Apply Snell's law $n_1 \sin \theta_1 = n_2 \sin \theta_2$ to predict the refraction of light at a boundary between two media, including the critical angle for total internal reflection, and explain dispersion in terms of frequency-dependent refractive index

What this dot point is asking

VCAA wants you to apply Snell's law to refraction at boundaries, calculate critical angles for total internal reflection, explain dispersion as the frequency dependence of the refractive index, and connect these phenomena to applications including fibre optics and prisms.

Refraction

Refraction is the change in direction of a wave passing from one medium to another with a different propagation speed. For light, refraction occurs when light enters or leaves a medium with a different refractive index.

The refractive index of a medium is:

where $c$ is the speed of light in vacuum and $v$ is the speed of light in the medium. By definition $n \geq 1$ for ordinary materials.

Typical values:

A higher $n$ means light travels slower in the medium and bends more sharply at the surface.

Snell's law

When light passes from medium 1 (refractive index $n_1$) to medium 2 (refractive index $n_2$) at angle $\theta_1$ from the normal, the refracted ray emerges at angle $\theta_2$ from the normal, where:

This is Snell's law.

Two cases:

• From less dense to more dense ($n_2 > n_1$). $\sin \theta_2 < \sin \theta_1$, so $\theta_2 < \theta_1$. The refracted ray bends toward the normal.
• From more dense to less dense ($n_2 < n_1$). $\sin \theta_2 > \sin \theta_1$, so $\theta_2 > \theta_1$. The refracted ray bends away from the normal.

Both incident and refracted rays, and the normal, lie in the same plane (the plane of incidence).

What happens at the boundary

When light hits a boundary, three things happen:

1. Some reflects back into medium 1 at the angle of incidence (law of reflection: $\theta_r = \theta_1$).
2. Some refracts into medium 2 at angle $\theta_2$.
3. Some is absorbed by the material (typically a small fraction unless the medium is strongly absorbing).

The proportions reflected vs refracted depend on the angle and on the refractive indices (Fresnel equations, beyond VCE scope). For VCE Physics, treat both reflected and refracted rays as present; their angles are determined by reflection and Snell's law.

Critical angle and total internal reflection

When light travels from a denser to a less dense medium ($n_1 > n_2$), Snell's law predicts $\theta_2 > \theta_1$. There is a specific angle of incidence at which $\theta_2 = 90^\circ$ (the refracted ray is along the boundary): this is the critical angle $\theta_c$.

From Snell's law with $\theta_2 = 90^\circ$:

For angles of incidence greater than $\theta_c$, no refraction is possible (the equation $\sin \theta_2 = (n_1 / n_2) \sin \theta_1 > 1$ has no real solution). All the light reflects back into the denser medium: total internal reflection (TIR).

Conditions for TIR

1. Light must travel from a denser medium to a less dense medium ($n_1 > n_2$).
2. The angle of incidence must exceed the critical angle ($\theta_1 > \theta_c$).

Standard critical angles

• Water (1.33) to air (1.00): $\theta_c = \arcsin(1.00 / 1.33) \approx 48.6^\circ$.
• Glass (1.50) to air: $\theta_c = \arcsin(1.00 / 1.50) \approx 41.8^\circ$.
• Diamond (2.42) to air: $\theta_c = \arcsin(1.00 / 2.42) \approx 24.4^\circ$.

Diamond's very small critical angle is the reason for its sparkle: light entering from above is internally reflected many times before exiting through specific facets at the bottom.

Applications of total internal reflection

Optical fibres

A thin glass or plastic core (high $n$) surrounded by a cladding (lower $n$). Light entering the core at sufficiently grazing angle reflects off the core-cladding boundary at angles greater than $\theta_c$, so all light remains trapped in the core as it travels along the fibre. Used for high-bandwidth communications (internet, telephony) and medical endoscopes.

Prisms

Right-angle prisms can be used as 100 percent efficient reflectors at angles where the light hits the back face at greater than $\theta_c$. Used in binoculars, periscopes, and reflex cameras.

Diamond brilliance

As above.

Mirages

Hot air near a road has lower density and lower $n$ than air above. Light from the sky bends as it traverses the density gradient, and at very grazing angles undergoes effective TIR off the hot air layer, creating an apparent puddle.

Dispersion

The refractive index of a real medium depends on the frequency of the light: $n = n(f)$. This frequency dependence is dispersion.

For most transparent materials, $n$ is larger for higher frequencies (shorter wavelengths). Violet light bends more than red light when entering a denser medium.

Prism dispersion

A glass prism refracts light entering one face and refracts it again on exit. Because different colours have different $n$, they bend by different amounts, and the prism separates white light into a spectrum.

Order (most bent to least): violet, blue, green, yellow, orange, red.

This is why a prism produces the familiar rainbow band of colours from a beam of white light. Newton (1665) used a prism to demonstrate that white light is composed of all the visible colours, and that the colours are not introduced by the prism.

Rainbows

A rainbow is dispersion in raindrops. Sunlight enters a spherical raindrop, refracts (with dispersion), reflects off the back of the drop, and exits refracting again. Different colours emerge at slightly different angles, producing the familiar arc with red on the outside (42 degrees from the antisolar point) and violet on the inside (40 degrees).

Chromatic aberration in lenses

Single-element lenses (like a magnifying glass) suffer from chromatic aberration: different colours focus at slightly different points because their refractive indices differ. Camera and microscope lenses correct this with multiple elements made of different glass types.

Examples in context

Example 1. Optical-fibre internet backbone from Hobart to Melbourne via Bass Strait. Optical fibres laid under Bass Strait carry telecom signals via total internal reflection. Core refractive index $n_{\rm core} = 1.50$, cladding $n_{\rm clad} = 1.48$. Critical angle is $\sin\theta_c = n_{\rm clad}/n_{\rm core} = 1.48/1.50 = 0.987$, giving $\theta_c = 80.6^\circ$. Light entering the fibre at angles less than $9.4^\circ$ from the axis bounces along the core by total internal reflection. Glass dispersion ($n$ varies slightly with wavelength) causes different colour components to travel at slightly different speeds; this is countered by using single-wavelength laser sources at $1550$ nm where attenuation in silica is minimal.

Example 2. Rainbow over the Yarra after a Melbourne storm. Sunlight entering a $1$ mm water droplet refracts at the front surface, reflects off the back, and refracts again exiting the front. Snell's law at each refraction with $n_{\rm water}$ varying from $1.331$ (red) to $1.343$ (violet) produces a $40$-$42^\circ$ angular spread. Red light emerges at $42.4^\circ$ from the antisolar point; violet at $40.2^\circ$. The $\approx 2^\circ$ angular spread is the rainbow's width. Secondary rainbows arise from two internal reflections at a wider angle ($51^\circ$), with reversed colour order.

Try this

Q1. State Snell's law and define the critical angle for total internal reflection. [2 marks]

• Cue. $n_1 \sin\theta_1 = n_2 \sin\theta_2$. Critical angle is $\theta_c$ such that $\sin\theta_c = n_2/n_1$ for light passing from denser to less dense medium.

Q2. A ray of light passes from glass ($n = 1.50$) into water ($n = 1.33$). For an incidence angle of $40^\circ$, calculate the refraction angle. [3 marks]

• Cue. $1.50 \sin 40^\circ = 1.33 \sin\theta_2$; $\sin\theta_2 = 0.725$; $\theta_2 = 46.5^\circ$.

Q3. Refer to optical fibres for Bass Strait telecoms. (a) Define total internal reflection. (b) Calculate the critical angle for $n_{\rm core} = 1.50$, $n_{\rm clad} = 1.48$. (c) Explain why dispersion can degrade a fibre-optic signal. [2+2+3 marks]

• Cue. (a) Light reflects entirely back into the denser medium at angles above $\theta_c$. (b) $80.6^\circ$. (c) Different wavelength components travel at different speeds, smearing out time-domain pulses.