OxfordAQA International AS and A-level Physics: Refraction, total internal reflection and optical fibres (9630) – Revision Notes

These notes condense section 3.5.8 (Refraction at a plane surface) of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4. This is shared AS and A-level content. The study guide. Check what to revise with the free diagnostics, tick the printable checklist and see the course hub. Test yourself with the practice questions.

Key definitions and equations

Quantity	Statement	Notes
Refractive index	n = c / cs	c is the speed in a vacuum, cs the speed in the substance. No unit. n of air is approximately 1
Snell’s law	n1 sin θ1 = n2 sin θ2	Angles from the normal. Medium 1 is the one the light leaves
Critical angle	sin θc = n2 / n1	Needs n1 greater than n2. For air as medium 2, sin θc = 1/n1
Wave speed	cs = fλ	f stays the same at a boundary
Wavelength in a medium	λ = λ(air) / n	Follows from constant f

Constants used in these notes: c = 3.00 × 10⁸ m s⁻¹, n for air = 1.

Refraction: method in steps

1. Draw the boundary and the normal. Mark both angles from the normal.
2. Label the medium the light leaves as 1 and the one it enters as 2.
3. Write n1 sin θ1 = n2 sin θ2 and rearrange for the unknown.
4. Check the direction: into a denser medium the ray bends towards the normal.
5. Quote the answer with a sensible number of significant figures.

Reminder. Air to a plastic with n = 1.47, incidence 35°. sin θ2 = sin 35° / 1.47 = 0.3902, so θ2 = 23.0°.

Must-know distinctions

Property at a boundary	Changes?
Frequency	No
Speed	Yes, cs = c/n
Wavelength	Yes, shorter in the denser medium
Direction	Yes, unless incidence is along the normal

Total internal reflection: method in steps

1. Check that the light is going from higher n to lower n. If not, there is no TIR.
2. Calculate θc from sin θc = n2/n1.
3. Compare the angle of incidence with θc. A larger angle gives TIR; a smaller one gives mostly refraction with a weak reflection; an equal one makes the refracted ray graze the surface.
4. For a ray at exactly θc, state that the angle of refraction is 90°.

Reminder. Glass n = 1.50 against air: sin θc = 1/1.50, so θc = 41.8°. Against water (n = 1.33): sin θc = 1.33/1.50 = 0.8867, so θc = 62.5°. A ray at 45° is totally reflected at a glass-air boundary but not at a glass-water boundary.

Step index optical fibre

Part	Property	Function
Core	Higher n (n1), thin	Carries the light pulses
Cladding	Lower n (n2)	Allows TIR at the core boundary; protects the core surface; stops cross-talk between fibres; adds strength

• Rays hitting the core boundary at more than θc stay in the core by repeated TIR.
• Only step index fibres are on the specification: n changes abruptly from n1 to n2.
• Rays within (90° − θc) of the axis are guided.

Pulse broadening and dispersion

Type	Cause	Reduced by
Modal dispersion	Different paths (modes) take different times	Narrow core (single mode); closer n1 and n2
Material dispersion	n depends on wavelength, so wavelengths travel at different speeds	Narrow range of wavelengths, such as a laser

• Pulse broadening means the output pulse is longer than the input pulse.
• Consequence: pulses overlap, so they must be sent further apart. The maximum data rate falls as the broadening increases.
• Modal path times for length L: axial L n1/c; at the critical angle L n1²/(c n2).

Reminder. L = 5.0 km, n1 = 1.50, n2 = 1.47. Axial time 2.50 × 10⁻⁵ s. Critical-angle ray 2.55 × 10⁻⁵ s. Spread 5.1 × 10⁻⁷ s.

Absorption

• Energy is absorbed and scattered along the fibre, so the pulse gets smaller (lower power and amplitude).
• Absorption does not lengthen a pulse. Dispersion does.
• A constant fraction of power is lost per unit length, so the fraction remaining after x km is (fraction passed per km) raised to the power x.
• Consequence: regenerators or amplifiers are needed at intervals on long links. A regenerator restores both size and shape.
• Reduced by purer glass and by using light the glass absorbs weakly.

Worked reminders

Speed and wavelength. Light of frequency 5.00 × 10¹⁴ Hz has wavelength c/f = 3.00 × 10⁸ / 5.00 × 10¹⁴ = 600 nm in air. In glass of n = 1.50 the frequency is unchanged, the speed is 2.00 × 10⁸ m s⁻¹ and the wavelength is 400 nm.

Finding n from a boundary. Light in glass of n = 1.60 meets air. At the critical angle sin θc = 1/1.60 = 0.625, so θc = 38.7°. Any ray in the glass meeting the surface at more than 38.7° stays inside.

Rays guided in the core. A ray must meet the core-cladding boundary at more than θc, so it must lie within (90° − θc) of the axis. For core n = 1.56 and cladding n = 1.52, θc = 77.0° and the limit is 13.0° to the axis.

Pulse spread. Time difference = (L n1 / c)(n1/n2 − 1). It grows with length and with the index difference, so a short, single-mode fibre has the smallest spread.

Practical: finding n with a block

1. Place the block on paper and draw round it. Draw a normal at the point of entry.
2. Direct a thin ray at the point of entry at a measured angle of incidence i.
3. Mark the emerging ray, remove the block, join the points, and measure the angle r inside from the normal.
4. Repeat for at least six values of i.
5. Graph sin i (y-axis) against sin r (x-axis); the slope equals n of the block, with air as medium 1.
6. Use points far apart on the line to find the gradient.

Check that the ray stays thin and sharp, and that the normal is at 90° to the surface. Cut the uncertainty by drawing a thin ray, marking points as far apart as possible and using a fine-scale protractor.

Quick self-test

1. The speed of light in a liquid is 2.25 × 10⁸ m s⁻¹. Find n.
2. Light of wavelength 640 nm in air enters glass of n = 1.60. Find its wavelength in the glass.
3. A ray in air meets a liquid (n = 1.36) at 50° to the normal. Find the angle of refraction.
4. Find the critical angle for a glass-air boundary when n = 1.62.
5. State the two conditions for total internal reflection.
6. A core has n = 1.50 and the cladding has n = 1.45. Find the critical angle.
7. Name the type of dispersion caused by rays taking different path lengths.
8. Why does a laser source reduce material dispersion more than an LED?
9. A fibre passes 90% of the power in each kilometre. What fraction remains after 3.0 km?
10. Give one effect of absorption and one effect of pulse broadening on a signal.

Answers

1. n = 3.00 × 10⁸ / 2.25 × 10⁸ = 1.33.
2. λ = 640 / 1.60 = 400 nm.
3. sin θ2 = sin 50° / 1.36 = 0.5633, so θ2 = 34.3°.
4. sin θc = 1/1.62 = 0.6173, so θc = 38.1°.
5. The light travels from a more dense to a less dense medium and the angle of incidence exceeds the critical angle.
6. sin θc = 1.45/1.50 = 0.9667, so θc = 75.2°.
7. Modal dispersion.
8. A laser emits a very narrow range of wavelengths, so there is little difference in speed within the pulse. An LED emits a wide range.
9. 0.90³ = 0.729, so 73%.
10. Absorption: the signal power falls, so it may become too weak to detect. Pulse broadening: pulses overlap, which limits the rate at which pulses can be sent.

Where marks are usually lost

• Taking angles from the boundary line instead of the normal.
• Swapping n1 and n2 in the critical angle formula, giving a sine above 1.
• Claiming TIR for light going from a less dense to a denser medium.
• Saying frequency changes on refraction.
• Giving the function of the cladding without mentioning its lower refractive index.
• Confusing modal and material dispersion, or saying absorption makes pulses longer.
• Describing a bent ray with no stated direction relative to the normal.
• Rounding the angle too early in a chain of calculations.
• Leaving the pulse broadening argument as “the signal gets blurred” without saying the pulses overlap and limit the data rate.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Section 3.5.8 Refraction at a plane surface.