.6 Total Internal Reflection and Critical Angle

AP Syllabus focus: 'Total internal reflection can occur from higher to lower refractive index; at the critical angle the refracted ray travels along the surface, and beyond it all light reflects.'

Total internal reflection explains how light can be trapped inside a material instead of passing through a boundary. To understand it, you must track the refractive indices, the incident angle, and the critical angle.

When Total Internal Reflection Occurs

Total internal reflection happens at a boundary between two transparent media. The light must start in the medium with the greater index of refraction and travel toward a medium with the smaller index. As the angle of incidence increases, the refracted ray bends farther away from the normal. This trend continues until the refracted ray reaches the boundary itself. If the angle is increased even more, the light no longer passes into the second medium.

Total internal reflection: When light striking a boundary is completely reflected back into the original medium.

Two conditions must both be satisfied.

  • The light must go from higher refractive index to lower refractive index.

  • The angle of incidence must be large enough.

If either condition fails, total internal reflection does not occur. For example, light going from air into glass can refract and reflect, but it cannot undergo total internal reflection at that boundary because it is entering the higher-index medium.

The Critical Angle

The special angle that separates ordinary refraction from total internal reflection is called the critical angle. It is measured in the higher-index medium, from the normal to the surface. At this angle, the refracted ray is not directed into the second medium. Instead, it travels along the boundary, making a 9090^\circ angle with the normal.

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Schematic of total internal reflection geometry at a boundary with n1>n2n_1>n_2. The diagram emphasizes that the critical angle θc\theta_c is measured in the higher-index medium from the normal, and that the refracted ray is tangent to the interface at θc\theta_c. This is the limiting case used to derive sinθc=n2n1\sin\theta_c=\dfrac{n_2}{n_1}. Source

Critical angle: The angle of incidence in the higher-index medium for which the refracted ray travels along the boundary.

For a boundary where n1>n2n_1>n_2, the critical angle can be found from the refractive indices.

sinθc=n2n1 \sin \theta_c = \dfrac{n_2}{n_1}

θc \theta_c = Critical angle, in degrees or radians

n1 n_1 = Index of refraction of the incident medium, unitless

n2 n_2 = Index of refraction of the second medium, unitless

This equation only applies when the incident medium has the larger refractive index. If n1n2n_1 \le n_2, total internal reflection cannot occur for that direction of travel, so a useful critical angle is not defined. A smaller critical angle means total internal reflection begins at a smaller angle of incidence.

Below, At, and Above the Critical Angle

At the same boundary, three different behaviors are possible depending on the incident angle:

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Ray diagrams for the same boundary as the incident angle increases. The middle case highlights the defining feature of the critical angle: the refracted ray travels along the interface (i.e., the refracted angle is 9090^\circ to the normal). Above this angle, the diagram shows only a reflected ray in the original (higher-index) medium. Source

  • Below the critical angle: Some light refracts into the second medium, and some may reflect. The transmitted ray bends away from the normal because it is entering a lower-index medium.

  • At the critical angle: The refracted ray travels exactly along the surface. This is the limiting case between refraction and total internal reflection.

  • Above the critical angle: No refracted ray enters the second medium. All the light is reflected back into the original medium, so the interaction is total internal reflection.

These three cases are especially important in ray diagrams. The critical angle marks the boundary between a situation with a transmitted ray and a situation with only a reflected ray.

Interpreting What "All Light Reflects" Means

Within the geometric optics model, all light reflects means no refracted ray is drawn beyond the boundary once the incident angle exceeds the critical angle. The boundary still matters because the light changes direction there, but the light stays entirely in the original medium. This is why total internal reflection can trap light inside transparent materials. The effect is not caused by a mirror coating. It comes from the change in refractive index at the boundary itself. For AP Physics 2 reasoning, once total internal reflection begins, the second medium no longer carries a transmitted ray in the ray diagram.

Interpreting Ray Diagrams and Problems

In problems, first identify which medium the light starts in. Then compare the two indices of refraction. If the light is moving toward a lower-index medium, total internal reflection is possible. Next compare the incident angle with the critical angle:

  • If θi<θc\theta_i<\theta_c, refraction occurs.

  • If θi=θc\theta_i=\theta_c, the refracted ray runs along the surface.

  • If θi>θc\theta_i>\theta_c, total internal reflection occurs.

A common mistake is to think that a large angle alone guarantees total internal reflection. It does not. The direction of travel matters just as much as the size of the angle. Another common mistake is measuring the angle from the surface instead of from the normal. In this topic, all relevant angles are measured from the normal.

FAQ

An optical fiber has a core with a higher refractive index than the surrounding cladding. Light inside the core can strike the boundary at angles greater than the critical angle, so it stays trapped inside the fiber.

This allows light signals to travel long distances with low loss. If the fiber bends too sharply or has imperfections, some light can still escape.

Diamond has a very high refractive index, which gives it a relatively small critical angle. That makes it easier for light inside the diamond to hit a surface above the critical angle and reflect internally many times.

A well-cut diamond is shaped to encourage this repeated internal reflection, so more light eventually leaves in bright directions that make the stone look brilliant.

Yes. The refractive index of a material can vary slightly with wavelength, so different colors can have slightly different critical angles.

That means red, green, and blue light may not all begin total internal reflection at exactly the same incident angle. In some materials, this contributes to color separation effects.

Yes. Total internal reflection is not limited to solids. It can occur at any boundary between transparent media if light travels from the higher-index medium to the lower-index medium and the incident angle is large enough.

In practice, the effect is often easier to observe in solids and liquids because their boundaries are usually more stable and their refractive index differences are often larger.

Frustrated total internal reflection happens when light would normally undergo total internal reflection, but another material is placed extremely close to the boundary. Under those conditions, some light can cross the gap.

This is a wave effect rather than a simple ray effect. In AP Physics 2 Algebra, total internal reflection is treated with rays, but this phenomenon shows that the ray model has limits in very thin-gap situations.

Practice Questions

A light ray travels inside water toward a water-air boundary. State the two conditions required for total internal reflection to occur.

  • 1 mark: States that light must travel from a higher refractive index medium to a lower refractive index medium.

  • 1 mark: States that the angle of incidence must be greater than the critical angle.

A ray of light in glass with index of refraction n1=1.50n_1=1.50 strikes a glass-air boundary where n2=1.00n_2=1.00.

(a) Calculate the critical angle.

(b) State what happens at the boundary for incident angles of 3030^\circ, 41.841.8^\circ, and 5050^\circ.

  • 1 mark: Uses sinθc=n2n1 \sin \theta_c=\dfrac{n_2}{n_1}

  • 1 mark: Substitutes correctly, sinθc=1.001.50 \sin \theta_c=\dfrac{1.00}{1.50}

  • 1 mark: Calculates θc41.8 \theta_c \approx 41.8^\circ

  • 1 mark: States that at 3030^\circ the ray refracts into the air, so total internal reflection does not occur.

  • 1 mark: States that at 41.841.8^\circ the refracted ray travels along the surface.

  • 1 mark: States that at 5050^\circ total internal reflection occurs.

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