Refraction and Total Internal Reflection

Refraction and Total Internal Reflection

Info: Board Coverage AQA Paper 2 | Edexcel CP2 | OCR (A) Paper 2 | CIE P2

Explore the simulation above to develop intuition for this topic.

Intuition

Light bends when it changes speed — just like a car turning on a road: Imagine a car driving from tarmac (fast) onto mud (slow) at an angle. The wheel that hits the mud first slows down while the other wheel is still on tarmac, causing the car to turn. Light does the same thing — when it enters a denser medium (higher refractive index), it slows down and bends toward the normal.

Why it matters: Refraction explains why pools look shallower than they are, why lenses work, how fibre optics guide light around corners, and why rainbows form. Understanding total internal reflection explains how diamonds sparkle and how undersea cables carry internet signals across oceans.

The key insight: Total internal reflection only happens when light travels from a denser to a less dense medium (high $n$ to low $n$) at an angle greater than the critical angle. This is why you can see the surface of a pool from underwater (total internal reflection at shallow angles) but can’t see the bottom from above at steep angles.

1. Refractive Index

Definition. The refractive index $n$ of a medium is the ratio of the speed of light in Vacuum to the speed of light in the medium:

$\boxed{n = \frac{c}{v}}$

Since $v < c$ for all material media, $n > 1$. The refractive index is a dimensionless quantity.

Intuition. Light slows down in a denser medium because the electromagnetic wave interacts with The electrons in the material. The denser the material (more electrons per unit volume), the slower The light, and the higher the refractive index.

Dispersion and the Refractive Index

The refractive index of a material is not constant — on the wavelength of light. Shorter Wavelengths (blue/violet) are refracted more than longer wavelengths (red). This is because shorter Wavelengths interact more strongly with the electrons in the material.

For glass, a typical empirical relationship (Cauchy’s equation) is:

$n(\lambda) = A + \frac{B}{\lambda^2}$

Where $A$ and $B$ are constants specific to the material and $\lambda$ is the wavelength in vacuum. This wavelength dependence is what causes white light to separate into a spectrum when passing Through a prism.

Info: Board Coverage AQA Paper 2 | Edexcel CP2 | OCR (A) Paper 2 | CIE P2 Application: Diamond Cutting. Diamond has an exceptionally high refractive index ($n = 2.42$) And large dispersion. Jewellers cut diamonds with many angled facets so that light entering the top Of the diamond strikes the internal facets at angles well above the critical angle ($\theta_c = 24.4^\circ$). The light is trapped inside by repeated TIR and eventually exits through The top, directing brightness back towards the viewer. The large dispersion also splits white light Into a rainbow of colours, creating the characteristic “fire” of a diamond. A well-cut diamond has a Specific facet geometry ( 57 facets in a round brilliant cut) optimised so that light Entering through the crown always hits pavilion facets at angles exceeding the critical angle.

2. Snell’s Law

Snell’s Law. At the boundary between two media with refractive indices $n_1$ and $n_2$:

$\boxed{n_1 \sin\theta_1 = n_2 \sin\theta_2}$

Where $\theta_1$ is the angle of incidence and $\theta_2$ is the angle of refraction, both measured From the normal.

Derivation from Wave Theory (Huygens’ Construction)

Consider a plane wavefront $AB$ arriving at the boundary between two media. Let the wave travel at Speed $v_1$ in medium 1 and $v_2$ in medium 2.

By the time point $B$ on the wavefront reaches the boundary at $B'$Point $A$ has already entered Medium 2 and travelled a distance $v_2 t$ as a secondary wavelet, where $t = BB'/v_1$.

The new wavefront is the tangent from $B'$ to the wavelet centred at $A$. The geometry gives:

$\frac{\sin\theta_1}{\sin\theta_2} = \frac{BB'/AB'}{v_2 t / AB'} = \frac{BB'}{v_2 t} = \frac{v_1 t}{v_2 t} = \frac{v_1}{v_2}$

Since $n = c/v$:

$\frac{\sin\theta_1}{\sin\theta_2} = \frac{n_2}{n_1} \implies n_1 \sin\theta_1 = n_2 \sin\theta_2$

$\square$

Derivation from Fermat’s Principle

Fermat’s Principle states that light travels between two points along the path that takes the least time.

Consider a ray travelling from point $P$ in medium 1 to point $Q$ in medium 2, crossing the Boundary. Let the boundary be the $x$-axis, with $P$ at $(0, h_1)$ and $Q$ at $(d, -h_2)$. The ray Hits the boundary at $(x, 0)$.

The total travel time is:

$t = \frac{\sqrt{x^2 + h_1^2}}{v_1} + \frac{\sqrt{(d-x)^2 + h_2^2}}{v_2}$

For the minimum time, $dt/dx = 0$:

$\frac{x}{v_1\sqrt{x^2 + h_1^2}} - \frac{d-x}{v_2\sqrt{(d-x)^2 + h_2^2}} = 0$

Noting that $\sin\theta_1 = x/\sqrt{x^2 + h_1^2}$ and $\sin\theta_2 = (d-x)/\sqrt{(d-x)^2 + h_2^2}$:

$\frac{\sin\theta_1}{v_1} = \frac{\sin\theta_2}{v_2}$

$\frac{n_1 \sin\theta_1}{c} = \frac{n_2 \sin\theta_2}{c} \implies n_1 \sin\theta_1 = n_2 \sin\theta_2$

$\square$

Intuition. When light enters a denser medium ($n_2 > n_1$), it bends towards the normal ($\theta_2 < \theta_1$). This is because one side of the wavefront slows down before the other, Causing the wavefront to pivot towards the normal.

Worked Example: Light through a Glass Slab

A ray of light enters a glass slab ($n = 1.50$) from air at $\theta_1 = 40^\circ$. The slab has Parallel faces.

Step 1: At the first surface (air to glass). $1.00 \times \sin 40° = 1.50 \times \sin\theta_2$. $\theta_2 = \arcsin(0.643/1.50) = 25.4^\circ$.

Step 2: At the second surface (glass to air). The ray hits the second surface at $\theta_3 = 25.4^\circ$ (equal to $\theta_2$ because the faces are parallel). $1.50 \times \sin 25.4° = 1.00 \times \sin\theta_4$. $\theta_4 = 40^\circ$.

Result. The emergent ray is parallel to the incident ray but laterally displaced. This lateral Displacement $d$ depends on the thickness $t$ of the slab:

$d = t \cdot \sin\theta_1 \left(1 - \frac{\cos\theta_1}{\sqrt{n^2 - \sin^2\theta_1}}\right)$

3. Total Internal Reflection

Condition for Total Internal Reflection

Total internal reflection (TIR) occurs when:

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

Derivation of the Critical Angle

At the critical angle, the refracted ray travels along the boundary ($\theta_2 = 90^\circ$):

$n_1 \sin\theta_c = n_2 \sin 90° = n_2$

$\boxed{\sin\theta_c = \frac{n_2}{n_1}}$

For light going from a medium of refractive index $n$ into air ($n_2 \approx 1$):

$\boxed{\sin\theta_c = \frac{1}{n}}$

Derivation of why TIR only occurs from denser to less dense. Snell’s law gives $\sin\theta_2 = \frac{n_1}{n_2}\sin\theta_1$. If $n_1 > n_2$Then $\frac{n_1}{n_2} > 1$ and $\sin\theta_2$ can exceed 1 for sufficiently large $\theta_1$Which is impossible — so the light is Entirely reflected. If $n_1 < n_2$Then $\frac{n_1}{n_2} < 1$ and $\sin\theta_2 < \sin\theta_1 \leq 1$ for all $\theta_1$So refraction always occurs.

Real-World Example: Mirages

On a hot day, the ground heats the air immediately above it. Hot air is less dense and has a Slightly lower refractive index than cooler air above. This creates a gradual decrease in refractive Index with height, forming a continuous gradient rather than a sharp boundary.

Light from the sky heading downward towards the ground encounters this gradient. The gradual bending (continuous refraction) can cause the light to curve upwards, eventually undergoing TIR-like Behaviour when the angle relative to the horizontal exceeds the critical angle for the hot-to-cool Air transition. An observer sees this light as if it came from the ground, interpreting it as a pool Of water (a “mirage”).

This is not true TIR (which requires a sharp boundary), but the principle is the same: light curves Away from regions of higher refractive index and can be totally reflected if the gradient is steep Enough.

Evanescent Wave

When light is incident at a boundary at exactly the critical angle, the transmitted wave travels Along the boundary. For angles just beyond the critical angle, an evanescent wave penetrates a Short distance into the second medium. This wave carries no energy away from the boundary and decays Exponentially:

$E = E_0 \, e^{-z/\delta}$

Where $\delta$ is the penetration depth and $z$ is the distance into the second medium. The Evanescent wave is exploited in technologies such as:

• Frustrated total internal reflection: placing another surface close to the boundary allows the evanescent wave to “tunnel” across the gap, converting TIR back into transmission.
• Optical fibre sensors: chemicals or biological molecules near the fibre surface can interact with the evanescent wave, changing the propagation and allowing detection.

Info: Board Coverage AQA Paper 2 | Edexcel CP2 | OCR (A) Paper 2 | CIE P2

4. Optical Fibres

Optical fibres use total internal reflection to guide light along a curved path.

Structure

An optical fibre consists of:

• Core: denser medium (higher $n$), glass or plastic
• Cladding: less dense medium (lower $n$), surrounding the core

Light enters the fibre and undergoes repeated TIR at the core-cladding boundary, propagating along The fibre with minimal loss.

Acceptance Angle and Numerical Aperture

Light entering the fibre at too large an angle will not satisfy the TIR condition at the Core-cladding boundary.

For light entering from air into a fibre with core index $n_1$ and cladding index $n_2$:

At the core-cladding boundary, TIR requires: $\sin\theta_2 \geq \sin\theta_c = n_2/n_1$ (where $\theta_2$ is measured from the normal to the core-cladding boundary).

At the air-core boundary (Snell’s law): $\sin\theta_{\max} = n_1\cos\theta_c = n_1\sqrt{1 - \sin^2\theta_c} = n_1\sqrt{1 - (n_2/n_1)^2} = \sqrt{n_1^2 - n_2^2}$.

The numerical aperture is:

$\mathrm{NA} = \sqrt{n_1^2 - n_2^2}$

Signal Degradation

Two main effects degrade the signal in optical fibres:

1. Absorption: some light energy is absorbed by the fibre material, reducing signal intensity.
2. Dispersion: different wavelengths travel at slightly different speeds, causing pulse broadening.

• Material dispersion: the refractive index depends on wavelength.
• Modal dispersion: rays entering at different angles travel different path lengths (in multimode fibres).

Types of Optical Fibre

Step-index multimode fibre. The core has a uniform refractive index $n_1$ and the cladding has a Lower uniform index $n_2$. Light rays travel in zigzag paths, reflecting off the core-cladding Boundary. Rays entering at different angles take different path lengths, causing modal Dispersion — different rays arrive at different times, broadening the signal pulse. Step-index Fibres are suitable for short-distance communication (e.g., within buildings or vehicles).

Graded-index fibre. The refractive index of the core decreases gradually from the centre axis to The cladding. Light rays follow curved (approximately sinusoidal) paths rather than sharp zigzags. Rays that travel further from the axis pass through regions of lower refractive index, where they Travel faster. This compensates for the longer path length, significantly reducing modal dispersion. Graded-index fibres are used for medium-distance links (e.g., LANs, cable television).

Single-mode fibre. The core is extremely narrow ( 8 to 10 micrometres), so only one Mode (the axial ray) can propagate. This eliminates modal dispersion entirely. Single-mode fibres Are used for long-distance telecommunications (e.g., undersea cables spanning thousands of Kilometres).

Applications of TIR and Optical Fibres

Endoscopy. Medical endoscopes use bundles of optical fibres (coherent and incoherent bundles) to View inside the body. A coherent bundle has the fibres arranged in the same spatial pattern at Both ends, so an image is faithfully transmitted. An incoherent bundle carries illumination Light from an external source to the internal area. The surgeon views the image through an eyepiece Or on a screen. TIR ensures light stays within each individual fibre without leaking into Neighbouring fibres, maintaining image quality.

Periscopes. Submarine periscopes and some military periscopes use prisms rather than mirrors. A Right-angled triangular prism arranged so that light enters perpendicular to the hypotenuse face Hits the two shorter faces at 45°. Since glass has a critical angle of about 42°, the 45° Angle exceeds the critical angle, producing TIR. Prism periscopes are preferred over mirror Periscopes because TIR gives 100% reflection (no absorption by a metallic coating), and the Reflective surface is protected inside the glass, making it resistant to scratches and tarnishing.

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Problem Set

Cross-References

• Wave Properties: Establishes the relationship between wave speed, frequency, and wavelength that governs refraction.
• Superposition and Interference: Shows how phase differences arise from path differences, which can be affected by refraction.
• Snell’s Law and Refraction: Provides the mathematical framework for calculating angles of refraction at boundaries.
• Total Internal Reflection: Explores the critical angle phenomenon and its applications in fiber optics and mirages.