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Refraction, Reflection & Polarisation

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Edexcel IAL Physics · Waves

Refraction, Reflection & Polarisation

When light crosses into a new material or bounces off a surface, its speed, direction, and even the way it wiggles can all change — and every rule in this chapter is really just describing those three things happening in different situations.

Summary — What This Chapter Covers

  • Intensity is power spread over an area (I = P/A), and it obeys an inverse-square law with distance for a point source.
  • Refraction happens when light crosses a boundary between materials — it changes speed, and (unless it hits the boundary straight on) it changes direction too.
  • Refractive index (n) tells you how much a material slows light down compared to a vacuum. Snell's Law (n₁sinθ₁ = n₂sinθ₂) connects refractive index to the angles either side of a boundary.
  • The critical angle is the special angle of incidence (inside a dense material) at which the refracted ray grazes along the boundary at exactly 90°.
  • Total internal reflection (TIR) happens beyond the critical angle, when going from a denser to a less dense material — 100% of the light reflects, none escapes.
  • The refractive index experiment uses a ray box, a perspex block, and careful angle-marking to measure n practically.
  • Plane polarisation restricts a transverse wave's oscillations to a single plane. Only transverse waves can be polarised — longitudinal waves never can.
  • Light can be polarised by filters, reflection, or refraction, and polarisation has real uses — from sunglasses to stress-testing materials to analysing sugar solutions.

1. Intensity of Radiation

Think about a wave — light, sound, whatever — as a courier carrying energy. Intensity just measures how much energy that courier delivers, per second, to each square metre of surface it hits. That's it. It's power density.

Core Formula
I = P / A
Intensity (W m⁻²) = Power (W) ÷ Area (m²)

Now here's the part that trips people up conceptually: intensity isn't just about how much energy the source is pumping out — it's about how concentrated that energy is when it lands. The same 100 W lightbulb feels blinding up close and barely noticeable from across a football field, even though it's emitting the exact same power the whole time. What's changing is the area that power is spread across.

Amplitude and Frequency

Intensity also depends on the wave's amplitude and frequency — specifically, it's proportional to the square of each:

Proportionality
I ∝ A²   and   I ∝ f²
Double the amplitude → intensity increases by a factor of 4. Double the frequency → same thing.
Analogy: Imagine shaking a rope. Shake it with twice the amplitude (bigger swings) and you're putting in noticeably more energy — not double, but quadruple, because energy in a wave scales with the square of how far the particles swing. Same logic applies to how fast you shake it (frequency).

Spherical Waves & the Inverse Square Law

A point source (like a bare lightbulb or a speaker) radiates energy equally in all directions — forming an expanding sphere of wavefronts. The area of that sphere at radius r is 4πr². So as the wave travels outward, the same total power P is spread across a bigger and bigger sphere:

Intensity from a Point Source
I = P / (4πr²)
This means I ∝ 1/r² — the famous inverse square law.
Source ---- r ----> [Area A] I | ---- 2r ---> [Area 4A] I/4 | ---- 3r ---> [Area 9A] I/9 Double the distance → area quadruples → intensity drops to 1/4
Why this matters This is not a linear relationship — a common trap. Moving twice as far away doesn't halve the intensity, it quarters it. Moving three times as far away doesn't give you a third of the intensity — it gives you a ninth.
Practice Question 1

A progressive wave Q has intensity I₀. Wave P has half the amplitude of Q, but double the frequency of Q. What is the intensity of wave P, in terms of I₀?

Practice Question 2

A point source of light has power 60 W. Calculate the intensity of the light at a distance of 3.0 m from the source, assuming no absorption.

2. Refraction & Refractive Index

Refraction occurs whenever light crosses a boundary between two transparent materials of different density — like going from air into glass, or from water into air. Two things happen simultaneously at that boundary: the light's speed changes, and (unless it hits the boundary dead-on) its direction changes too.

Here's the crucial insight: the bending isn't some separate, mysterious effect — it's a direct consequence of the speed change. One side of the wavefront hits the new material and slows down before the rest of the wavefront does, so the whole wavefront pivots, like a marching band changing direction when one end walks onto muddy ground.

Analogy: Picture a supermarket trolley rolling from smooth tiles onto thick carpet at an angle. The wheel that hits the carpet first slows down immediately, while the other wheel is still speeding along on the tiles. That mismatch swings the whole trolley — bending its path toward the carpet side. That's exactly what happens to a light wave entering a denser material at an angle.

The Bending Rules

SituationWhat Happens
Entering a more dense medium (e.g. air → glass)Light slows down and bends towards the normal
Entering a less dense medium (e.g. glass → air)Light speeds up and bends away from the normal
Light travels along the normal (perpendicular)No change in direction — but speed still changes!
Easy to miss Even when light hits a boundary straight-on (along the normal) and doesn't bend at all, it is still slowing down or speeding up. Direction and speed are two separate things — a change in one doesn't require a change in the other.

Refractive Index — the Formula

The refractive index, n, is a number that tells you exactly how much slower light travels in a material compared to in a vacuum.

Refractive Index
n = c / v
c = speed of light in a vacuum (3.0 × 10⁸ m s⁻¹)  ·  v = speed of light in the material

Because light can never travel faster in a material than it does in a vacuum, v is always less than c — which means n is always greater than 1. A "high refractive index" material (like diamond, n ≈ 2.4) is called optically dense — it slows light down a lot. Air is so close to a vacuum that we just treat its refractive index as approximately 1 in calculations.

Snell's Law

Snell's Law is the equation that links the angle of incidence and angle of refraction to the refractive indices of the two materials involved:

Snell's Law
n₁ sin θ₁ = n₂ sin θ₂
n₁ = refractive index of the material the ray starts in  ·  n₂ = refractive index of the material it enters
θ₁, θ₂ = angles measured from the normal, in materials 1 and 2 respectively
NORMAL | \ θ1 | \ | \ | ----------\-------|-------------- BOUNDARY \θ2 | \ | \ | v | (n1 = material the ray STARTS in — always first) (n2 = material the ray ENTERS — always second)
Keep it straight Material 1 is always where the ray comes from. Material 2 is always where the ray is going to. Mixing these up is the #1 cause of "why did I get n less than 1?!" panic in the exam.
Practice Question

A light ray hits a glass surface from air at an angle of incidence of 42°. The angle of refraction inside the glass is 26°. Show that the refractive index of the glass is about 1.5, and calculate the speed of light inside the glass.

3. Critical Angle

Now imagine you're inside a dense material (like glass) shining light out towards a less dense material (like air), and you slowly increase the angle of incidence. As you do, the angle of refraction increases too — and it increases faster than the angle of incidence, because you're going from dense to less dense (bending away from the normal).

Eventually, at some specific angle of incidence, the refracted ray gets bent so far that it skims exactly along the boundary itself — refracted at 90°. That special angle of incidence is called the critical angle, C.

Critical Angle Formula
sin(C) = 1/n
where n is the refractive index of the denser material the light is travelling in

This formula drops straight out of Snell's Law. Set n₁ = n (the dense material), θ₁ = C, n₂ = 1 (air), and θ₂ = 90° (refraction along the boundary):

Derivation
n sin(C) = 1 × sin(90°) = 1   →   sin(C) = 1/n
Analogy: Think of a torch shone up through the surface of a swimming pool from underwater, tilted more and more towards the horizontal. At a certain tilt, the beam just grazes along the water's surface instead of poking up into the air at all. Tilt it any further and — as you'll see in the next section — it doesn't escape the water at all anymore.
Practice Question

A block of diamond has a refractive index of 2.42. Calculate the critical angle for light travelling from diamond into air.

4. Total Internal Reflection (TIR)

So what happens if you increase the angle of incidence beyond the critical angle? The refracted ray can't bend past 90° — there's no way for it to "escape" the boundary anymore. Instead, 100% of the light reflects back into the denser material. None of it transmits through. This is total internal reflection.

Two Conditions BOTH Required for TIR
θ₁ > C   AND   n₁ > n₂
1. The angle of incidence must exceed the critical angle
2. Light must be travelling from a MORE dense material into a LESS dense one
θ < C θ = C θ > C (refraction) (critical angle) (TOTAL INTERNAL REFLECTION) ↗ weak refracted ---→ refracted / ray escapes ray grazes NO ray escapes / along surface ALL light reflects ● ● ● /|\ bright reflected /|\ bright /|\ ray fully | ray also present | reflected | reflects back DENSE MEDIUM DENSE MEDIUM DENSE MEDIUM
TermExplanation
Refractionθ < C — light bends and exits, weaker partial reflection also happens
Critical Angleθ = C — refracted ray travels exactly along the boundary at 90°
Total Internal Reflectionθ > C — no light escapes at all, 100% reflects internally
Exam wording trap Always write out the full name — "Total Internal Reflection" — not just "internal reflection" or "reflection." Examiners specifically look for all three words, because partial reflection can happen at any angle; it's the total part (100%, no transmission) combined with internal (staying inside the denser medium) that defines TIR.
Practice Question

A ray of light travels inside an optical fibre (n = 1.50) and strikes the fibre wall (surrounded by air) at an angle of 55° to the normal. Will total internal reflection occur? Justify your answer with a calculation.

5. Measuring Refractive Index — Required Practical

Aim: To investigate the refraction of light through a perspex (or glass) block and determine its refractive index.

Equipment

  • Ray box — provides a narrow beam of light
  • Perspex/glass block — the material being investigated
  • Protractor — measures the angles
  • Sheet of paper, pencil, ruler — for marking the ray's path

Method (Step by Step)

  1. Place the block on paper and draw around its outline with a pencil.
  2. Direct a ray of light from the ray box at the side face of the block.
  3. Mark small crosses (×) at: a point on the incoming ray, where the ray enters the block, where it exits the block, and a point on the exiting ray about 5 cm beyond the block.
  4. Draw a dashed normal line (at right angles to the surface) at the point of entry.
  5. Remove the block and join up the marked points with straight ruled lines.
  6. Replace the block in its outline and repeat for several different angles of incidence.
incoming ray \ \ θ1 (angle of incidence) \ ---------●------------------ surface of block \ \ θ2 (angle of refraction, θ2 < θ1) \ [inside the block]

Analysing the Results

SituationRule
Light entering the blockBends towards the normal: i > r
Light exiting the blockBends away from the normal: i < r
Angle of incidence = 90°No bending occurs at all: i = r

Once you have several pairs of (angle of incidence, angle of refraction) readings, you calculate n for each pair using n = sinθ₁/sinθ₂, or — for better accuracy — plot a graph of sin(θ₁) against sin(θ₂). Because Snell's Law says n₁sinθ₁ = n₂sinθ₂, rearranging gives sinθ₁ = n·sinθ₂ (with air as material 2, n₂=1). This means the gradient of that straight-line graph equals n — much more reliable than trusting a single pair of readings.

Safety Considerations • The ray box bulb gets hot — could burn if touched (run any burns under cold water for 5+ minutes)
• Never look directly into the beam — stand behind the ray box
• Keep liquids away from the electrical equipment
• Handle the perspex block carefully — scratches or damage affect your results

6. Plane Polarisation

This topic is really about one question: which directions is a wave allowed to wiggle in?

A transverse wave (like light) oscillates perpendicular to its direction of travel. But "perpendicular" doesn't mean just one direction — it means any direction within the flat plane perpendicular to travel. Picture looking straight down the barrel of a wave travelling towards you: the oscillations could point up-down, left-right, or any diagonal in between. Ordinary light is a jumble of waves oscillating in all of these directions at once — this is called unpolarised light.

Definition — Plane Polarisation
Particle oscillations restricted to ONE single plane
...perpendicular to the direction of wave propagation
Analogy: Imagine wiggling a rope tied to a wall, and you can wiggle your hand in any direction — up-down, side-to-side, circles, anything — that's like unpolarised light. Now thread the rope through a picket fence with vertical gaps. Only up-down wiggles can pass through; anything sideways gets blocked by the fence posts. What comes out the other side only wiggles up-down. That's polarisation, and the fence is acting like a polarising filter.
UNPOLARISED VERTICALLY HORIZONTALLY WAVE POLARISED POLARISED ↑↗ ↑ ← ✦ → ---> | ---> ←------→ ↙↓ ↓ (oscillates in (oscillates ONLY (oscillates ONLY every direction) vertically) horizontally)
Critical exam fact Only transverse waves can be polarised (light, radio waves, water waves, S-waves). Longitudinal waves (sound, P-waves) can never be polarised, because their particles already only oscillate along the single direction of travel — there's no "other plane" to restrict them to.

Method 1 — Polarising Filters

A polarising filter has a transmission axis — only oscillations aligned with that axis get through. Pass unpolarised light through one filter, and you get plane-polarised light out the other side (whatever survives lines up with the transmission axis). Now if you place a second filter after it, rotated 90° to the first, absolutely no light gets through — the first filter already restricted the light to one plane, and the second filter demands a perpendicular plane, so nothing satisfies both.

Method 2 — Polarisation by Reflection

When unpolarised light reflects off a non-metallic surface (like water, glass, or a wet road), it becomes partially polarised — the reflected light oscillates more in the plane parallel to the reflecting surface than in other planes.

This is exactly why polaroid sunglasses work: if the reflecting surface (road, lake) is horizontal, the glare reflecting off it is partially polarised horizontally. Sunglasses with a vertical transmission axis block most of that horizontal glare, letting through the vertically-oscillating light that carries the useful image of what's underneath.

Method 3 — Polarisation by Refraction

Light can also become partially polarised when it refracts (transmits) from one medium into another. This time, the refracted light is polarised in the plane perpendicular to the transmitting surface — the opposite orientation to the reflected polarisation. So at any boundary, the reflected beam and the refracted beam end up partially polarised at right angles to each other.

Practice Question

Explain why polaroid sunglasses are effective at reducing glare from a wet road, but do not significantly reduce the brightness of the sky when looking straight up.

Real-World Uses of Polarisation

ApplicationHow It Works
Stress analysisA transparent sample (e.g. plastic) is placed between two crossed polarising filters (90° apart). Where the material is stressed, it rotates the plane of polarisation of light passing through, letting some light through the second filter — producing a colourful interference pattern. Brighter regions = more stress; darker regions = less stress.
Chemical analysisCertain solutions (like sugar solutions) rotate the plane of polarisation of light passing through them. The higher the concentration of the solution, the greater the angle of rotation — so measuring that angle reveals the concentration.
Polaroid sunglassesBlock horizontally-polarised glare reflected from horizontal surfaces like water and roads.
LCD screensUse polarising filters combined with liquid crystals that can rotate polarised light to control which pixels appear lit.

What to Memorise

Term / FormulaMeaning
I = P/AIntensity = power per unit area
I ∝ A², I ∝ f²Intensity depends on the square of amplitude and frequency
I = P/(4πr²)Intensity from a point source — inverse square law
n = c/vRefractive index = speed in vacuum ÷ speed in material
n₁sinθ₁ = n₂sinθ₂Snell's Law
sin(C) = 1/nCritical angle formula (dense material into air)
TIR conditionsθ > C, AND travelling from more dense → less dense medium
Optically denseA material with a high refractive index — slows light a lot
Plane polarisationOscillations restricted to a single plane perpendicular to travel
Reflection polarises...parallel to the reflecting surface
Refraction polarises...perpendicular to the transmitting surface

Concepts Checklist

Exam Tips & Common Mistakes

Check your refractive index isn't less than 1. If you calculate n < 1, you've almost certainly swapped material 1 and material 2 in Snell's Law, or mixed up sin values. Always sanity-check this before moving on.
Angles are always measured from the normal — never from the surface of the boundary itself. The normal line isn't physically "there"; it's a construction line drawn perpendicular to the surface purely so you have a consistent reference point to measure angles from.
"Show that" questions need extra precision. If the question gives you a value to 2 significant figures and asks you to "show that" something equals it, calculate your own answer to at least 3 significant figures first — this proves you actually did the calculation rather than just copying the given value.
Total Internal Reflection needs its full name and both conditions. Examiners want to see you state BOTH: (1) angle of incidence exceeds critical angle, AND (2) light travels from a denser to a less dense medium. Missing either condition loses marks — TIR cannot happen going from less dense into more dense material, no matter the angle.
Intensity is inverse-square, not inverse-linear. A very common error is assuming "twice the distance = half the intensity." It's actually a quarter. Always square the distance ratio.
Polarisation only happens to transverse waves. If a question asks you to justify why a certain wave type can or can't be polarised, always link your answer back to the direction of oscillation relative to the direction of travel — that's the actual physics, not just a rule to recite.
Reflection polarises parallel; refraction polarises perpendicular — to the surface, in both cases. These are commonly mixed up. A quick way to remember: reflected light "stays close to" the surface plane (parallel), while refracted light "dives into" the material (perpendicular).
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  • 2. Refraction & Refractive Index
  • Exam Tips & Common Mistakes
  • Spherical Waves & the Inverse Square Law
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