Free Cheatsheet · Topic C.3 · SL + HL

IB Physics HL Wave Phenomena — Complete Cheatsheet

Every formula, derivation, trick, and trap for IB Physics HL Topic C.3 Wave Phenomena — refraction, Snell's law, total internal reflection, double-slit interference, single-slit diffraction (HL), and gratings (HL).

Topic: C.3 Wave Phenomena Syllabus: SL + HL (single-slit & gratings HL only) Read time: ~14 minutes Last updated: Apr 2026

Wave Phenomena (Topic C.3) is the most heavily examined wave topic in IB Physics HL. SL students need refraction, Snell's law, total internal reflection, the principle of superposition and the Young's double-slit formula. HL students extend this with single-slit diffraction ($\theta = \lambda/b$), the double-slit envelope effect, missing orders, and the diffraction grating equation $n\lambda = d \sin\theta$.

This cheatsheet condenses every Topic C.3 formula and exam trap onto one page. The most common HL pitfalls — claiming frequency changes at a boundary, misidentifying the TIR direction, mixing up double-slit and single-slit formulas, forgetting $\sin\theta \le 1$ for the maximum grating order — are flagged in red. Scroll to the bottom for the printable PDF and the full Photon Academy library.

§1 — Wavefronts & Refraction C.3 SL + HL

Key concepts

  • Wavefront: a surface of equal phase (e.g. all the crests).
  • Ray: perpendicular to the wavefront; shows the direction of energy flow.
  • Plane waves have parallel rays (distant or far-field source).
  • Point source gives circular (2-D) or spherical (3-D) wavefronts and diverging rays.

At a boundary

  • Reflection: $\theta_i = \theta_r$ (both measured from the normal).
  • Refraction: direction changes, frequency stays the same.
  • $v$ changes $\Rightarrow$ $\lambda$ changes; $f$ unchanged.
TrickIn a denser (higher $n$) medium, wavefronts are closer together (smaller $\lambda$) and the ray bends towards the normal.
TrapNever say frequency changes at a boundary — only speed and wavelength change. Frequency is set by the source.

§2 — Snell's Law & Total Internal Reflection C.3 SL + HL

Refractive index

Definition:$n = \dfrac{c}{v}$   ($n \ge 1$ always; larger $n$ = slower light).

Snell's law

Snell:$n_1 \sin\theta_1 = n_2 \sin\theta_2$
Equivalents:$\dfrac{n_1}{n_2} = \dfrac{\sin\theta_2}{\sin\theta_1} = \dfrac{v_2}{v_1} = \dfrac{\lambda_2}{\lambda_1}$

All angles are measured from the normal to the boundary, never from the surface.

Critical angle & TIR

Critical angle:$\sin\theta_c = \dfrac{n_2}{n_1}$   (with $n_1 > n_2$).

TIR conditions: (1) light travels from denser to less dense medium; (2) angle of incidence $\theta > \theta_c$.

TrickFor light entering medium $Q$ from air, $n_Q = \sin\theta_\text{air} / \sin\theta_Q$. This is the fastest way to extract a refractive index from an experimental data table.
TrapTIR can only occur going from a denser to a less dense medium. It cannot happen for light travelling air $\to$ glass.
NoteApplication: optical fibres exploit TIR to transmit light along bent paths with negligible loss. The cladding has a slightly lower refractive index than the core.

§3 — Superposition & Interference C.3 SL + HL

Principle of superposition

$y_\text{total} = y_1 + y_2$ — the algebraic sum of displacements at every point. After meeting, the two waves continue unchanged.

Conditions for stable interference

Sources must be coherent: same frequency and a constant phase difference. Then, assuming the sources are in phase to start with:

  • Constructive (bright fringe): path difference $= n\lambda$.
  • Destructive (dark fringe): path difference $= (n + \tfrac{1}{2}) \lambda$.

Young's double-slit formula

Fringe spacing:$s = \dfrac{\lambda D}{d}$

where $s$ is fringe spacing, $\lambda$ is wavelength, $D$ is slit-to-screen distance, and $d$ is slit separation.

ChangeEffect on $s$
$\lambda$ increases$s$ increases
$D$ increases$s$ increases
$d$ increases$s$ decreases
TrickIf $\lambda$, $D$ and $d$ are all multiplied by the same factor $k$, the fringe spacing $s$ is unchanged.
TrapFor constructive interference the path difference must be a whole number of wavelengths. If you write $(n + \tfrac{1}{2})\lambda$ for constructive you have swapped the two conditions.

§4 — Single-Slit Diffraction C.3 HL only

First minimum:$\theta = \dfrac{\lambda}{b}$   ($b$ = slit width, $\theta$ in radians).

Angular width of the central maximum $= 2\lambda/b$. On a screen at distance $D$, the central maximum has width $2\lambda D/b$.

  • Narrower slit $\Rightarrow$ wider, dimmer central maximum.
  • Wider slit $\Rightarrow$ narrower, brighter central maximum.

Envelope effect & missing orders

Double-slit fine fringes (spacing $\approx \lambda/d$) are modulated by the single-slit envelope (width $\approx \lambda/b$). A double-slit maximum of order $n$ is missing when $n = k\,d/b$ for integer $k$ — the envelope minimum coincides with a double-slit maximum.

Trick$n_\text{missing} = d/b$ for the first missing order. If $d = 4b$, the $n = 4$ fringe is missing; the $n = 8$ fringe is also missing (and so on).

§5 — Diffraction Gratings C.3 HL only

Grating equation:$n\lambda = d \sin\theta$

where $d = 1/N$ is the slit spacing ($N$ = lines per metre) and $n$ is the order. Maximum order: $n_\max = \lfloor d/\lambda \rfloor$.

  • More slits $\Rightarrow$ sharper, brighter principal maxima (and more secondary minima between them).
  • For the same $d$, a grating gives the maxima at the same angles as a double slit — but with much greater contrast.
ComparisonEffect
Same $d$ as double slitSame maximum positions
More slits than double slitNarrower, brighter maxima
Many slitsMore secondary minima between principal maxima
TrickConvert lines/mm to grating spacing $d$: 600 lines/mm $\Rightarrow d = 1/(600 \times 10^3)\ \mathrm{m} = 1.67\ \mathrm{\mu m}$.
TrapThe maximum order is limited by $\sin\theta \le 1$. Always check $n\lambda \le d$. If $d/\lambda = 3.7$ then only 3 orders are visible on each side.
NoteWhite light through a grating disperses into spectra; in each order, violet (short $\lambda$) appears at smaller angles and red at larger angles.

§6 — Formula Summary C.3 SL + HL

FormulaUse
$n = c/v$Refractive index
$n_1 \sin\theta_1 = n_2 \sin\theta_2$Snell's law
$\sin\theta_c = n_2/n_1$Critical angle
$\Delta d = n\lambda$Constructive interference
$\Delta d = (n + \tfrac{1}{2})\lambda$Destructive interference
$s = \lambda D / d$Young's fringe spacing
$\theta = \lambda / b$Single-slit 1st minimum (HL)
$n\lambda = d \sin\theta$Diffraction grating (HL)
$d = 1/N$Grating spacing from lines/m
$n_\max = \lfloor d/\lambda \rfloor$Max grating order (HL)

§7 — Exam Attack Plan All sections

When you see this in the question — reach for that:

Question triggerReach for
"Find the angle of refraction"Snell's law: $n_1 \sin\theta_1 = n_2 \sin\theta_2$.
"Will TIR occur?"Check (1) denser → less dense, (2) $\theta > \theta_c$.
"Find fringe spacing"$s = \lambda D / d$ (Young's).
"What changes at a boundary?"Speed and wavelength change; frequency stays the same.
"Path difference for bright fringe"$n\lambda$ (whole wavelengths).
"Find central max width" (HL)$2\lambda/b$ angular, $2\lambda D/b$ on screen.
"Missing order in double slit" (HL)$n = k\,d/b$.
"Maximum grating order" (HL)$n_\max = \lfloor d/\lambda \rfloor$.
"Number of orders visible" (HL)$2n_\max + 1$ (counting both sides + central).
"Lines per mm to $d$"$d = 1/N$ in metres (HL).

Worked Example — IB-Style HL Diffraction Grating Problem

Question (HL Paper 2 style — 7 marks)

Monochromatic light of wavelength 589 nm is normally incident on a diffraction grating with 500 lines per mm. Calculate (a) the spacing $d$ of the grating, (b) the angle at which the first-order maximum appears, (c) the maximum order that can be observed.

Solution

  1. Spacing from $d = 1/N$: $N = 500\ \text{lines/mm} = 5.00 \times 10^5\ \text{lines/m}$.  (M1)
  2. $d = 1 / (5.00 \times 10^5) = 2.00 \times 10^{-6}\ \mathrm{m} = 2.00\ \mathrm{\mu m}$.  (A1)
  3. First-order maximum from $n\lambda = d \sin\theta$ with $n = 1$: $\sin\theta = \lambda/d$.  (M1)
  4. $\sin\theta = (589 \times 10^{-9}) / (2.00 \times 10^{-6}) = 0.2945 \Rightarrow \theta = 17.1^\circ$.  (A1)
  5. Maximum order from $n_\max = \lfloor d/\lambda \rfloor$: $d/\lambda = 2.00 \times 10^{-6} / 589 \times 10^{-9} = 3.40$.  (M1)
  6. Take floor: $n_\max = 3$. So orders $n = 0, \pm 1, \pm 2, \pm 3$ are visible (7 in total).  (A1)(R1)

Examiner's note: The single most common error is forgetting that $\sin\theta$ cannot exceed 1, leading students to "compute" a non-existent fourth-order maximum. Always check $n\lambda \le d$ before reporting an answer.

Common Student Questions

Why does frequency stay constant during refraction?
Frequency is set by the source. At a boundary, wavefronts must match up on both sides — the same number of wavefronts per second arrive on the second medium as left the first. So the frequency cannot change. Only the wave speed and (because $v = f\lambda$) the wavelength change at a boundary. Saying frequency changes is one of the most common mark losses in IB Topic C.3 questions.
When can total internal reflection occur?
TIR requires two conditions: (1) the wave is travelling from a denser (higher $n$) medium to a less dense (lower $n$) medium, and (2) the angle of incidence exceeds the critical angle $\theta_c$, where $\sin\theta_c = n_2/n_1$. TIR cannot happen going from air to glass — only glass to air or water to air. Optical fibres exploit TIR to transmit light along bent paths with negligible loss.
What conditions are needed for stable interference fringes?
The two sources must be coherent: same frequency AND a constant phase difference. They should also have similar amplitudes for high-contrast fringes. Constructive interference (bright fringes) occurs where the path difference is a whole number of wavelengths, $n\lambda$. Destructive interference (dark fringes) occurs where the path difference is $(n + \tfrac{1}{2})\lambda$. This assumes the sources are in phase to start with.
What happens to the central maximum width if I narrow the slit?
From the single-slit formula $\theta = \lambda/b$ (HL), narrowing the slit $b$ increases $\theta$ — so the central maximum becomes wider (and dimmer because the same energy spreads over more area). Widening the slit makes the central maximum narrower and brighter. The angular width of the central maximum is $2\lambda/b$ on either side of the centre, giving a screen width of $2\lambda D/b$.
How do I find the maximum order $n_\max$ for a diffraction grating?
From the grating equation $n\lambda = d\sin\theta$, the maximum value of $\sin\theta$ is 1. So $n_\max = \lfloor d/\lambda \rfloor$ — take the integer part of $d/\lambda$. Always check this in HL grating questions: the question may ask "how many orders are visible?" and the answer is $2 n_\max + 1$ (counting both sides plus the central maximum $n = 0$).

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