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Educerie · IB Diploma · Physics
Theme C Wave behaviour · C.3 Wave phenomena
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Describe waves in two and three dimensions with wavefronts and rays | SL, HL | "Draw the wavefronts after…" (2 marks) |
| Describe reflection, refraction and transmission at a boundary | SL, HL | Label incident, reflected and transmitted rays; explain refraction by a change of speed |
| Draw wavefront–ray diagrams of refraction and diffraction | SL, HL | Sketch wavefronts crossing a boundary or passing a gap (2–3 marks) |
| Use Snell's law, n₁/n₂ = sin θ₂/sin θ₁ = v₂/v₁ | SL, HL | Calculate an angle, a refractive index or a speed (2–3 marks); Paper 1B sin–sin graph |
| Explain and calculate the critical angle and total internal reflection | SL, HL | "Calculate the critical angle" (2 marks); state the two conditions for TIR |
| Describe diffraction round a body and through an aperture | SL, HL | "Explain why sound is heard round a corner but light is not seen" (2 marks) |
| Apply the principle of superposition to waves and pulses | SL, HL | Sketch the resultant of two overlapping pulses |
| Explain why double-source interference needs coherent sources | SL, HL | "Outline why two lamps do not give fringes" (2 marks) |
| Use path difference = nλ and (n + ½)λ | SL, HL | "Determine whether P is a point of maximum or minimum intensity" (2–3 marks) |
| Use s = λD/d for Young's double slit | SL, HL | Calculate λ or s; Paper 1B s against D |
| Describe single-slit diffraction and use θ = λ/b | HL only | Width of the central maximum; effect of slit width (2–3 marks) |
| Explain how the single-slit pattern modulates the double-slit pattern | HL only | Sketch the modulated pattern; explain a missing order |
| Use nλ = d sin θ for gratings, with monochromatic and white light | HL only | Angle of an order, highest order, overlapping spectra (2–4 marks) |
Before you start
You need C.2: wavelength, frequency, speed, v = fλ, and the rule that the source sets f while the medium sets v. You need trigonometry in degrees for Snell's law, and radians and the small-angle idea for the HL diffraction formula. Have the data booklet open at C.3.
1The idea in one paragraph
A wave spreading across a pond or through space is drawn with wavefronts, lines joining points in step, and rays, arrows showing which way the energy goes. When a wave reaches a boundary between two media, some of it reflects and some is transmitted; the transmitted part changes speed, and if it arrives at an angle it changes direction, which is refraction, governed by Snell's law. Going from a slow medium into a faster one, beyond a critical angle, nothing is transmitted at all. When a wave passes an edge or a gap it spreads out, which is diffraction, strongest when the gap is about one wavelength. Where two waves overlap their displacements add, which is superposition; two coherent sources then produce fixed places of reinforcement and cancellation, decided by the path difference. Young's double slit turns that into bright and dark fringes spaced s = λD/d. At HL, a single slit makes its own pattern, which shapes the double-slit pattern, and many slits make the sharp maxima of a diffraction grating.
2Wavefronts and rays
Figure 1 shows the two tools for drawing waves in two and three dimensions.
A wavefront is a line (in 2D) or surface (in 3D) joining neighbouring points that are in phase: all at a crest at the same moment, for example. Adjacent wavefronts drawn crest to crest are one wavelength apart.
A ray is a line showing the direction in which the wave, and its energy, travels. Rays are always perpendicular to the wavefronts.
Far from a small source, or from a straight source, the wavefronts are straight and parallel: plane wavefronts, with parallel rays (panel a). Close to a point source they are circles, or spheres in three dimensions, with rays spreading out from the source (panel b).
3At a boundary: reflection, refraction and transmission
When a wave reaches a boundary between two media, three things can happen, and usually all three happen at once. Figure 2 shows them.
Angles are always measured from the normal, the line at right angles to the boundary at the point where the ray arrives.
Reflection. Part of the wave bounces back into the first medium. The angle of reflection equals the angle of incidence. The reflected wave is in the same medium, so its speed, frequency and wavelength are unchanged.
Transmission. Part of the wave crosses into the second medium. Most boundaries transmit some energy and reflect the rest.
Refraction. The transmitted wave travels at a different speed in the new medium. If it arrives at an angle, it changes direction. Panel (b) shows why. The wavefront is slanted, so one end of it reaches the boundary first and slows down while the other end is still moving fast in the first medium. The wavefront swings round, like a line of marchers where one end steps onto mud. Entering a slower medium, the ray bends towards the normal and the wavelength shortens; entering a faster medium it bends away from the normal and the wavelength lengthens. Throughout, the frequency does not change, just as in C.2.
A ray arriving along the normal (angle of incidence zero) changes speed and wavelength but not direction.
4Snell's law and the refractive index
How far the ray bends depends on how much the speed changes. The refractive index, n, of a medium measures how much slower light is in it than in a vacuum: n = c/v, so a vacuum has n = 1 exactly, air about 1.00, water 1.33 and glass around 1.5. The data booklet gives Snell's law as:
n₁/n₂ = sin θ₂/sin θ₁ = v₂/v₁
θ₁ is the angle between the ray and the normal in medium 1, and θ₂ the same angle in medium 2. The most useful arrangement is n₁ sin θ₁ = n₂ sin θ₂: the side with the bigger n has the smaller angle.
Light passes from air (n = 1.00) into glass (n = 1.50) at 40° to the normal.
Finding n from data. In Paper 1B you may be given pairs of angles for light entering a block. Rearranged, Snell's law is sin θ₁ = n × sin θ₂ when medium 1 is air. So a graph of sin θ₁ (in air) against sin θ₂ (in the block) is a straight line through the origin whose gradient is n. Figure 3 is an invented data set drawn that way.
The gradient, taken from a large triangle, is 0.745 ÷ 0.50 = 1.49, so the block's refractive index is 1.49. Plotting sines, not angles, is what makes the line straight; a graph of θ₁ against θ₂ curves.
5Critical angle and total internal reflection
Now send light the other way: from glass into air, from the slower medium into the faster one. The ray bends away from the normal, so the refracted angle is always bigger than the incident angle. Increase the incident angle and the refracted ray swings further out until it skims along the boundary at 90°. Figure 4 shows the three stages.
The incident angle at which the refracted angle is exactly 90° is the critical angle, θc. Put θ₂ = 90° into Snell's law, with sin 90° = 1:
For any incident angle bigger than θc, Snell's law would need sin θ₂ greater than 1, which is impossible. No wave is transmitted and all of it is reflected back into the first medium. This is total internal reflection (TIR). It needs two conditions, and exam answers must state both:
- the wave travels from a medium of higher refractive index towards one of lower refractive index (slower to faster), and
- the angle of incidence is greater than the critical angle.
Optical fibres use it: light in a glass core meets the lower-index cladding above θc and reflects all the way along.
6Diffraction
Diffraction is the spreading of a wave as it passes through a gap (an aperture) or round the edge of an obstacle. Figure 5 shows the three cases the course uses.
Through a wide gap (panel a), much wider than the wavelength, most of the wave goes straight on and only the edges curve. Through a gap about one wavelength wide (panel b), the wave spreads into almost semicircular wavefronts, as if the gap were a new point source. Round an edge (panel c), the wave bends into the region that a straight-line picture says should be in shadow.
The rule is about size compared with wavelength: diffraction is significant when the gap or obstacle is similar in size to λ, and slight when it is much larger. That explains an everyday puzzle. You can hear someone talking round a corner but you cannot see them. Speech has wavelengths from tens of centimetres to a few metres, similar to a doorway, so it diffracts strongly through it. Light has wavelengths around 500 nm, a million times smaller than the doorway, so its diffraction there is negligible.
What does not change: diffraction happens in one medium, so speed, frequency and wavelength are all unchanged. On a diagram, the wavefronts after the gap must have the same spacing as those before it.
7Superposition
When two waves meet at a point, they pass through each other, and while they overlap the displacement there is the sum of the two. This is the principle of superposition: the resultant displacement at a point is the vector sum of the displacements of the individual waves at that point. Figure 6 shows it with two pulses on a rope.
Two pulses on the same side (panel a) add to a pulse twice as tall while they overlap: constructive interference. A pulse and an equal pulse on the other side (panel b) cancel to a flat rope for an instant: destructive interference. In both cases each pulse carries on afterwards exactly as it was. Waves do not bounce off each other, and neither is damaged.
In panel (b) the rope is flat for an instant but still moving, so the energy has not vanished.
8Interference from two sources
Two sources sending out waves of the same frequency create an interference pattern: fixed places where the waves always reinforce and fixed places where they always cancel. Figure 7 draws the crests and troughs from two sources S₁ and S₂.
Where a crest meets a crest, or a trough a trough, the waves are in step and interfere constructively. Where a crest meets a trough they cancel. What decides which? The path difference: the extra distance travelled by the wave from the further source. For two sources that oscillate in phase:
constructive: path difference = nλ destructive: path difference = (n + ½)λ (n = 0, 1, 2, …)
A path difference of a whole number of wavelengths puts the waves back in step; an extra half-wavelength puts them exactly out of step. The teal lines in Figure 7 join points of constant path difference nλ, and the clay lines points of (n + ½)λ.
Two loudspeakers connected to the same signal generator emit 680 Hz in phase. The speed of sound is 340 m s⁻¹.
Why the sources must be coherent. Two sources are coherent when they have the same frequency and a constant phase difference. Only then do the loud and quiet places stay put, so that a pattern can be seen or measured. Two ordinary lamps are not coherent: each emits light in short, random bursts whose phase changes millions of times a second, so the pattern shifts far too fast to observe and you see only even illumination. The fix, which Young used, is to take one source and split its light in two, so that whatever phase jumps happen, happen to both halves together. Two speakers driven by one signal generator are coherent for the same reason.
9Young's double-slit experiment
Monochromatic light, today usually a laser, falls on two narrow slits a small distance d apart. Each slit diffracts the light, so the two beams spread and overlap on a screen a distance D away, and the overlap shows a row of equally spaced bright and dark fringes. Figure 8 shows the arrangement.
The centre of the pattern is equidistant from both slits, so the path difference is zero and there is a bright fringe there, the central maximum. Moving along the screen, the path difference grows steadily; each extra λ gives the next bright fringe. For a point at angle θ, the path difference is d sin θ, and when the angle is small this is close to d × (y/D), where y is the distance from the centre. Setting it equal to nλ gives bright fringes at y = nλD/d, one every λD/d. So the fringe separation, the distance between the centres of adjacent bright fringes, is:
s = λD/d
Longer wavelength, wider fringes. Screen further away, wider fringes. Slits closer together, wider fringes.
A laser of wavelength 633 nm shines on slits 0.25 mm apart, with the screen 2.0 m away.
Run backwards, this is how the wavelength of light is measured. A student with slits 0.30 mm apart and a screen 1.50 m away measures 29.5 mm across 10 fringe spacings.
Count spacings, not bright fringes: from the 1st bright fringe to the 11th is 10 spacings.
Why it matters: no particle model of light could explain dark bands where two beams overlap, so Young's fringes, in the early 1800s, became the classic evidence that light is a wave. Over a century later electron beams produced the same kind of patterns: evidence that particles have a wavelength too.
10HLSingle-slit diffraction
SL students can skip to section 12.
Light through one narrow slit of width b does not make a single bright line. It makes a wide, bright central maximum with much fainter secondary maxima either side, separated by dark minima. Figure 9 plots the intensity.
Why is there a dark direction at all? Split the slit into a top half and a bottom half. Pair each point in the top half with the point b/2 below it in the bottom half. In the direction where those two paths differ by λ/2, every pair cancels, and the whole slit sends no light that way. That happens when (b/2) sin θ = λ/2, that is b sin θ = λ. For small angles sin θ ≈ θ in radians, which gives the data booklet form for the first minimum:
θ = λ/b (θ in radians, measured from the centre of the pattern)
So the central maximum runs from −λ/b to +λ/b: it is twice as wide as each secondary maximum, and it carries most of the light.
For light of 600 nm through a slit 0.12 mm wide, onto a screen 2.0 m away:
The effect of slit width, qualitatively. A narrower slit gives a wider central maximum (θ = λ/b grows) and a dimmer one, since less light gets through and it is spread over more angle. The dashed curve in Figure 9 is the same light through a slit half as wide. A wider slit gives a narrower, brighter central maximum. A longer wavelength, like a narrower slit, widens the pattern: red light spreads more than blue.
The single slit shapes the double-slit pattern. In Young's experiment each of the two slits has a finite width b, so each one is also a single slit, and it can only send light where its own diffraction pattern allows. The double-slit fringes therefore appear inside the single-slit pattern, as Figure 10 shows: equally spaced fringes whose brightness follows the single-slit curve, called the envelope. The fringes are said to be modulated by it.
Two consequences. The fringes are not equally bright, as the simple SL picture suggests; they fade away from the centre. And where an interference maximum falls exactly on a single-slit minimum, it is missing. In Figure 10 the slit separation is 3b: the third-order interference maximum, at sin θ = 3λ/d = λ/b, lands on the first single-slit minimum and disappears, leaving five bright fringes inside the central envelope.
11HLMultiple slits and diffraction gratings
Add more equally spaced slits and something useful happens. The bright maxima stay in exactly the same places, but they become much sharper and much brighter, with only very faint light between them. Figure 11(a) compares two slits with ten at the same spacing.
Why sharper: a little away from a maximum, two slits' waves are only slightly out of step, so the light fades gradually; with many slits, waves from slits far apart are well out of step and together cancel almost everything. Only exactly at the maxima do all the waves add. Brighter: more slits let more light through.
A diffraction grating is a plate with hundreds of slits (lines) per millimetre. The directions of its maxima are given by:
nλ = d sin θ
where d is the spacing between adjacent slits, θ is measured from the straight-through direction, and n is the order of the maximum. Gratings are usually described by lines per millimetre, so the first step is always to find d.
A grating has 300 lines per mm and is lit normally with 550 nm light.
No angle has a sine bigger than 1, so orders stop at the largest whole number below d/λ. Always round down.
White light and several wavelengths. The zero-order maximum, straight through, has zero path difference for every wavelength, so it stays white. In every other order each wavelength goes to its own angle, longer wavelengths further out, so each order becomes a spectrum, violet nearest the centre and red furthest out. Higher orders are spread wider, and they start to overlap. For the 300-line grating, Figure 11(b) shows the numbers for 400 nm to 700 nm:
| Order | Violet (400 nm) | Red (700 nm) |
|---|---|---|
| 1 | 6.9° | 12.1° |
| 2 | 13.9° | 24.8° |
| 3 | 21.1° | 39.1° |
The second-order red (24.8°) lies beyond the third-order violet (21.1°), so the second and third spectra overlap. For a mix of a few monochromatic wavelengths, such as the lines from a gas lamp, each order shows the lines separately, which is why gratings are used to measure wavelengths precisely: sharp maxima give precise angles.
12Where marks are lost
Measuring angles from the surface. Every angle in Snell's law, reflection and TIR is measured from the normal. An angle to the boundary gives the wrong sine and scores nothing.
Saying the frequency changes in refraction or diffraction. Frequency is set by the source. Refraction changes speed and wavelength; diffraction changes neither.
Giving only one condition for TIR. It needs both: the wave travels towards a lower refractive index, and the angle of incidence exceeds the critical angle.
Explaining diffraction as "the wave bends round". Say it spreads, and say when: it is significant when the gap is comparable to the wavelength. Drawn wavefronts after a gap must keep the same spacing.
Calling any two sources coherent. Same frequency is not enough; the phase difference must be constant. Two lamps, or two separate lasers, do not give a stable pattern.
Counting fringes instead of spacings. Ten bright fringes span nine spacings. And s is between adjacent bright (or adjacent dark) fringes, not from bright to dark.
Mixing up d, D and s, and their units. d is the slit separation (fractions of a millimetre), D the slit-to-screen distance (metres), s the fringe separation (millimetres). Convert everything to metres before substituting.
HL: using θ = λ/b in degrees, or rounding the highest order up. θ = λ/b gives radians. And n from d/λ = 6.06 is 6, never 7.
13Draw it right
- Wavefronts are evenly spaced one wavelength apart, and rays cross them at right angles, with arrows.
- Refraction wavefronts: closer together in the slower medium, continuous across the boundary, and the ray bent towards the normal on the slow side. Draw the normal as a dashed line.
- Reflection: equal angles either side of the normal, both measured from it.
- TIR sequence: below θc a refracted ray plus a weak reflection; at θc the refracted ray along the surface; above θc reflection only.
- Diffraction: same wavelength after the gap as before it; nearly semicircular wavefronts for a gap about λ wide, mostly straight ones for a wide gap.
- Superposition: add displacements point by point; after the overlap, draw each pulse unchanged and moving on.
- Double slit (SL): equally spaced fringes with the central maximum at zero path difference. HL: the fringes sit inside the single-slit envelope, the central envelope twice as wide as the side ones, and grating maxima narrow and tall.
14Try it
Marks in brackets. Answers and marker's notes are at the end.
Q1. Light passes from glass (n = 1.50) into water (n = 1.33). What happens to its speed, frequency and wavelength? 1 mark
A. speed increases, frequency unchanged, wavelength increases
B. speed increases, frequency increases, wavelength unchanged
C. speed decreases, frequency unchanged, wavelength decreases
D. speed unchanged, frequency decreases, wavelength increases
Q2. A ray of light in water (n = 1.33) reaches the water–air surface at 35° to the normal.
(a) Calculate the angle of refraction in the air. 2 marks
(b) Calculate the critical angle for the water–air boundary, and state what happens to a ray that reaches the surface at 55°. 2 marks
Q3. State what is meant by coherent sources, and explain why two separate filament lamps do not produce an observable interference pattern. 3 marks
Q4. Two loudspeakers connected to the same signal generator emit sound of wavelength 0.40 m. A microphone is 3.20 m from one speaker and 3.80 m from the other. Determine whether the microphone is at a point of maximum or minimum loudness. 3 marks
Q5. In a double-slit experiment with green laser light of wavelength 532 nm, a student measures the fringe separation s for several slit-to-screen distances D. Her results (invented) are:
| D / m | 0.80 | 1.20 | 1.60 | 2.00 | 2.40 |
|---|---|---|---|---|---|
| s / mm | 1.05 | 1.61 | 2.12 | 2.67 | 3.18 |
(a) Explain why she finds s by measuring across several fringes. 1 mark
(b) The gradient of her graph of s against D is 1.33 × 10⁻³. Determine the slit separation. 2 marks
(c) The green laser is replaced by a red one. State and explain the effect on the fringes. 2 marks
Q6 (HL). Light of wavelength 633 nm passes through a single slit 0.050 mm wide onto a screen 1.80 m away.
(a) Calculate the width of the central maximum on the screen. 2 marks
(b) Describe how the central maximum changes if the slit is made narrower. 2 marks
Q7 (HL). A diffraction grating has 600 lines per mm and is illuminated normally with light of wavelength 520 nm.
(a) Calculate the angle of the second-order maximum. 2 marks
(b) Determine the highest order that can be observed. 2 marks
(c) Explain why the maxima from the grating are sharper than those from a double slit of the same spacing. 2 marks
15In one breath
Wavefronts join points in phase; rays, perpendicular to them, show the energy's direction. At a boundary a wave partly reflects and partly transmits; the transmitted wave changes speed and wavelength, not frequency, bending towards the normal in a slower medium: n₁/n₂ = sin θ₂/sin θ₁ = v₂/v₁, angles from the normal. From higher to lower n, beyond the critical angle where sin θc = n₂/n₁, everything reflects: total internal reflection. Diffraction is spreading through a gap or round an edge, significant when the gap is about λ, with λ unchanged. Overlapping waves add their displacements. Coherent sources, same frequency and constant phase difference, give a fixed pattern: path difference nλ is constructive, (n + ½)λ destructive. Young's double slit gives fringes s = λD/d. HL: a single slit has its first minima at θ = λ/b and a central maximum twice the width of the others, narrower slits giving wider, dimmer patterns; that envelope modulates the double-slit fringes and can remove an order; gratings obey nλ = d sin θ with sharp, bright maxima, and white light makes overlapping spectra, violet inside, red outside.
Answers
Q1. A. Water has the lower refractive index, so light is faster in it; frequency is set by the source and does not change; with v = fλ and f fixed, the wavelength increases. C reverses the speed change. B and D change the frequency, which never happens at a boundary.
Q2. (a) n₁ sin θ₁ = n₂ sin θ₂ gives 1.33 × sin 35° = 1.00 × sin θ₂, so sin θ₂ = 0.763 and θ₂ = 49.7°. (b) sin θc = 1.00 ÷ 1.33, so θc = 48.8°. The ray at 55° exceeds the critical angle, so it is totally internally reflected back into the water. (a) M1 for Snell's law the right way round, A1 for 49.7°; (b) A1 for 48.8°, A1 for TIR because 55° > θc. 25.5° in (a) divides by 1.33 and scores 0.
Q3. Coherent sources have the same frequency and a constant phase difference. A filament lamp emits light in short random bursts, so the phase of each lamp changes randomly and very rapidly. The phase difference between two lamps is therefore not constant, the positions of constructive and destructive interference shift faster than the eye can follow, and only uniform illumination is seen. 1 for same frequency and constant phase difference, 1 for random phase changes in each lamp, 1 for the pattern changing too fast to observe. "Different colours" alone scores 0 for the explanation.
Q4. Path difference = 3.80 − 3.20 = 0.60 m. In wavelengths, 0.60 ÷ 0.40 = 1.5, so the path difference is (1 + ½)λ. The waves arrive in antiphase and interfere destructively, so the microphone is at a point of minimum loudness. M1 for the path difference, M1 for 1.5λ, A1 for minimum with the reason. "Minimum" with no working scores 0.
Q5. (a) The fringe separation is small, so measuring across many fringes and dividing by the number of spacings reduces the percentage uncertainty in s. (b) From s = λD/d, the gradient of s against D is λ/d, with s converted to metres. So d = λ ÷ gradient = 532 × 10⁻⁹ ÷ 1.33 × 10⁻³ = 4.0 × 10⁻⁴ m, that is 0.40 mm. (c) Red light has a longer wavelength, and s is proportional to λ, so the fringes become further apart. (a) 1 for reduced uncertainty; (b) M1 for gradient = λ/d, A1 for 4.0 × 10⁻⁴ m; (c) 1 for further apart, 1 for longer wavelength with s ∝ λ. An answer of 0.40 m in (b) mishandles the millimetres and scores M1 only.
Q6 (HL). (a) θ = λ ÷ b = 633 × 10⁻⁹ ÷ 0.050 × 10⁻³ = 0.0127 rad. The central maximum spans 2θ, so its width is 2 × 1.80 × 0.0127 = 0.046 m, about 46 mm. (b) It becomes wider, because θ = λ/b increases as b decreases, and dimmer, because less light passes through the slit and it is spread over a larger area. (a) M1 for θ = λ/b and doubling, A1 for 0.046 m; (b) 1 for wider with the reason, 1 for dimmer. Forgetting to double gives 23 mm and scores M1 only.
Q7 (HL). (a) d = 1 ÷ 600 mm = 1.67 × 10⁻⁶ m. sin θ = 2 × 520 × 10⁻⁹ ÷ 1.67 × 10⁻⁶ = 0.624, so θ = 38.6°. (b) n ≤ d ÷ λ = 1.67 × 10⁻⁶ ÷ 520 × 10⁻⁹ = 3.2, so the highest order is 3. (c) With many slits, a direction slightly away from a maximum puts waves from widely separated slits out of phase, and the many waves cancel almost completely; only exactly at the maxima do all of them arrive in phase. With two slits the cancellation near a maximum is only partial, so the intensity falls off gradually. (a) M1 for d in metres and substitution, A1 for 38.6°; (b) M1 for d/λ or sin θ ≤ 1, A1 for 3, not 4; (c) 1 for near-total cancellation just off a maximum, 1 for all in phase only at the maxima. "More light gets through" explains brightness, not sharpness: 0.
Educerie · written from the published IB Diploma Programme Physics guide, first assessment 2025, section C.3 Wave phenomena. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.