Why is "Swapping d (slit separation) and D (screen distance) in s = λD/d" a common mistake in Wave phenomena?

Why it happens: Both symbols are distances and look similar, so students plug them into the wrong places in the fringe equation. How to avoid it: Remember d is the small distance *between the two slits* (on the bottom of the fraction) and D is the large distance to the screen (on the top). Check the magnitudes: d is usually a fraction of a millimetre, D is metres. Source: IB Physics Theme C subject reports

Why is "Using the wrong path-difference condition for bright and dark fringes" a common mistake in Wave phenomena?

Why it happens: The two conditions (nλ and (n+½)λ) are easily confused under exam pressure. How to avoid it: Bright (constructive) = a whole number of wavelengths, nλ. Dark (destructive) = an extra half wavelength, (n+½)λ, so a crest meets a trough. Anchor it: zero path difference (n=0) is the central bright fringe. Source: IB Physics Theme C subject reports

Why is "Getting the direction of refraction wrong (towards vs away from the normal)" a common mistake in Wave phenomena?

Why it happens: Students memorise 'refraction bends light' without linking the direction to the change in speed. How to avoid it: Into a denser (slower, higher-n) medium the ray bends TOWARDS the normal; into a less dense (faster) medium it bends AWAY. Check your answer: entering glass from air, θ₂ should be smaller than θ₁. Source: IB Physics Theme C subject reports

Why is "Mis-stating the conditions for total internal reflection" a common mistake in Wave phenomena?

Why it happens: Students quote only one condition (usually 'angle greater than the critical angle') and forget the medium requirement. How to avoid it: State BOTH conditions every time: the light must travel from a denser to a less dense medium AND the angle of incidence must exceed the critical angle. TIR is impossible going from less dense to denser, whatever the angle. Source: IB Physics Theme C subject reports

Why is "Thinking the frequency changes when a wave is refracted" a common mistake in Wave phenomena?

Why it happens: Because the speed and wavelength both change, students assume the frequency does too. How to avoid it: On refraction the frequency is UNCHANGED (it is fixed by the source). Only the speed and wavelength change, in proportion, so that v = fλ still holds with the same f in each medium. Source: IB Physics Theme C subject reports

Why is "Forgetting that interference needs coherent, monochromatic light" a common mistake in Wave phenomena?

Why it happens: Students describe the fringes and the maths but omit why a stable pattern forms at all. How to avoid it: Always state that the sources must be COHERENT — a constant phase difference and the same frequency (monochromatic) — otherwise the fringe positions shift randomly and no steady pattern is seen. In practice one source is split by the two slits. Source: IB Physics Theme C subject reports

IB DP Physics (100452)

Wave phenomena

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Detailed Study Notes

Detailed notes on Wave behaviour for IB DP Physics, covering key concepts, explanations, examples, and exam-focused revision points.

C.3 Wave phenomena — IB Physics SL Study Notes (Theme C: Wave behaviour)

What waves do when they meet boundaries, gaps and each other: reflection, refraction and Snell's law, refractive index and wave speed, total internal reflection and the critical angle, diffraction, the principle of superposition, constructive and destructive interference, and Young's double-slit experiment with fringe separation s = λD/d.

At a glance

Reflection: angle of incidence = angle of reflection, both measured from the normal.
Refraction is the bending of a wave when it changes speed crossing a boundary. Entering a denser (slower) medium, it bends towards the normal.
Snell's law: $n_1\sin\theta_1 = n_2\sin\theta_2$ , with refractive index $n = c/v$ . A bigger $n$ means an optically denser, slower medium.
Total internal reflection happens only going from a denser to a less dense medium and when the angle exceeds the critical angle $\theta_c$ , where $\sin\theta_c = n_2/n_1$ .
Diffraction is the spreading of a wave through a gap or around an obstacle — most noticeable when the gap ≈ the wavelength (kept qualitative at SL).
Superposition: where waves overlap, their displacements add. Constructive interference at path difference $n\lambda$ ; destructive at $(n+\tfrac{1}{2})\lambda$ .
Young's double slit: evenly spaced bright/dark fringes with separation $s = \lambda D/d$ ( $d$ = slit separation, $D$ = slit-to-screen distance).
Refraction changes a wave's speed and wavelength but NOT its frequency; a stable interference pattern needs coherent (constant phase difference), monochromatic light.

What you’ll learn

Mapped to the 100452 subject guide (2025-onwards).

State the law of reflection and describe refraction as the bending of a wave caused by a change in wave speed at a boundary.
Apply Snell's law $n_1\sin\theta_1 = n_2\sin\theta_2$ and the refractive index $n = c/v = \sin\theta_1/\sin\theta_2$ , and predict the direction of bending.
Explain total internal reflection, derive and use the critical angle condition $\sin\theta_c = n_2/n_1$ , and describe applications such as optical fibres and prisms.
Describe diffraction qualitatively as the spreading of waves through a gap or around an obstacle, most significant when the gap is comparable to the wavelength.
Apply the principle of superposition, state the conditions for constructive ( $n\lambda$ ) and destructive ( $(n+\tfrac{1}{2})\lambda$ ) interference, and use Young's double-slit relation $s = \lambda D/d$ , including the need for coherent light.

Reflection and refraction: what happens at a boundary

Reflection: i = r from the normal. Refraction: the wave changes speed and bends.

When a wave meets the boundary between two media, part of it is reflected back and part is transmitted (refracted) into the second medium. All angles in this topic are measured from the normal — the line drawn perpendicular to the surface at the point where the wave strikes.

Reflection obeys a simple law: $\theta_{\text{incidence}} = \theta_{\text{reflection}}$ The reflected ray leaves at the same angle to the normal as the incident ray arrives, and stays in the same plane. This is true for light, water waves, sound — any wave.

Refraction is the more interesting behaviour. When a wave crosses into a medium in which it travels at a different speed, its direction changes:

Entering a medium where it travels slower (an optically denser medium), the wave bends towards the normal.
Entering a medium where it travels faster (less dense), it bends away from the normal.

The key physical idea — and a favourite examiner point — is that refraction is caused by the change in wave speed, not by the change in direction itself. Think of a line of marchers stepping from firm ground onto mud at an angle: the first foot onto the mud slows first, so the line pivots. In the same way one edge of a wavefront slows before the other, swinging the wavefront round.

$\text{Wave speed and wavelength change on refraction; the } \textbf{frequency stays the same}.$

Because $v = f\lambda$ and the source sets the frequency $f$ , a slower speed $v$ in the new medium means a shorter wavelength $\lambda$ there, while $f$ is unchanged.

Going from air into glass the ray slows and bends towards the normal, so the angle of refraction θ₂ is smaller than the angle of incidence θ₁. Reverse the direction and the ray bends away from the normal.

All angles are measured from the normal, not the surface.
Reflection: angle of incidence = angle of reflection.
Refraction is caused by a change in wave speed; towards the normal = into a slower (denser) medium.

Snell's law and the refractive index

n₁ sinθ₁ = n₂ sinθ₂, with n = c/v. Bigger n = slower, optically denser.

The exact relationship between the angles is Snell's law: $n_1\sin\theta_1 = n_2\sin\theta_2$ where $n_1, n_2$ are the refractive indices of the two media and $\theta_1, \theta_2$ are the angles (from the normal) in each.

The refractive index of a medium measures how much it slows light down compared with a vacuum: $n = \frac{c}{v}$ where $c = 3.00\times10^{8}\,\text{m s}^{-1}$ is the speed of light in a vacuum and $v$ is its speed in the medium. Because $v \le c$ , the refractive index is always $n \ge 1$ . A vacuum has $n = 1$ exactly; air is very close, $n \approx 1.00$ .

Medium	Refractive index $n$	Speed of light $v$
Vacuum	1.00	$3.00\times10^{8}\,\text{m s}^{-1}$
Air	≈ 1.00	≈ $3.00\times10^{8}\,\text{m s}^{-1}$
Water	1.33	$2.26\times10^{8}\,\text{m s}^{-1}$
Glass	≈ 1.50	$2.00\times10^{8}\,\text{m s}^{-1}$

A larger $n$ means an optically denser medium in which light travels more slowly — and towards which the ray bends when it enters.

When light passes from air (or a vacuum) into a medium, Snell's law simplifies because $n_1 \approx 1$ , giving the useful form: $n = \frac{\sin\theta_1}{\sin\theta_2}$ with $\theta_1$ the angle in air and $\theta_2$ the angle in the medium.

Worked micro-example: light hits a water surface ( $n=1.33$ ) from air at $50^\circ$ to the normal. Then $1.00\sin50^\circ = 1.33\sin\theta_2 \Rightarrow \sin\theta_2 = \frac{0.766}{1.33} = 0.576 \Rightarrow \theta_2 = 35.2^\circ.$ The ray bends towards the normal (from $50^\circ$ to $35^\circ$ ) exactly as expected for entering a denser medium.

Linking speed and wavelength. Since $n = c/v$ and frequency is unchanged, you can also write $n_1 v_1 = n_2 v_2$ rearranges to $\dfrac{n_2}{n_1} = \dfrac{v_1}{v_2} = \dfrac{\lambda_1}{\lambda_2}$ : the wavelength is shorter in the medium with the higher refractive index.

Snell's law: n₁ sinθ₁ = n₂ sinθ₂ (angles from the normal).
Refractive index n = c/v ≥ 1; bigger n = slower light = optically denser.
From air: n = sinθ₁/sinθ₂; wavelength is shorter in the higher-n medium.

See the full worked example for wave phenomena →

Total internal reflection and the critical angle

Going from denser to less dense, above the critical angle the wave is fully reflected.

When light travels from a denser medium to a less dense one (e.g. glass to air), it bends away from the normal. As the angle of incidence increases, the refracted ray bends further and further, until at one special angle — the critical angle $\theta_c$ — the refracted ray travels along the boundary ( $\theta_2 = 90^\circ$ ).

Increase the angle beyond $\theta_c$ and there is no refracted ray at all: 100% of the light is reflected back into the denser medium. This is total internal reflection (TIR).

Two conditions must both be met for TIR:

The light must be travelling from an optically denser medium into a less dense one (from higher $n$ to lower $n$ ).
The angle of incidence must be greater than the critical angle ( $\theta > \theta_c$ ).

The critical angle is found by setting $\theta_2 = 90^\circ$ in Snell's law (derived in the next section): $\sin\theta_c = \frac{n_2}{n_1}$ and, when the second medium is air or a vacuum ( $n_2 \approx 1$ ): $\sin\theta_c = \frac{1}{n_1}.$

For glass ( $n = 1.50$ ) to air, $\sin\theta_c = 1/1.50 = 0.667$ , so $\theta_c = 41.8^\circ$ . Any ray inside the glass striking the surface at more than $41.8^\circ$ is totally internally reflected.

Applications examiners love:

Optical fibres: light enters the glass core and strikes the core–cladding wall at more than the critical angle, so it is totally internally reflected again and again, travelling along the fibre with almost no loss — the basis of high-speed internet and endoscopes.
Prisms: a 45°–45°–90° glass prism turns light through 90° or 180° by TIR (as in binoculars and periscopes), because 45° exceeds glass's critical angle of about 42°.

As the angle inside the denser glass increases, the refracted ray bends further from the normal until, at the critical angle, it grazes along the surface. Beyond that angle the light is totally internally reflected.

TIR needs BOTH: denser → less dense, AND angle > critical angle.
Critical angle: sinθc = n₂/n₁ (= 1/n₁ when the second medium is air/vacuum).
Applications: optical fibres and reflecting prisms.

See the full worked example for wave phenomena →

Deriving the critical angle from Snell's law (derivation)

Set the angle of refraction to 90° in Snell's law and the critical-angle formula drops out.

You are not asked to memorise a separate formula for the critical angle — it comes straight out of Snell's law in one line, which means you can never forget it. Understanding this derivation is exactly the kind of "reason from first principles" that IB rewards.

Start from Snell's law for light going from medium 1 (denser, index $n_1$ ) into medium 2 (less dense, index $n_2$ ): $n_1\sin\theta_1 = n_2\sin\theta_2.$

The critical angle $\theta_c$ is defined as the angle of incidence in the denser medium for which the refracted ray just grazes along the boundary — that is, the angle of refraction reaches its maximum possible value of $\theta_2 = 90^\circ$ .

Substitute $\theta_1 = \theta_c$ and $\theta_2 = 90^\circ$ : $n_1\sin\theta_c = n_2\sin90^\circ.$

Since $\sin90^\circ = 1$ , this rearranges directly to: $\boxed{\sin\theta_c = \frac{n_2}{n_1}}.$

When the less dense medium is air or a vacuum, $n_2 \approx 1$ , and writing $n_1 = n$ for the denser medium gives the version you will use most often: $\boxed{\sin\theta_c = \frac{1}{n}}.$

Why this guarantees $n_1 > n_2$ . For a real angle to exist we need $\sin\theta_c \le 1$ , which requires $n_2/n_1 \le 1$ , i.e. $n_2 \le n_1$ . In other words, a critical angle only exists when light goes from a denser to a less dense medium — the mathematics itself enforces the first condition for total internal reflection. If you ever try to compute a critical angle going the wrong way, you get $\sin\theta_c > 1$ , which has no solution — a neat check that you have the media the right way round.

Numerical check. For a water–air boundary ( $n_1 = 1.33$ , $n_2 = 1.00$ ): $\sin\theta_c = \frac{1.00}{1.33} = 0.752 \Rightarrow \theta_c = 48.8^\circ,$ which is why an underwater diver looking up sees the whole sky squeezed into a bright circular "window" — beyond 48.8° from the vertical, the surface acts like a mirror.

Critical angle = the incidence angle for which refraction reaches θ₂ = 90°.
Put θ₂ = 90° into Snell's law → sinθc = n₂/n₁ (and 1/n into air).
sinθc ≤ 1 requires n₂ ≤ n₁ — the maths forces the denser-to-less-dense rule.

Diffraction: waves spreading through gaps

Waves spread out through gaps and around obstacles — most when the gap ≈ the wavelength.

Diffraction is the spreading out of a wave as it passes through a gap or around the edge of an obstacle. At SL this is treated qualitatively — you describe and draw it, but you do not calculate diffraction-minimum angles (that is HL).

The amount of spreading depends on how the size of the gap compares with the wavelength $\lambda$ :

If the gap is much larger than the wavelength, the wave passes almost straight through with only slight spreading at the edges.
If the gap is comparable to the wavelength ( $\text{gap} \approx \lambda$ ), the wave spreads out strongly into semicircular wavefronts — diffraction is most significant here.

This is why you can hear someone around a corner but not see them: sound waves have wavelengths of order metres, comparable to a doorway, so they diffract strongly; light has a wavelength of only about $5\times10^{-7}\,\text{m}$ , far smaller than any doorway, so it barely spreads.

Plane wavefronts arriving at a gap about one wavelength wide emerge as curved wavefronts that spread into the region beyond — the wave "bends round" the edges of the gap.

Diffraction also explains why waves bend around obstacles and why a single wave from a gap can then overlap with another to produce interference — the topic of the next section.

Diffraction = spreading of a wave through a gap or around an obstacle.
Strongest when the gap is about the same size as the wavelength.
SL treats diffraction qualitatively (wavefront diagrams), not with equations.

Superposition, interference and Young's double slit

Displacements add. Path difference nλ = bright, (n+½)λ = dark. Fringes: s = λD/d.

When two (or more) waves overlap, the principle of superposition tells us the resultant displacement at any point and instant is the vector sum of the individual displacements. Where the waves are in step they reinforce; where they are out of step they cancel. This is interference.

Constructive interference (a bright fringe / loud spot) occurs where the two waves arrive in phase. This happens when the path difference between them is a whole number of wavelengths: $\text{path difference} = n\lambda, \quad n = 0, 1, 2, \ldots$

Destructive interference (a dark fringe / quiet spot) occurs where the waves arrive exactly out of phase (a crest meets a trough). This happens when the path difference is an odd number of half-wavelengths: $\text{path difference} = \left(n + \tfrac{1}{2}\right)\lambda, \quad n = 0, 1, 2, \ldots$

Coherence. For a stable, observable interference pattern the two sources must be coherent — they must have a constant phase difference (and therefore the same frequency), which in practice means monochromatic light from a single source split into two. Ordinary light bulbs give random, ever-changing phase, so no steady pattern forms.

Young's double-slit experiment is the classic demonstration. Monochromatic light passes through two narrow, closely spaced slits. Each slit diffracts the light so the two beams overlap, and on a distant screen they interfere to give a pattern of evenly spaced bright and dark fringes. The distance between adjacent bright (or adjacent dark) fringes is the fringe separation $s$ : $s = \frac{\lambda D}{d}$ where:

$\lambda$ = wavelength of the light,
$D$ = distance from the slits to the screen,
$d$ = separation of the two slits (the small distance between the slits).

Young's double slit: light from the two slits (separation d) overlaps on a screen a distance D away. Bright fringes appear where the path difference is a whole number of wavelengths, and neighbouring fringes are separated by s = λD/d.

Reading the formula the right way round. The bright fringes are wider apart when the wavelength is longer, the screen is further away, or the slits are closer together. Watch the roles: $d$ (slit separation) is on the bottom and $D$ (screen distance) is on the top — mixing them up is the single most common error in this whole topic.

Micro-example. Red light of wavelength $650\,\text{nm}$ passes through slits $0.40\,\text{mm}$ apart onto a screen $2.0\,\text{m}$ away: $s = \frac{\lambda D}{d} = \frac{(650\times10^{-9})(2.0)}{0.40\times10^{-3}} = 3.25\times10^{-3}\,\text{m} = 3.25\,\text{mm}.$

Superposition: overlapping waves' displacements add.
Constructive: path difference = nλ; destructive: (n+½)λ.
Young's fringes: s = λD/d, needs coherent (constant phase) monochromatic light.

See the full worked example for wave phenomena →

Quick recap

Reflection: angle of incidence = angle of reflection from the normal. Refraction: a wave changes speed at a boundary and bends (towards the normal into a denser, slower medium).
Snell's law n₁ sinθ₁ = n₂ sinθ₂ and refractive index n = c/v; a bigger n means optically denser and slower, and refraction changes speed and wavelength but not frequency.
Total internal reflection needs light going from denser to less dense AND an angle above the critical angle, where sinθc = n₂/n₁ (= 1/n into air); used in optical fibres and prisms.
Diffraction is the spreading of waves through a gap or around an obstacle, most significant when the gap is comparable to the wavelength (qualitative at SL).
Superposition adds displacements: constructive interference at path difference nλ, destructive at (n+½)λ.
Young's double slit gives evenly spaced fringes of separation s = λD/d and requires coherent, monochromatic light.

Memorise this

Verbatim phrases, formulae and definitions IB DP mark schemes credit (key for AO1 knowledge marks on Paper 1).

Reflection: angle of incidence = angle of reflection (both measured from the normal).
Snell's law: n₁ sinθ₁ = n₂ sinθ₂.
Refractive index: n = c/v (and n = sinθ₁/sinθ₂ from air); bigger n = denser = slower.
Into a denser (slower) medium a wave bends TOWARDS the normal; into a less dense one, AWAY.
Critical angle: sinθc = n₂/n₁ (= 1/n into air/vacuum). TIR needs denser→less dense AND θ > θc.
Constructive interference: path difference = nλ; destructive: path difference = (n+½)λ.
Young's fringe separation: s = λD/d (d = slit separation, D = slit-to-screen distance).
Refraction changes speed and wavelength but NOT frequency; a stable pattern needs coherent, monochromatic light.

How it’s examined

Wave phenomena appears across all three papers of IB Physics. Paper 1A (MCQ): identifying which way a ray bends at a boundary, selecting the correct critical-angle or Snell's-law expression, recognising the conditions for TIR, and reading off how the fringe separation changes when λ, D or d is altered — no calculator, so the numbers are chosen to be clean. Paper 1B (data-based): using measured angles to determine a refractive index (e.g. a graph of sinθ₁ against sinθ₂ whose gradient is n), or fringe-spacing measurements to find a wavelength. Paper 2: multi-step problems (3–6 marks) combining refraction, the critical angle and TIR (often in the context of an optical fibre or a glass block), and Young's double-slit calculations using s = λD/d, plus 'describe/explain' questions on why coherence is needed or why diffraction is strongest when the gap ≈ λ. Command terms: state, determine, calculate, show that, describe, explain. Examiner reports for Theme C repeatedly flag: measuring angles from the surface instead of the normal, stating only one of the two TIR conditions, swapping d and D in the fringe equation, using the wrong path-difference condition for bright/dark fringes, and forgetting that refraction leaves the frequency unchanged. Always show full working and keep units and significant figures consistent — method marks are awarded even when the arithmetic slips.

Sources: IB Diploma Programme Physics Guide (first assessment 2025) — Theme C: Wave behaviour (C.3 Wave phenomena); IB Physics Data Booklet (2025); IB Physics subject reports and specimen papers (2023–2025). Last reviewed 2026-07-21.

Take this whole topic with you

Step-by-step worked examples — Wave phenomena

Step-by-step solutions to past-paper-style questions on wave phenomena, written exactly the way a tutor would explain them at the board.

Question type:

1Applying Snell's law at an air–glass boundary
Getting startedDirect calculation• refraction, Snell's law, AO2
Question
A ray of light travels from air (n = 1.00) into a glass block of refractive index 1.50, striking the surface at an angle of 40° to the normal. Calculate the angle of refraction inside the glass. (3 marks)
Step-by-step solution
1. Step 1
  Write Snell's law with medium 1 = air and medium 2 = glass.
  $n_1\sin\theta_1 = n_2\sin\theta_2$
2. Step 2
  Substitute the known values and rearrange for $\sin\theta_2$ .
  $\sin\theta_2 = \frac{n_1\sin\theta_1}{n_2} = \frac{1.00\sin40^\circ}{1.50} = \frac{0.643}{1.50} = 0.429$
3. Step 3
  Take the inverse sine to find the angle of refraction.
  $\theta_2 = \sin^{-1}(0.429) = 25.4^\circ$
Answer
The angle of refraction is 25.4° (to 3 s.f.).
Examiner tip
Mark scheme: (1) correct use of Snell's law; (2) correct substitution and rearrangement; (3) θ₂ = 25.4° with a degree symbol. Because the light enters a denser medium the ray bends towards the normal, so θ₂ (25°) < θ₁ (40°) — a quick check the answer is sensible.
2Refractive index and the speed of light in a medium
Getting startedDirect calculation• refractive index, AO2
Question
The refractive index of water is 1.33. The speed of light in a vacuum is 3.00 × 10⁸ m s⁻¹. Calculate the speed of light in water. (2 marks)
Step-by-step solution
1. Step 1
  Refractive index relates the vacuum speed to the speed in the medium. Rearrange $n = c/v$ for $v$ .
  $v = \frac{c}{n}$
2. Step 2
  Substitute the values.
  $v = \frac{3.00\times10^{8}}{1.33} = 2.26\times10^{8}\,\text{m s}^{-1}$
Answer
The speed of light in water is 2.26 × 10⁸ m s⁻¹.
Examiner tip
Mark scheme: (1) correct rearrangement v = c/n; (2) v = 2.26 × 10⁸ m s⁻¹ with unit. Since n > 1 the speed must be less than c — an answer larger than 3 × 10⁸ signals the fraction has been inverted.
3Finding the critical angle for glass
Getting startedDirect calculation• total internal reflection, critical angle, AO2
Question
A glass block has a refractive index of 1.50. Calculate the critical angle for light passing from the glass into air. (2 marks)
Step-by-step solution
1. Step 1
  The second medium is air (n ≈ 1.00), so use the critical-angle condition $\sin\theta_c = 1/n$ .
  $\sin\theta_c = \frac{1}{n} = \frac{1}{1.50} = 0.667$
2. Step 2
  Take the inverse sine.
  $\theta_c = \sin^{-1}(0.667) = 41.8^\circ$
Answer
The critical angle is 41.8°.
Examiner tip
Mark scheme: (1) correct condition sinθc = 1/n (or n₂/n₁); (2) θc = 41.8°. Remember this only applies going from the denser glass into the less dense air — a critical angle for air into glass does not exist.
4Does total internal reflection occur?
Building confidenceWord problem• total internal reflection, critical angle, AO2, AO3
Question
A ray of light travelling inside water (n = 1.33) reaches the flat water–air surface at an angle of 50° to the normal. Determine whether the ray is totally internally reflected or refracts out into the air. (3 marks)
Step-by-step solution
1. Step 1
  The light is going from denser water to less dense air (condition 1 for TIR is met). Find the critical angle to test condition 2.
  $\sin\theta_c = \frac{1}{1.33} = 0.752 \Rightarrow \theta_c = 48.8^\circ$
2. Step 2
  Compare the angle of incidence with the critical angle.
  $\theta = 50^\circ > \theta_c = 48.8^\circ$
3. Step 3
  Since the angle exceeds the critical angle (and the ray is going from denser to less dense), both conditions for TIR are satisfied.
Answer
The angle of incidence (50°) is greater than the critical angle (48.8°), so total internal reflection occurs — the ray does not refract out into the air.
Examiner tip
Mark scheme: (1) critical angle 48.8°; (2) comparison 50° > 48.8°; (3) correct conclusion that TIR occurs. Both conditions must be checked: the ray is denser→less dense AND the angle beats the critical angle.
5Fringe separation in Young's double slit
Building confidenceDirect calculation• interference, double slit, AO2
Question
In a Young's double-slit experiment, light of wavelength 600 nm passes through two slits 0.50 mm apart. The fringes are observed on a screen 2.0 m away. Calculate the fringe separation. (3 marks)
Step-by-step solution
1. Step 1
  Identify the quantities in SI units: $\lambda = 600\times10^{-9}\,\text{m}$ , $d = 0.50\times10^{-3}\,\text{m}$ , $D = 2.0\,\text{m}$ . Use $s = \lambda D/d$ .
  $s = \frac{\lambda D}{d}$
2. Step 2
  Substitute the values.
  $s = \frac{(600\times10^{-9})(2.0)}{0.50\times10^{-3}} = \frac{1.2\times10^{-6}}{5.0\times10^{-4}}$
3. Step 3
  Evaluate.
  $s = 2.4\times10^{-3}\,\text{m} = 2.4\,\text{mm}$
Answer
The fringe separation is 2.4 mm.
Examiner tip
Mark scheme: (1) correct formula with d as the slit separation on the bottom; (2) all quantities converted to metres; (3) s = 2.4 mm. Forgetting to convert nm and mm to metres is the classic slip that gives a wildly wrong power of ten.
6Finding a wavelength from a fringe pattern
Building confidenceMulti-step problem• interference, double slit, AO2, AO3
Question
A double-slit pattern has a fringe separation of 1.5 mm on a screen 1.8 m from slits that are 0.60 mm apart. Calculate the wavelength of the light used. (3 marks)
Step-by-step solution
1. Step 1
  Rearrange $s = \lambda D/d$ to make $\lambda$ the subject.
  $\lambda = \frac{s\,d}{D}$
2. Step 2
  Convert to SI units: $s = 1.5\times10^{-3}\,\text{m}$ , $d = 0.60\times10^{-3}\,\text{m}$ , $D = 1.8\,\text{m}$ .
  $\lambda = \frac{(1.5\times10^{-3})(0.60\times10^{-3})}{1.8} = \frac{9.0\times10^{-7}}{1.8}$
3. Step 3
  Evaluate.
  $\lambda = 5.0\times10^{-7}\,\text{m} = 500\,\text{nm}$
Answer
The wavelength is 5.0 × 10⁻⁷ m (500 nm) — green light.
Examiner tip
Mark scheme: (1) correct rearrangement λ = sd/D; (2) consistent SI substitution; (3) λ = 500 nm. A visible-light answer between about 400 and 700 nm is the sanity check that the units were handled correctly.
7Speed, frequency and wavelength on refraction
StretchMulti-step problem• refraction, wave speed, AO2
Question
Light of wavelength 500 nm in air (n = 1.00) enters glass of refractive index 1.50. Taking c = 3.00 × 10⁸ m s⁻¹, calculate (a) the frequency of the light, (b) its speed in the glass and (c) its wavelength in the glass. (4 marks)
Step-by-step solution
1. Step 1
  (a) Frequency is fixed by the source and is the same in both media. In air, $v = c$ , so use $f = c/\lambda$ .
  $f = \frac{c}{\lambda} = \frac{3.00\times10^{8}}{500\times10^{-9}} = 6.0\times10^{14}\,\text{Hz}$
2. Step 2
  (b) Speed in the glass from $n = c/v$ .
  $v = \frac{c}{n} = \frac{3.00\times10^{8}}{1.50} = 2.0\times10^{8}\,\text{m s}^{-1}$
3. Step 3
  (c) Frequency is unchanged, so the wavelength in the glass is $\lambda' = v/f$ (equivalently $\lambda' = \lambda/n$ ).
  $\lambda' = \frac{v}{f} = \frac{2.0\times10^{8}}{6.0\times10^{14}} = 3.33\times10^{-7}\,\text{m} = 333\,\text{nm}$
Answer
(a) f = 6.0 × 10¹⁴ Hz; (b) v = 2.0 × 10⁸ m s⁻¹; (c) λ′ = 333 nm.
Examiner tip
Mark scheme: (1) f = 6.0 × 10¹⁴ Hz; (2) v = 2.0 × 10⁸ m s⁻¹; (3) frequency unchanged stated or used; (4) λ′ = 333 nm. The assessed idea is that on refraction the frequency stays constant while the speed and wavelength both fall by the factor n.
8Total internal reflection in an optical fibre
StretchMulti-step problem• total internal reflection, optical fibre, AO2, AO3
Question
The glass core of an optical fibre has a refractive index of 1.52 and is surrounded by air. (a) Calculate the critical angle for the core–air boundary. (b) A ray inside the core strikes the wall at 75° to the normal. State and explain what happens to it, and why this makes optical fibres useful. (4 marks)
Step-by-step solution
1. Step 1
  (a) The ray goes from the denser core into less dense air, so use $\sin\theta_c = 1/n$ .
  $\sin\theta_c = \frac{1}{1.52} = 0.658 \Rightarrow \theta_c = 41.1^\circ$
2. Step 2
  (b) Compare the angle of incidence with the critical angle.
  $75^\circ > \theta_c = 41.1^\circ$
3. Step 3
  The angle exceeds the critical angle and the ray is going from denser to less dense, so it is totally internally reflected. This repeats along the fibre, so light is carried with almost no loss.
Answer
(a) θc = 41.1°. (b) 75° > 41.1°, so the ray undergoes total internal reflection; repeated TIR guides the light along the fibre with very little energy loss, which is why optical fibres carry data and images efficiently.
Examiner tip
Mark scheme: (1) critical angle 41.1°; (2) comparison 75° > 41.1°; (3) TIR identified; (4) link to guiding light with low loss along the fibre. Naming both TIR conditions and connecting to the application secures full marks.

Model Answers — Wave phenomena

High-scoring sample answers for wave phenomena on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). (a) Define the refractive index of a medium. (b) State Snell's law for light passing from medium 1 into medium 2.
Model answer
(a) The refractive index $n$ of a medium is the ratio of the speed of light in a vacuum to the speed of light in that medium: $n = \frac{c}{v}.$ It is a dimensionless number, always $\ge 1$ .

(b) Snell's law: $n_1\sin\theta_1 = n_2\sin\theta_2$ , where $\theta_1$ and $\theta_2$ are the angles between the ray and the normal in media 1 and 2 respectively.
Why this scores
Why this scores 2/2. (1) refractive index defined as c/v (with the idea it compares speeds); (2) Snell's law stated correctly with angles measured from the normal. Defining n vaguely as 'how much a material bends light' is not precise enough for the mark.
Question 2
3 marks
Q (3 marks). Diamond has a refractive index of 2.42. Calculate the critical angle for light passing from diamond into air, and state one reason a large refractive index makes diamond sparkle.
Model answer
Using $\sin\theta_c = 1/n$ for a diamond–air boundary: $\sin\theta_c = \frac{1}{2.42} = 0.413$ $\theta_c = \sin^{-1}(0.413) = 24.4^\circ.$

Because the critical angle is so small (only 24.4°), light entering a cut diamond strikes many internal faces at angles greater than this, so it is totally internally reflected repeatedly before finally leaving — this traps and redirects the light, giving diamond its characteristic brilliance and sparkle.
Why this scores
Why this scores 3/3. (1) correct condition sinθc = 1/n; (2) θc = 24.4°; (3) links the small critical angle to repeated total internal reflection. The physics point is that a large n gives a small critical angle, making TIR easy to achieve.
Question 3
3 marks
Q (3 marks). (a) State the principle of superposition. (b) State the condition, in terms of path difference, for (i) constructive interference and (ii) destructive interference between two waves.
Model answer
(a) The principle of superposition states that when two or more waves overlap at a point, the resultant displacement is the sum of the individual displacements of the waves at that point.

(b) For two coherent waves of wavelength $\lambda$ :

(i) Constructive interference occurs when the path difference is a whole number of wavelengths: $\text{path difference} = n\lambda$ (with $n = 0, 1, 2, \ldots$ ).

(ii) Destructive interference occurs when the path difference is an odd number of half-wavelengths: $\text{path difference} = (n + \tfrac{1}{2})\lambda$ .
Why this scores
Why this scores 3/3. (1) superposition = displacements add; (2) constructive at nλ; (3) destructive at (n+½)λ. A frequent error is to swap these two conditions — bright fringes are the whole-number-of-wavelengths case.
Question 4
4 marks
Q (4 marks). A ray of light passes from air into diamond (n = 2.42), striking the surface at 30° to the normal. Taking c = 3.00 × 10⁸ m s⁻¹, calculate (a) the angle of refraction and (b) the speed of light in the diamond. (c) State what happens to the frequency of the light as it enters the diamond.
Model answer
(a) Angle of refraction (Snell's law, air → diamond): $1.00\sin30^\circ = 2.42\sin\theta_2 \Rightarrow \sin\theta_2 = \frac{0.500}{2.42} = 0.207$ $\theta_2 = \sin^{-1}(0.207) = 11.9^\circ.$

(b) Speed in diamond: $v = \frac{c}{n} = \frac{3.00\times10^{8}}{2.42} = 1.24\times10^{8}\,\text{m s}^{-1}.$

(c) The frequency is unchanged — refraction alters the speed and wavelength of the light, but its frequency stays the same because it is fixed by the source.
Why this scores
Why this scores 4/4. (1) correct Snell substitution; (2) θ₂ = 11.9° (strong bending towards the normal for such a high n); (3) v = 1.24 × 10⁸ m s⁻¹; (4) frequency unchanged. The frequency statement is a routine examiner discriminator that many candidates miss.
Question 5
4 marks
Q (4 marks). In a double-slit experiment the slits are 0.25 mm apart and the screen is 1.2 m away. The bright fringes are measured to be 3.0 mm apart. (a) Calculate the wavelength of the light. (b) State and explain what happens to the fringe separation if the slits are moved closer together.
Model answer
(a) Rearranging $s = \lambda D/d$ for the wavelength: $\lambda = \frac{s\,d}{D} = \frac{(3.0\times10^{-3})(0.25\times10^{-3})}{1.2} = \frac{7.5\times10^{-7}}{1.2} = 6.25\times10^{-7}\,\text{m}.$ So $\lambda = 625\,\text{nm}$ (red light).

(b) From $s = \lambda D/d$ , the fringe separation is inversely proportional to the slit separation $d$ . Moving the slits closer together decreases $d$ , so the fringe separation $s$ increases — the fringes spread further apart.
Why this scores
Why this scores 4/4. (1) correct rearrangement λ = sd/D; (2) λ = 625 nm; (3) recognises s ∝ 1/d; (4) fringes get wider apart. The relationship between s and d (inverse) is exactly what examiners probe with the second part.
Question 6
5 marks
Q (5 marks). An optical fibre has a glass core of refractive index 1.50 surrounded by air. (a) Calculate the critical angle for the core. (b) Explain, with reference to the critical angle, how light is transmitted along the fibre and why this is described as total internal reflection.
Model answer
(a) For light going from the denser core into less dense air: $\sin\theta_c = \frac{1}{n} = \frac{1}{1.50} = 0.667 \Rightarrow \theta_c = 41.8^\circ.$

(b) Explanation:

Light travelling down the fibre strikes the core wall at a large angle to the normal — greater than the critical angle of 41.8°.

Because the light is going from an optically denser medium (the glass core) into a less dense medium and the angle of incidence exceeds the critical angle, there is no refracted ray: all of the light energy is reflected back into the core.

This is called total internal reflection because 100% of the light is reflected (unlike an ordinary mirror, which absorbs some). The process repeats at each bounce, so the light is guided along the fibre with very little loss even when it bends.
Why this scores
Why this scores 5/5. (a) 1 mark for θc = 41.8°. (b) 4 marks: light hits the wall above the critical angle (1); denser-to-less-dense condition (1); no refraction, all light reflected (1); 'total' because 100% reflected and repeats along the fibre (1). Stating BOTH TIR conditions is essential.
Question 7
6 marks
Q (6 marks). Monochromatic light is shone through two narrow, closely spaced slits and a pattern of evenly spaced bright and dark fringes appears on a distant screen. (a) Explain how the bright and dark fringes are formed. (b) Explain why the light must be coherent for a clear, stable pattern to be seen. (c) State one change that would increase the spacing of the fringes.
Model answer
(a) Each slit diffracts the light, so the two emerging beams overlap on the screen. By the principle of superposition their displacements add:

Where the two waves arrive in phase (path difference a whole number of wavelengths, $n\lambda$ ), they interfere constructively to give a bright fringe.

Where they arrive exactly out of phase (path difference $(n+\tfrac{1}{2})\lambda$ ), they interfere destructively and cancel, giving a dark fringe.

(b) For a stable pattern the two sources must be coherent — they must maintain a constant phase difference (and therefore the same frequency, i.e. monochromatic light). If the phase relationship changed randomly, the positions of the bright and dark fringes would shift about too fast to see and the pattern would wash out. Using a single source split by the two slits guarantees this coherence.

(c) Any one of: increase the wavelength of the light, increase the distance $D$ from the slits to the screen, or decrease the slit separation $d$ — since $s = \lambda D/d$ .
Why this scores
Why this scores 6/6. (a) 3 marks: diffraction/overlap, constructive at nλ → bright, destructive at (n+½)λ → dark. (b) 2 marks: coherence = constant phase difference, otherwise the pattern is not stable. (c) 1 mark: a valid change consistent with s = λD/d. The coherence explanation must mention constant phase difference, not just 'same colour'.
Question 8
6 marks
Q (6 marks). A ray of light passes from air into a semicircular glass block of refractive index 1.60, entering the flat face at 45° to the normal. (a) Calculate the angle of refraction inside the glass. (b) Calculate the speed of light in the glass. (c) Calculate the critical angle for the glass–air boundary and hence state whether a ray inside the glass striking a surface at 35° to the normal would be totally internally reflected.
Model answer
(a) Angle of refraction (Snell's law, air → glass): $1.00\sin45^\circ = 1.60\sin\theta_2 \Rightarrow \sin\theta_2 = \frac{0.707}{1.60} = 0.442$ $\theta_2 = \sin^{-1}(0.442) = 26.2^\circ.$

(b) Speed of light in the glass: $v = \frac{c}{n} = \frac{3.00\times10^{8}}{1.60} = 1.88\times10^{8}\,\text{m s}^{-1}.$

(c) Critical angle: $\sin\theta_c = \frac{1}{1.60} = 0.625 \Rightarrow \theta_c = 38.7^\circ.$ A ray striking a surface at $35^\circ$ has an angle less than the critical angle ( $35^\circ < 38.7^\circ$ ), so it is not totally internally reflected — it refracts out of the glass into the air.
Why this scores
Why this scores 6/6. (a) 2 marks: Snell substitution and θ₂ = 26.2°; (b) 1 mark: v = 1.88 × 10⁸ m s⁻¹; (c) 3 marks: critical angle formula, θc = 38.7°, and the correct comparison/conclusion that 35° < θc so no TIR. Testing the angle against the critical angle is the assessed decision.
Question 9
11 marks
Q (11 marks — extended response). A beam of monochromatic light of wavelength 590 nm in air is used in an optics experiment. It enters a rectangular glass block of refractive index 1.52. Taking c = 3.00 × 10⁸ m s⁻¹: (a) The light strikes the flat top surface at 35° to the normal. Calculate the angle of refraction inside the glass. (b) Calculate the speed and the wavelength of the light inside the glass. (c) Calculate the critical angle for the glass–air boundary. (d) Inside the block the ray then strikes a side face at 44° to the normal. State and explain whether it undergoes total internal reflection. (e) After leaving the block, the light passes through a double slit of separation 0.40 mm and forms fringes on a screen 2.5 m away. Calculate the fringe separation.
Model answer
(a) Angle of refraction on entering the glass (Snell's law, air → glass): $1.00\sin35^\circ = 1.52\sin\theta_2 \Rightarrow \sin\theta_2 = \frac{0.574}{1.52} = 0.377$ $\theta_2 = \sin^{-1}(0.377) = 22.2^\circ.$

(b) Speed and wavelength in the glass. The frequency is fixed by the source: $f = \frac{c}{\lambda} = \frac{3.00\times10^{8}}{590\times10^{-9}} = 5.08\times10^{14}\,\text{Hz}.$ Speed in the glass: $v = \frac{c}{n} = \frac{3.00\times10^{8}}{1.52} = 1.97\times10^{8}\,\text{m s}^{-1}.$ Wavelength in the glass (frequency unchanged, so $\lambda' = v/f = \lambda/n$ ): $\lambda' = \frac{590\,\text{nm}}{1.52} = 388\,\text{nm}.$

(c) Critical angle for the glass–air boundary: $\sin\theta_c = \frac{1}{1.52} = 0.658 \Rightarrow \theta_c = 41.1^\circ.$

(d) Total internal reflection at the side face. The ray travels from the denser glass into less dense air (condition 1 met) and strikes the face at $44^\circ$ , which is greater than the critical angle of $41.1^\circ$ (condition 2 met). Because both conditions are satisfied, the ray is totally internally reflected back into the glass — none of it escapes through the side face.

(e) Fringe separation. Once back in air the wavelength is again $590\,\text{nm}$ . Using $s = \lambda D/d$ with $d = 0.40\times10^{-3}\,\text{m}$ and $D = 2.5\,\text{m}$ : $s = \frac{(590\times10^{-9})(2.5)}{0.40\times10^{-3}} = \frac{1.475\times10^{-6}}{4.0\times10^{-4}} = 3.7\times10^{-3}\,\text{m} = 3.7\,\text{mm}.$
Why this scores
Why this scores 11/11. (a) Snell substitution + θ₂ = 22.2° (2); (b) v = 1.97 × 10⁸ m s⁻¹ (1), frequency unchanged used, λ′ = 388 nm (2); (c) θc = 41.1° (2); (d) both TIR conditions stated and 44° > 41.1° → TIR (2); (e) s = λD/d with the air wavelength → 3.7 mm (2). This is a model IB Paper 2 extended response: the grade-9 discriminators are (i) knowing the frequency stays constant so λ falls by the factor n inside the glass, (ii) checking BOTH TIR conditions at the side face, and (iii) reverting to the air wavelength (590 nm, not 388 nm) once the light has left the block for the fringe calculation.

Key Formulae — Wave phenomena

The formulae you need to memorise for wave phenomena on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Snell's law of refraction
$n_1\sin\theta_1 = n_2\sin\theta_2$
$n_1$
refractive index of medium 1
$n_2$
refractive index of medium 2
$\theta_1$
angle to the normal in medium 1
$\theta_2$
angle to the normal in medium 2
When to use
Whenever a ray crosses a boundary between two media — to find an unknown angle or refractive index. Keep each n with the angle in the same medium.
Example
Air (1.00) into glass (1.50) at 40°: sinθ₂ = sin40°/1.50 = 0.429 → θ₂ = 25.4°.
Refractive index and wave speed
$n = \frac{c}{v}$
$n$
refractive index (dimensionless, ≥ 1)
$c$
speed of light in a vacuum, 3.00 × 10⁸ m s⁻¹
$v$
speed of light in the medium (m s⁻¹)
When to use
To convert between a refractive index and the speed of light in a medium; a larger n means slower light and an optically denser medium.
Example
Water n = 1.33: v = c/n = 3.00 × 10⁸ / 1.33 = 2.26 × 10⁸ m s⁻¹.
Critical angle for total internal reflection
$\sin\theta_c = \frac{n_2}{n_1}$
$\theta_c$
critical angle in the denser medium
$n_1$
refractive index of the denser medium (light starts here)
$n_2$
refractive index of the less dense medium
When to use
For light going from a denser to a less dense medium. When the second medium is air/vacuum, n₂ ≈ 1 so sinθc = 1/n₁. TIR occurs when the angle of incidence exceeds θc.
Example
Glass (1.50) to air: sinθc = 1/1.50 = 0.667 → θc = 41.8°.
Young's double-slit fringe separation
$s = \frac{\lambda D}{d}$
$s$
fringe separation — distance between adjacent bright (or dark) fringes (m)
$\lambda$
wavelength of the light (m)
$D$
distance from the slits to the screen (m)
$d$
separation of the two slits (m)
When to use
For the evenly spaced fringes of a two-slit interference pattern with coherent, monochromatic light. Note d (slit separation) is on the bottom and D (screen distance) on the top.
Example
λ = 600 nm, D = 2.0 m, d = 0.50 mm: s = (600×10⁻⁹)(2.0)/(0.50×10⁻³) = 2.4 mm.
Wave equation
$v = f\lambda$
$v$
wave speed (m s⁻¹)
$f$
frequency (Hz) — unchanged on refraction
$\lambda$
wavelength (m)
When to use
To link speed, frequency and wavelength. On refraction the frequency is constant, so a change in speed produces a proportional change in wavelength.
Example
Light of f = 6.0 × 10¹⁴ Hz in glass (v = 2.0 × 10⁸ m s⁻¹): λ = v/f = 333 nm.
Interference path-difference conditions
$\text{constructive: } n\lambda \qquad \text{destructive: } \left(n+\tfrac{1}{2}\right)\lambda$
$n$
an integer, n = 0, 1, 2, …
$\lambda$
wavelength of the waves (m)
When to use
To decide whether two overlapping coherent waves reinforce (bright/loud) or cancel (dark/quiet) at a point, from the difference in the distances travelled.
Example
Path difference of exactly 1.5λ = (1+½)λ → destructive interference (a dark fringe).

Key Definitions and Keywords — Wave phenomena

Definitions to memorise and the exact keywords mark schemes credit for wave phenomena answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Reflection
Examiner keyword
The bouncing back of a wave at a boundary. The angle of incidence equals the angle of reflection, both measured from the normal.
Refraction
Examiner keyword
The change in direction of a wave when it crosses a boundary and changes speed. It bends towards the normal entering a slower (denser) medium and away from the normal entering a faster one.
Refractive index (n)
Examiner keyword
A measure of how much a medium slows light: n = c/v, where c is the vacuum speed and v the speed in the medium. Always ≥ 1; a larger n means an optically denser, slower medium.
Snell's law
Examiner keyword
The relationship n₁ sinθ₁ = n₂ sinθ₂ linking the refractive indices and the angles (from the normal) on either side of a boundary.
Normal
Examiner keyword
The line drawn perpendicular to a surface at the point where a wave strikes it. All angles of incidence, reflection and refraction are measured from the normal.
Angle of incidence
The angle between an incoming ray and the normal at the boundary.
Angle of refraction
The angle between a refracted (transmitted) ray and the normal in the second medium.
Optically denser medium
Examiner keyword
A medium with a higher refractive index, in which light travels more slowly. A wave bends towards the normal when it enters an optically denser medium.
Total internal reflection (TIR)
Examiner keyword
The complete reflection of a wave back into the denser medium, occurring when light travels from a denser to a less dense medium at an angle greater than the critical angle. No light is transmitted.
Critical angle (θc)
Examiner keyword
The angle of incidence in the denser medium for which the refracted ray travels along the boundary (angle of refraction = 90°), given by sinθc = n₂/n₁ (= 1/n into air/vacuum).
Diffraction
Examiner keyword
The spreading out of a wave as it passes through a gap or around an obstacle. It is most significant when the size of the gap is comparable to the wavelength.
Principle of superposition
Examiner keyword
When two or more waves overlap, the resultant displacement at any point is the sum (vector sum) of the individual displacements of the waves at that point.
Constructive interference
Examiner keyword
The reinforcement of two waves that arrive in phase (path difference nλ), producing a larger amplitude — a bright fringe or loud spot.
Destructive interference
Examiner keyword
The cancellation of two waves that arrive out of phase (path difference (n+½)λ), producing a smaller or zero amplitude — a dark fringe or quiet spot.
Path difference
Examiner keyword
The difference in the distances travelled by two waves from their sources to a given point. It determines whether interference at that point is constructive or destructive.
Coherence
Examiner keyword
The property of two wave sources that maintain a constant phase difference (and hence the same frequency). Coherent, monochromatic light is required to observe a stable interference pattern.

Common Mistakes and Misconceptions — Wave phenomena

The traps other students keep falling into on wave phenomena questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Swapping d (slit separation) and D (screen distance) in s = λD/d
IB Physics Theme C subject reports
Why it happens
Both symbols are distances and look similar, so students plug them into the wrong places in the fringe equation.
How to avoid it
Remember d is the small distance between the two slits (on the bottom of the fraction) and D is the large distance to the screen (on the top). Check the magnitudes: d is usually a fraction of a millimetre, D is metres.
✕Using the wrong path-difference condition for bright and dark fringes
IB Physics Theme C subject reports
Why it happens
The two conditions (nλ and (n+½)λ) are easily confused under exam pressure.
How to avoid it
Bright (constructive) = a whole number of wavelengths, nλ. Dark (destructive) = an extra half wavelength, (n+½)λ, so a crest meets a trough. Anchor it: zero path difference (n=0) is the central bright fringe.
✕Getting the direction of refraction wrong (towards vs away from the normal)
IB Physics Theme C subject reports
Why it happens
Students memorise 'refraction bends light' without linking the direction to the change in speed.
How to avoid it
Into a denser (slower, higher-n) medium the ray bends TOWARDS the normal; into a less dense (faster) medium it bends AWAY. Check your answer: entering glass from air, θ₂ should be smaller than θ₁.
✕Mis-stating the conditions for total internal reflection
IB Physics Theme C subject reports
Why it happens
Students quote only one condition (usually 'angle greater than the critical angle') and forget the medium requirement.
How to avoid it
State BOTH conditions every time: the light must travel from a denser to a less dense medium AND the angle of incidence must exceed the critical angle. TIR is impossible going from less dense to denser, whatever the angle.
✕Thinking the frequency changes when a wave is refracted
IB Physics Theme C subject reports
Why it happens
Because the speed and wavelength both change, students assume the frequency does too.
How to avoid it
On refraction the frequency is UNCHANGED (it is fixed by the source). Only the speed and wavelength change, in proportion, so that v = fλ still holds with the same f in each medium.
✕Forgetting that interference needs coherent, monochromatic light
IB Physics Theme C subject reports
Why it happens
Students describe the fringes and the maths but omit why a stable pattern forms at all.
How to avoid it
Always state that the sources must be COHERENT — a constant phase difference and the same frequency (monochromatic) — otherwise the fringe positions shift randomly and no steady pattern is seen. In practice one source is split by the two slits.

Wave phenomena

Detailed Study Notes

At a glance

What you’ll learn

Reflection and refraction: what happens at a boundary

Snell's law and the refractive index

Total internal reflection and the critical angle

Deriving the critical angle from Snell's law (derivation)

Diffraction: waves spreading through gaps

Superposition, interference and Young's double slit

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Applying Snell's law at an air–glass boundary

2Refractive index and the speed of light in a medium

3Finding the critical angle for glass

4Does total internal reflection occur?

5Fringe separation in Young's double slit

6Finding a wavelength from a fringe pattern

7Speed, frequency and wavelength on refraction

8Total internal reflection in an optical fibre

Snell's law of refraction

Refractive index and wave speed

Critical angle for total internal reflection

Young's double-slit fringe separation

Wave equation

Interference path-difference conditions

Reflection

Refraction

Refractive index (n)

Snell's law

Normal

Angle of incidence

Angle of refraction

Optically denser medium

Total internal reflection (TIR)

Critical angle (θc)

Diffraction

Principle of superposition

Constructive interference

Destructive interference

Path difference

Coherence

✕Swapping d (slit separation) and D (screen distance) in s = λD/d

✕Using the wrong path-difference condition for bright and dark fringes

✕Getting the direction of refraction wrong (towards vs away from the normal)

✕Mis-stating the conditions for total internal reflection

✕Thinking the frequency changes when a wave is refracted

✕Forgetting that interference needs coherent, monochromatic light