Unit 4: How have new ideas and ways of thinking developed our understanding of the physical world?

VICPhysicsSyllabus dot point

How is scientific inquiry used to investigate fields, motion or light?

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

A focused answer to the VCE Physics Unit 4 dot point on refraction. Snell's law, refractive index, the critical angle for total internal reflection, and dispersion as the frequency dependence of refractive index. Includes worked examples and the fibre-optics context.

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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:

n=cvn = \frac{c}{v}

where cc is the speed of light in vacuum and vv is the speed of light in the medium. By definition n1n \geq 1 for ordinary materials.

Typical values:

Medium IMATH_7 (visible light)
Vacuum 1.0000 (exact)
Air at STP 1.0003 (effectively 1.00)
Water 1.33
Crown glass 1.52
Diamond 2.42

A higher nn 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 n1n_1) to medium 2 (refractive index n2n_2) at angle θ1\theta_1 from the normal, the refracted ray emerges at angle θ2\theta_2 from the normal, where:

n1sinθ1=n2sinθ2n_1 \sin \theta_1 = n_2 \sin \theta_2

This is Snell's law.

Two cases:

  • From less dense to more dense (n2>n1n_2 > n_1). sinθ2<sinθ1\sin \theta_2 < \sin \theta_1, so θ2<θ1\theta_2 < \theta_1. The refracted ray bends toward the normal.
  • From more dense to less dense (n2<n1n_2 < n_1). sinθ2>sinθ1\sin \theta_2 > \sin \theta_1, so θ2>θ1\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: θr=θ1\theta_r = \theta_1).
  2. Some refracts into medium 2 at angle θ2\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 (n1>n2n_1 > n_2), Snell's law predicts θ2>θ1\theta_2 > \theta_1. There is a specific angle of incidence at which θ2=90\theta_2 = 90^\circ (the refracted ray is along the boundary): this is the critical angle θc\theta_c.

From Snell's law with θ2=90\theta_2 = 90^\circ:

n1sinθc=n2sin90=n2n_1 \sin \theta_c = n_2 \sin 90^\circ = n_2

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

For angles of incidence greater than θc\theta_c, no refraction is possible (the equation sinθ2=(n1/n2)sinθ1>1\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 (n1>n2n_1 > n_2).
  2. The angle of incidence must exceed the critical angle (θ1>θc\theta_1 > \theta_c).

Standard critical angles

  • Water (1.33) to air (1.00): θc=arcsin(1.00/1.33)48.6\theta_c = \arcsin(1.00 / 1.33) \approx 48.6^\circ.
  • Glass (1.50) to air: θc=arcsin(1.00/1.50)41.8\theta_c = \arcsin(1.00 / 1.50) \approx 41.8^\circ.
  • Diamond (2.42) to air: θc=arcsin(1.00/2.42)24.4\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 nn) surrounded by a cladding (lower nn). Light entering the core at sufficiently grazing angle reflects off the core-cladding boundary at angles greater than θc\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 θc\theta_c. Used in binoculars, periscopes, and reflex cameras.

Diamond brilliance. As above.

Mirages. Hot air near a road has lower density and lower nn 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)n = n(f). This frequency dependence is dispersion.

For most transparent materials, nn 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 nn, 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.

Worked examples

Example 1. Light entering glass

Light in air (n1=1.00n_1 = 1.00) hits glass (n2=1.50n_2 = 1.50) at 45 degrees.

1.00×sin45=1.50×sinθ21.00 \times \sin 45^\circ = 1.50 \times \sin \theta_2

sinθ2=0.707/1.50=0.471\sin \theta_2 = 0.707 / 1.50 = 0.471

θ2=28.1\theta_2 = 28.1^\circ.

The ray bends toward the normal (entering a denser medium).

Example 2. Light exiting water at critical angle

Light in water (n1=1.33n_1 = 1.33) hits the surface at 50 degrees.

1.33×sin50=1.00×sinθ21.33 \times \sin 50^\circ = 1.00 \times \sin \theta_2

sinθ2=1.33×0.766=1.019\sin \theta_2 = 1.33 \times 0.766 = 1.019.

This is greater than 1: no real solution. Total internal reflection occurs. The critical angle for water-to-air is about 48.6 degrees, and 50 is beyond it.

Example 3. Dispersion in glass

A glass block has nred=1.50n_{\text{red}} = 1.50 and nviolet=1.55n_{\text{violet}} = 1.55. Light enters at 45 degrees. Find the angles of refraction for red and violet.

Red: sinθr=sin45/1.50=0.471\sin \theta_r = \sin 45^\circ / 1.50 = 0.471, θr=28.1\theta_r = 28.1^\circ.

Violet: sinθv=sin45/1.55=0.456\sin \theta_v = \sin 45^\circ / 1.55 = 0.456, θv=27.1\theta_v = 27.1^\circ.

The violet bends 1 degree more than red. After a few cm of glass, the two colours are visibly separated. This is the mechanism behind a prism.

Common errors

Snell's law inverted. n1sinθ1=n2sinθ2n_1 \sin \theta_1 = n_2 \sin \theta_2. A common slip is to write sinθ1/sinθ2=n2/n1\sin \theta_1 / \sin \theta_2 = n_2 / n_1 (inverted ratio).

Critical angle confused with incidence angle. θc\theta_c is computed from sinθc=n2/n1\sin \theta_c = n_2 / n_1 where n1>n2n_1 > n_2. It is the threshold angle of incidence at which TIR begins.

TIR in wrong direction. TIR only occurs going from denser to less dense (n1>n2n_1 > n_2). It cannot occur going from air into glass.

Angles from the wrong reference. Snell's law uses angles from the normal (perpendicular to the surface), not from the surface itself.

Dispersion direction. For ordinary glass, blue / violet light bends more than red. The rainbow has red on the outside, violet on the inside. Reversing this is a common slip.

Forgetting to take inverse sine. After computing sinθ2\sin \theta_2, you need θ2=sin1()\theta_2 = \sin^{-1}(\ldots) for the angle.

In one sentence

Snell's law n1sinθ1=n2sinθ2n_1 \sin \theta_1 = n_2 \sin \theta_2 predicts the refraction of light at a boundary between two media; when light travels from a denser to a less dense medium at an angle greater than the critical angle θc=sin1(n2/n1)\theta_c = \sin^{-1}(n_2 / n_1), total internal reflection occurs (used in fibre optics and prisms); and because nn varies with frequency, different colours refract differently, producing dispersion in prisms and rainbows.

Past exam questions, worked

Real questions from past VCAA papers on this dot point, with our answer explainer.

2024 VCAA4 marksLight travels from air (refractive index 1.00) into a glass block (refractive index 1.50) at an angle of incidence of $30^\circ$. (a) Calculate the angle of refraction. (b) Calculate the speed of light in the glass.
Show worked answer →

(a) Angle of refraction. Apply Snell's law.

n1sinθ1=n2sinθ2n_1 \sin \theta_1 = n_2 \sin \theta_2

1.00×sin30=1.50×sinθ21.00 \times \sin 30^\circ = 1.50 \times \sin \theta_2

sinθ2=0.5/1.50=0.333\sin \theta_2 = 0.5 / 1.50 = 0.333

θ2=sin1(0.333)19.5\theta_2 = \sin^{-1}(0.333) \approx 19.5^\circ.

(b) Speed in glass. v=c/n=3.0×108/1.50=2.0×108v = c / n = 3.0 \times 10^8 / 1.50 = 2.0 \times 10^8 m s1^{-1}.

Markers reward correct Snell's law set-up with refractive indices on the right side, an angle less than the incidence angle (bending toward the normal entering a denser medium), and the speed reduction by factor nn.

2023 VCAA3 marksLight in a glass fibre (refractive index 1.50) reaches an interface with air. (a) Calculate the critical angle. (b) Explain what happens to light incident at angles greater than the critical angle, and how this is used in fibre-optic communication.
Show worked answer →

(a) Critical angle. At the critical angle, the refracted ray is along the boundary (θ2=90\theta_2 = 90^\circ).

n1sinθc=n2sin90=n2n_1 \sin \theta_c = n_2 \sin 90^\circ = n_2

sinθc=n2/n1=1.00/1.50=0.667\sin \theta_c = n_2 / n_1 = 1.00 / 1.50 = 0.667

θc=sin1(0.667)41.8\theta_c = \sin^{-1}(0.667) \approx 41.8^\circ.

(b) Above the critical angle. Light incident at an angle greater than the critical angle undergoes total internal reflection: all the light is reflected back into the denser medium; none is transmitted into the less dense medium.

Fibre-optic use. Optical fibres use total internal reflection to trap light inside a glass or polymer core. Light travelling through the core is incident on the core-cladding boundary at angles greater than the critical angle (by design of the fibre geometry), so it reflects back into the core repeatedly without loss as it propagates. This allows light to travel many kilometres along a curved path with minimal attenuation. Modern undersea data cables use this principle to carry intercontinental internet traffic.

Markers reward correct sinθc=n2/n1\sin \theta_c = n_2 / n_1 derivation, the all-reflected statement (no transmitted ray), and the fibre-optic application.

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