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.
Cross-reference with the official IB Physics subject page for the full syllabus.
§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.
Refraction at a boundary. Going into the denser (slower) medium the ray bends toward the normal and the wavefronts crowd closer — only $v$ and $\lambda$ change, not $f$.
What examiners actually test on this concept:
§2 — Snell's Law & Total Internal Reflection C.3 SL + HL
Refractive index
Snell's law
All angles are measured from the normal to the boundary, never from the surface.
Critical angle & TIR
TIR conditions: (1) light travels from denser to less dense medium; (2) angle of incidence $\theta > \theta_c$.
Total internal reflection. From a denser medium: below $\theta_c$ the ray refracts out, at $\theta_c$ it grazes the surface, and above $\theta_c$ it is totally internally reflected.
What examiners actually test on this concept:
§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
where $s$ is fringe spacing, $\lambda$ is wavelength, $D$ is slit-to-screen distance, and $d$ is slit separation.
| Change | Effect on $s$ |
|---|---|
| $\lambda$ increases | $s$ increases |
| $D$ increases | $s$ increases |
| $d$ increases | $s$ decreases |
Young's double slit. Coherent light through two slits (separation $d$) overlaps to give bright and dark fringes on a screen distance $D$ away, spaced $s = \lambda D/d$.
What examiners actually test on this concept:
§4 — Single-Slit Diffraction C.3 HL only
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.
Single-slit diffraction. A broad central maximum with first minima at $\theta=\lambda/b$ (central width $2\lambda/b$), flanked by much dimmer side maxima — a narrower slit spreads it wider.
What examiners actually test on this concept:
§5 — Diffraction Gratings C.3 HL only
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.
| Comparison | Effect |
|---|---|
| Same $d$ as double slit | Same maximum positions |
| More slits than double slit | Narrower, brighter maxima |
| Many slits | More secondary minima between principal maxima |
Diffraction grating. Sharp bright orders at $n\lambda = d\sin\theta$ — the zero order straight ahead and higher orders at ever-larger angles.
What examiners actually test on this concept:
§6 — Formula Summary C.3 SL + HL
| Formula | Use |
|---|---|
| $n = c/v$ | 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 trigger | Reach 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
- Spacing from $d = 1/N$: $N = 500\ \text{lines/mm} = 5.00 \times 10^5\ \text{lines/m}$. (M1)
- $d = 1 / (5.00 \times 10^5) = 2.00 \times 10^{-6}\ \mathrm{m} = 2.00\ \mathrm{\mu m}$. (A1)
- First-order maximum from $n\lambda = d \sin\theta$ with $n = 1$: $\sin\theta = \lambda/d$. (M1)
- $\sin\theta = (589 \times 10^{-9}) / (2.00 \times 10^{-6}) = 0.2945 \Rightarrow \theta = 17.1^\circ$. (A1)
- Maximum order from $n_{\max} = \lfloor d/\lambda \rfloor$: $d/\lambda = 2.00 \times 10^{-6} / 589 \times 10^{-9} = 3.40$. (M1)
- 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?
When can total internal reflection occur?
What conditions are needed for stable interference fringes?
What happens to the central maximum width if I narrow the slit?
How do I find the maximum order $n_{\max}$ for a diffraction grating?
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