Refraction at a Plane Surface

Snell's law, total internal reflection, and optical fibre applications

SPEC 3.3.2.3 • REFRACTION

What is Refraction?

Definition

Refraction is the change in direction of a wave when it passes from one medium to another, caused by a change in wave speed.

When light crosses a boundary between two transparent materials, its speed changes because light interacts differently with atoms in different materials. Since frequency remains constant, the wavelength must change (from v = fλ).

Light bends towards the normal when entering a more optically dense medium (n₂ > n₁)

Direction of Bending

Towards normal: Light enters more optically dense medium (higher n, slower speed)

Away from normal: Light enters less optically dense medium (lower n, faster speed)

Refractive Index

The refractive index (n) of a material tells us how much light slows down in that material compared to in a vacuum.

Absolute Refractive Index

n = \frac{c}{c _{s}}

where c = speed of light in vacuum (3.00 × 10⁸ m s⁻¹)

c_s = speed of light in the substance

MaterialRefractive Index (n)Speed of Light

Vacuum1.00 (exactly)3.00 × 10⁸ m s⁻¹

Air1.0003 ≈ 1.00≈ 3.00 × 10⁸ m s⁻¹

Water1.332.26 × 10⁸ m s⁻¹

Glass1.50 (typical)2.00 × 10⁸ m s⁻¹

Diamond2.421.24 × 10⁸ m s⁻¹

Key Points

Higher refractive index = slower light speed = more "optically dense"

Refractive index is always ≥ 1 (nothing travels faster than c)

Air's refractive index is so close to 1 that we approximate c_air ≈ c

Refractive Index Between Materials

When light passes between two materials (neither being a vacuum), we can relate the refractive index to their speeds:

Between Two Materials

n of material 2 = \frac{c _{1}}{c _{2}}

Refractive index of material 2 relative to material 1

Example Calculation

Light passes from air (n = 1.00) into diamond (n = 2.42). The speed of light in air is 3.0 × 10⁸ m s⁻¹.

Speed in diamond = c/n = (3.0 × 10⁸)/2.42 = 1.24 × 10⁸ m s⁻¹

Exam Tip

The absolute refractive index of a material is its refractive index compared with a vacuum and is the value usually quoted simply as "the refractive index".

Snell's Law

Snell's law relates the angles of incidence and refraction to the refractive indices of the two media.

Snell's Law

n_{1} sin θ_{1} = n_{2} sin θ_{2}

The ray begins in medium with n₁; θ₁ is angle of incidence, θ₂ is angle of refraction

Angle Definitions

Angle of incidence (θ₁): The angle between the incident ray and the normal at the point of incidence.

Angle of refraction (θ₂): The angle between the refracted ray and the normal at the point of refraction.

Critical

Angles are ALWAYS measured from the NORMAL (perpendicular to the surface), not from the boundary surface itself!

Worked Example01

Light travels at 1.97 × 10⁸ m s⁻¹ in a type of glass. Calculate the refractive index of this glass.

1

Write down known values: c_s = 1.97 × 10⁸ m s⁻¹, c = 3.00 × 10⁸ m s⁻¹

2

Use the refractive index formula:

n = \frac{c}{c _{s}}

3

Substitute:

n = \frac{3.00 \times 1 0 ^{8}}{1.97 \times 1 0 ^{8}}

4

Calculate: n = 1.52

Answer: n = 1.52

Worked Example02

A beam of light passes from water (n₁ = 1.33) into diamond (n₂ = 2.42). If the angle of incidence in water is 35°, calculate the angle of refraction in the diamond.

1

Identify values: n₁ = 1.33 (water), n₂ = 2.42 (diamond), θ₁ = 35°

2

Apply Snell's law: n₁ sin θ₁ = n₂ sin θ₂

3

Rearrange: sin θ₂ = (n₁/n₂) × sin θ₁

4

Substitute: sin θ₂ = (1.33/2.42) × sin 35° = 0.550 × 0.574 = 0.316

5

Find θ₂: θ₂ = sin⁻¹(0.316) = 18.4°

Answer: θ₂ = 18.4°

Worked Example03

A beam of light travelling in glycerol passes into a diamond (n = 2.42). The angle of incidence in glycerol is 35° and the angle of refraction in diamond is 20.4°. Calculate the refractive index of glycerol.

1

Identify: n₁ = ? (glycerol), n₂ = 2.42 (diamond), θ₁ = 35°, θ₂ = 20.4°

2

Apply Snell's law: n₁ sin θ₁ = n₂ sin θ₂

3

Rearrange: n₁ = (n₂ sin θ₂) / sin θ₁

4

Substitute: n₁ = (2.42 × sin 20.4°) / sin 35°

5

Calculate: n₁ = (2.42 × 0.349) / 0.574 = 0.845 / 0.574 = 1.47

Answer: n = 1.47

Critical Angle

When light travels from a more optically dense to a less optically dense medium, there is a maximum angle of incidence that allows refraction.

Critical Angle

The critical angle (θc or c) is the angle of incidence in the denser medium that produces an angle of refraction of exactly 90°.

sin θ_{c} = \frac{n _{2}}{n _{1}}

(where n₁ > n₂, i.e., light going from denser to less dense)

(a) θ < θc: refraction (b) θ = θc: ray along boundary (c) θ > θc: total internal reflection

Total Internal Reflection

Definition

Total Internal Reflection (TIR) occurs when light is completely reflected back into the denser medium, with no light being refracted.

Two Conditions for TIR

1. Light must travel from a material with higher refractive index to one with lower refractive index (n₁ > n₂)

2. The angle of incidence must be greater than or equal to the critical angle (θ ≥ θc)

Exam Tip

TIR only occurs when the light travelling in one material meets a boundary with a material that has a lower refractive index. It can NEVER occur when light travels from less dense to more dense.

Deriving the Formula

At the critical angle, θ₂ = 90° and sin 90° = 1

From Snell's law: n₁ sin θc = n₂ × 1 = n₂

Therefore:

sin θ_{c} = \frac{n _{2}}{n _{1}}

Worked Example04

The refractive indices from air to glass and from air to water are 1.50 and 1.33 respectively. Calculate the critical angle for a water-glass surface.

1

First find the refractive index for light passing from water to glass:

_wn_g = n_glass/n_water = 1.50/1.33 = 1.13

2

For TIR, light goes from glass (higher n) to water (lower n). Critical angle is in the glass.

3

Using: _wn_g = 1/sin c (for light going glass → water)

4

sin c = 1/1.13 = 0.885

5

c = sin⁻¹(0.885) = 62.3°

Answer: Critical angle = 62.3°

Worked Example05

Light travels from glass (n = 1.50) towards air. Determine whether total internal reflection occurs if the angle of incidence is (a) 35° (b) 50°.

1

First, find critical angle: sin θc = n₂/n₁ = 1.00/1.50 = 0.667

θc = sin⁻¹(0.667) = 41.8°

2

Check condition 1: Is n₁ > n₂? Yes (1.50 > 1.00) ✓

3

(a) θ = 35°: Is 35° ≥ 41.8°? NO

TIR does NOT occur. Light refracts into the air.

4

(b) θ = 50°: Is 50° ≥ 41.8°? YES

TIR DOES occur. All light reflects back into the glass.

Answer: (a) No TIR - refraction (b) Yes - TIR occurs

Fibre Optics

An important application of total internal reflection is in fibre optics. Light is transmitted along thin glass fibres by repeated TIR.

Modern optical fibres have a core surrounded by cladding:

Structure

Core: High refractive index glass (n ≈ 1.55) - carries the light

Cladding: Lower refractive index glass (n ≈ 1.45) - surrounds the core

Functions of Cladding

1. Ensures TIR Occurs

The cladding has lower refractive index than core, ensuring n₁ > n₂ for TIR at the core-cladding boundary.

2. Increases Critical Angle (Reduces Modal Dispersion)

Without cladding (glass-air): sin θc = 1/1.55 = 0.645 → θc ≈ 40°

With cladding: sin θc = 1.45/1.55 = 0.935 → θc ≈ 69°

A larger critical angle means only rays very close to the axis undergo TIR. This reduces the difference in path lengths (modal dispersion).

3. Protects the Core

Cladding protects the delicate core from physical damage and contamination.

4. Prevents Signal Crossover

When fibres are bundled, cladding prevents light leaking between adjacent fibres.

Worked Example06

The refractive indices from air to the glass of the core and from air to the glass of the cladding are 1.55 and 1.45 respectively. Calculate the critical angle for the core-cladding surface.

1

Refractive index for light from core to cladding:

n_core/n_cladding = 1.55/1.45 = 1.069

2

For TIR: light goes from core (higher n) to cladding (lower n) ✓

3

Critical angle: _coren_cladding = 1/sin c

4

sin c = 1/1.069 = 0.935

5

c = sin⁻¹(0.935) = 69.3°

Answer: Critical angle = 69.3°

Pulse Broadening: Material Dispersion

When signals travel through optical fibres, they can become degraded. One cause is material dispersion.

Material Dispersion

Material dispersion occurs because the refractive index of glass varies with wavelength. Different wavelengths travel at different speeds, arriving at different times.

Solution

Use monochromatic light (single wavelength from a laser). With only one wavelength, all light travels at the same speed - no dispersion.

Absorption and Signal Boosting

Pulse Absorption

Absorption occurs when light energy is absorbed by the glass material. This reduces the signal's amplitude (intensity) as it travels.

Solution: Repeaters/Regenerators

Repeaters are placed at intervals along the fibre to boost the signal back to full strength.

Worked Example07

Light passes from water (n = 1.33) into glass (n = 1.50), then from glass into air. If the initial angle of incidence at the water-glass boundary is 40°, determine the final angle of refraction in air.

1

Water to glass: 1.33 × sin 40° = 1.50 × sin θ₂

sin θ₂ = (1.33 × 0.643)/1.50 = 0.570 → θ₂ = 34.8°

2

Check for TIR: Critical angle = sin⁻¹(1/1.50) = 41.8°

Is 34.8° ≥ 41.8°? NO, so refraction occurs.

3

Glass to air: 1.50 × sin 34.8° = 1.00 × sin θ₃

sin θ₃ = 1.50 × 0.571 = 0.856

4

θ₃ = sin⁻¹(0.856) = 58.9°

Answer: Angle of refraction in air = 58.9°

Practice

Question 1: The speed of light in a transparent material is 2.25 × 10⁸ m s⁻¹. Calculate the refractive index of this material.

1

Use n = c/c_s where c = 3.00 × 10⁸ m s⁻¹

2

n = (3.00 × 10⁸)/(2.25 × 10⁸) = 1.33

Answer: n = 1.33 (this is water)

Practice

Question 2: Calculate the critical angle for a water-diamond interface if the refractive indices are 1.33 and 2.42 respectively.

1

For TIR, light goes from diamond (higher n) to water (lower n).

2

sin θc = n₂/n₁ = 1.33/2.42 = 0.550

3

θc = sin⁻¹(0.550) = 33.4°

Answer: θc = 33.4°

Practice

Question 3: Explain why using monochromatic light in an optical fibre reduces signal degradation.

1

Material dispersion occurs because the refractive index varies with wavelength.

2

Different wavelengths travel at different speeds through the fibre.

3

This causes parts of the signal to arrive at different times, broadening the pulse.

4

Monochromatic light has only one wavelength, so all light travels at the same speed.

5

This eliminates material dispersion and prevents pulse broadening.

Answer: Monochromatic light eliminates material dispersion (single wavelength = single speed)

Summary

Key Formulas

ConceptFormula

Refractive Index

n = \frac{c}{c _{s}}

or

n_{2} = \frac{c _{1}}{c _{2}}

Snell's Law

n_{1} sin θ_{1} = n_{2} sin θ_{2}

Critical Angle

sin θ_{c} = \frac{n _{2}}{n _{1}}

(where n₁ > n₂)

Key Exam Points

Angles ALWAYS measured from the normal

TIR requires: (1) n₁ > n₂ AND (2) θ ≥ θc

Cladding: ensures TIR, increases θc, protects core, prevents crossover

Material dispersion: different λ → use monochromatic light

Modal dispersion: different paths → use cladding/monomode fibre

Absorption: signal weakens → use repeaters