C.3 Wave phenomena
- Syllabus
- First assessment 2025
- Topic
- —
- Level
- SL
Define a wavefront
A wavefront is a line or surface joining points that are in phase. Adjacent wavefronts are separated by one wavelength. In a uniform medium, the wavefronts are perpendicular to the direction of propagation.
Use rays to show propagation
A ray is a line showing the direction in which the wave transfers energy. Draw rays perpendicular to the local wavefronts; straight, parallel wavefronts give parallel rays, while circular wavefronts from a point source give radial rays.
Read the geometry
When a wavefront diagram changes direction at a boundary, compare the ray direction and the spacing of wavefronts on each side. The wavefront construction helps distinguish a change in speed from a change in frequency.
Common trap
Do not draw rays parallel to wavefronts. A ray follows energy propagation and is normal to the wavefront at each point.
The packet contains one directly relevant ray-diagram item and one unrelated polarization item. Use the ray evidence for diagram construction; no reliable frequency claim is made for the unrelated item.
Sketch / Label
Define wavefronts as in-phase lines or surfaces and draw rays perpendicular to them in the propagation direction. For a diagram question, label the image or ray construction only after identifying the correct normal direction.
Drawing rays along wavefronts instead of perpendicular to them, or treating the mixed polarization evidence as a wavefront question.
Representative question
Sketch two appropriate rays on the diagram to show the formation of the image. Label the image with the letter I.
ii
one correct ray
second correct ray that allows the image to be located
image drawn
Write the boundary relationship
For a ray crossing from medium 1 to medium 2, Snell’s law is n1sinθ1=n2sinθ2. Each angle is measured between the ray and the normal, not between the ray and the surface.
Solve the geometry first
Draw or identify the normal at the boundary, label incident and refracted angles, then substitute the refractive indices and sines. If the question asks for speed, combine the result with n=c/v.
Check the bend
Entering a higher-index medium decreases speed and bends the ray toward the normal. Entering a lower-index medium increases speed and bends it away from the normal; the frequency remains fixed at a stationary boundary.
Common trap
Do not use the angle between a wavefront and the normal as if it were the ray angle. A wavefront is perpendicular to the ray, so convert the angle when the diagram labels wavefronts.
Questions ask you to calculate light speed in water or select the correct refractive-index expression from a wavefront diagram. The evidence rewards the normal-angle convention and correct sine ratio.
Calculate / What is
Measure each angle from the normal, write n1 sinθ1 = n2 sinθ2, and substitute the correct refractive indices. If speed is requested, use n = c/v and keep significant figures appropriate.
Using angles measured from the surface, or reversing n1 and n2 when applying Snell’s law.
Representative question
Calculate the speed of light in the water. State the answer to an appropriate number of significant figures.
vwater =2.0×108×sin55∘sin71∘2.3×108<m s−1> Any answer to 2 s.f.
Use of Snell's Law for MP1
Award [3] for BCA
Define refractive index
The refractive index of a medium is n=c/v, where c is the speed of light in vacuum and v is its speed in the medium. A larger n means a lower light speed in that medium.
Relate speed and wavelength
At a stationary boundary the frequency is unchanged. Since v=fλ, a lower speed means a shorter wavelength. For two media, n2/n1=λ1/λ2 when the frequency is common.
Use a graph or measurement
If a graph’s gradient represents n, state that interpretation before reading the value. For uncertainty, use the spread from suitable maximum and minimum lines or the specified uncertainty method.
Common trap
Do not use n=v/c, and do not assume wavelength stays fixed when light enters a different medium. Frequency is the quantity that remains fixed at a stationary boundary.
Questions ask for a refractive index from a graph or compare refractive indices using wavelengths in two media. The evidence rewards the inverse speed relationship and correct wavelength ratio.
Determine
Use n = c/v, or compare wavelengths through n2/n1 = λ1/λ2 when frequency is unchanged. If a graph is used, identify what its gradient represents and report the refractive index with appropriate absolute uncertainty.
Using the speed ratio in the wrong direction or reporting a graph gradient without explaining that it represents n.
Representative question
Determine the value of the refractive index of the glass with its absolute uncertainty.
States gradient gives the value of the refractive index
OR
n=1.5n=1.5±0.2
MP1 can be shown as an equation.
Candidates may calculate the uncertainty by using the gradient of the line found in cii) or finding the average of the max and min lines of best fit.
Look for working leading to 0.1≤Δn≤0.2.
Check the two conditions
Total internal reflection can occur only when a wave travels from a higher-index medium to a lower-index medium, and the incidence angle is greater than the critical angle. At the critical angle, the refracted ray travels along the boundary: θ2=90∘.
Calculate the critical angle
From Snell’s law, sinθc=n2/n1 for n1>n2. For a dense medium to air, n2≈1, so sinθc=1/n1.
Use the boundary picture
For incidence below θc, there is a refracted ray. At θc, it grazes the boundary. Above θc, no refracted ray propagates into the lower-index medium and all the light is reflected back into the denser medium.
Common trap
Do not use the critical-angle equation when light travels from lower to higher refractive index, and do not measure the critical angle from the surface rather than the normal.
Questions ask you to calculate a critical angle or infer a medium’s light speed from θc. The evidence rewards the Snell’s-law boundary condition and correct inverse-sine calculation.
Calculate / What is
Confirm that the ray goes from higher n1 to lower n2, set the refracted angle to 90° at the threshold, and use sin θc = n2/n1. For a dense medium to air, use sin θc = 1/n1 and check that incidence above θc gives total internal reflection.
Using n1/n2 instead of n2/n1 in sin θc, or applying total internal reflection when the ray travels into the higher-index medium.
Representative question
Calculate the critical angle for the plastic-water interface.
sinrsini=1.601.33 and sinr=1i=⋖sin−10.831»=56<∘≫
Accept 0.98 rad (unit required)
Add overlapping displacements
When waves overlap, the resultant displacement at a point is the algebraic sum of the individual displacements: yresultant=y1+y2. The waves then continue propagating after the overlap.
Keep the signs
Displacements on the same side of equilibrium add; opposite displacements partially or completely cancel. Equal opposite pulses can produce zero displacement at an instant without destroying either wave.
Connect to interference
Repeated superposition of coherent waves can create stable maxima and minima. A diffraction pattern extending beyond a geometrical shadow is evidence that wave overlap and interference are involved.
Common trap
Do not add amplitudes as positive magnitudes only, and do not treat destructive interference as permanent disappearance of the waves.
Questions ask for a resultant displacement or explain why a diffraction pattern supports the wave model. The evidence rewards signed addition and an explicit link between maxima/minima and interference.
Explain / What is
Add the signed displacements at the specified point and time. Use constructive addition for same-sign displacements and cancellation for opposite-sign displacements; then connect maxima/minima to interference or diffraction evidence.
Adding amplitudes as magnitudes and ignoring the sign of each displacement at the stated time.
Representative question
Early theories of light suggest that a geometrical shadow of the slit will be observed on the screen. Explain how the diffraction pattern formed on the screen provides evidence for the wave theory of light.
observed pattern goes beyond the rectangular shape/geometrical shadow OR observed pattern shows maxima/minima
«this is explained by» interference/superposition of waves
Marking guidance:
Accept any correct description of the diffraction pattern for MP1.
Define coherence
Two waves are coherent if they have the same frequency and a constant phase difference. The phase relationship does not drift with time.
Connect coherence to a pattern
When coherent waves overlap, the locations of constructive and destructive interference remain fixed, producing a stable interference pattern. An ordinary pair of independent light sources usually has a changing phase relationship and does not produce a stable pattern.
Use one source when needed
A single source split into two paths can provide a common frequency and phase relationship. The resulting secondary sources can then act coherently for a double-source interference experiment.
Common trap
Same frequency alone is not enough. The phase difference must also remain constant; otherwise bright and dark locations move or wash out over time.
Questions ask why two sources need to be coherent. The evidence repeatedly rewards constant phase difference and the resulting fixed pattern, often with a single source split into two paths.
Explain
State both conditions for coherence: same frequency and constant phase difference. Then link the fixed phase relationship to a stable bright/dark interference pattern, and explain why independent sources usually fail.
Mentioning only equal frequency and omitting the requirement that phase difference remains constant.
Representative question
Explain why the two sources need to be coherent for the interference pattern to be observed.
Light comes from a single source
Waves need to have a constant phase difference / in phase
«To produce» a fixed/stable/clear/constant pattern «over time»
OR
Only coherent light has this property/produces this pattern
Define path difference
Path difference is the difference between the distances travelled by two waves from their sources to the same observation point. For in-phase coherent sources, it determines whether the waves arrive in phase or out of phase.
Apply the conditions
Constructive interference occurs when path difference =nλ. Destructive interference occurs when path difference =(n+21)λ, where n is a whole number.
Count fringes carefully
Start from the central bright fringe when the path difference is zero. Each additional bright fringe changes the path difference by λ; dark fringes lie halfway between adjacent bright conditions.
Common trap
Do not assign λ/2 to every dark point. The first dark condition is λ/2, then 3λ/2, 5λ/2, and so on.
Questions ask for path difference at a dark fringe after a stated number of bright or dark fringes. The evidence rewards counting from the central maximum and choosing the half-integer condition for darkness.
What is
Identify the central bright condition as zero path difference, count the bright/dark fringes between the reference and target points, and apply nλ for bright or (n + 1/2)λ for dark.
Choosing an integer multiple of λ for a dark fringe or losing the extra half-cycle when counting intervening fringes.
Representative question
In a double-slit experiment using coherent light of wavelength λ, the central bright fringe is observed on a screen at point P. A point of destructive interference occurs at point Q. Only one point of constructive interference is observed between P and Q.
What is the path difference at Q ?
2λ
λ
23λ
2λ
C
Set up two-source interference
Two coherent sources emit waves with the same frequency and a constant phase difference. At each observation point, compare the two source-to-point distances to find the path difference.
Map bright and dark regions
For in-phase sources, path difference nλ gives constructive interference and a bright or high-amplitude region. Path difference (n+21)λ gives destructive interference and a dark or low-amplitude region.
Read the pattern
Points equidistant from the two sources have zero path difference and form a central constructive line. Further maxima and minima occur where the path difference changes by half-wavelength steps.
Common trap
Do not use source-to-source separation as the path difference. It is the difference between the two travel distances to the same observation point.
Questions ask for a possible wavelength from a sound minimum or ask you to explain a bright/dark screen pattern. The evidence rewards identifying the phase at arrival and applying the correct half- or whole-wavelength condition.
What is / Explain
Find the two source-to-point distances and subtract them to obtain path difference. For in-phase coherent sources, use nλ for constructive interference and (n + 1/2)λ for destructive interference; explain the observed bright/dark pattern.
Using the path length to one source instead of subtracting the two paths, or assigning a bright fringe to a half-integer path difference.
Representative question
Two loudspeakers are driven in phase and emit sound of the same frequency. A minimum intensity of sound is detected at point P.
P is 4.0 m from one loudspeaker and 4.6 m from the other.
What is a possible wavelength of the sound?
20 cm
30 cm
40 cm
60 cm
C
Relate fringe spacing to the apparatus
For Young’s double-slit interference, fringe separation is s=λD/d, where λ is wavelength, D is slit-to-screen distance and d is slit separation.
Rearrange before substituting
Use λ=sd/D, d=λD/s, or D=sd/λ as needed. Measure the separation between adjacent bright or dark fringe centres; if several fringes are measured, divide the total width by the number of intervals.
Check the trends
Fringes spread farther apart when wavelength or screen distance increases, and become closer when slit separation increases. The small-angle model assumes D≫d and approximately plane wavefronts normal to the slits.
Common trap
Do not use the total width across several fringes as s without dividing by the number of fringe spacings, and do not confuse slit separation d with screen distance D.
Questions ask you to calculate wavelength from a measured pattern or identify which colour gives the largest fringe separation. The evidence rewards λ = sd/D and the correct wavelength trend.
Calculate / What is
Use s = λD/d, rearrange to the requested variable, convert units, and divide a measured multi-fringe width by its number of intervals. Check that longer wavelength gives larger fringe separation.
Using total pattern width as one fringe spacing or reversing d and D in λ = sd/D.
Representative question
Calculate, in nm,λ.
s=0.15/8=0.0188 m.
Use λ=ds/D.
λ=450 nm.
A boundary can split wave energy
When a travelling wave reaches a boundary, part of its energy may be reflected back into the first medium and part may be transmitted into the second. The transmitted wave is refracted when its speed changes and it crosses the boundary at a non-zero angle.
| Behaviour | What happens | Direction rule |
|---|---|---|
| Reflection | wave remains in medium 1 | angle of reflection equals angle of incidence |
| Transmission | wave enters medium 2 | continues across the boundary |
| Refraction | transmitted wave changes direction because speed changes | toward the normal if speed decreases; away if speed increases |
Read rays and wavefronts together
Angles are measured from the normal. Rays show energy-transfer direction and remain perpendicular to wavefronts. Frequency stays fixed at a stationary boundary; a change in speed therefore changes wavelength and wavefront spacing.
Boundary cases
At normal incidence the transmitted ray does not bend, although its speed and wavelength may change. Unless absorption is stated, reflected and transmitted energy together account for the incident energy; their amplitudes do not generally add directly.
Questions ask for a missing reflected wavelength or the minimum film thickness for constructive reflection. The evidence rewards including refractive index and the correct phase-shift condition.
Determine / What is
Draw or identify the top and bottom reflected rays, calculate the optical path contribution 2nt at normal incidence, and count phase reversals before selecting the constructive condition. For the air–film–air minimum-thickness case, use t = λ/(4n).
Using 2t instead of 2nt, or applying the air–film–air quarter-wave result without checking the phase changes at the two surfaces.
Representative question
The refractive index of the coating is 1.63 and the refractive index of the glass is 1.52 .
The thickness of the coating is 143 nm .
Determine the wavelength, in nm , that is missing in the light reflected to the girl assuming that the light is incident normally on the window.
Use of 2dn=mλ
Use of n=1.63
470 «nm»
Diffraction is wave spreading
A wave diffracts when its wavefront bends into the region behind an obstacle or spreads after passing through an aperture. The wave remains in the same medium, so diffraction itself does not require a change of speed or frequency.
Compare wavelength with the opening or body
Spreading is most noticeable when the aperture width or obstacle size is comparable to the wavelength. An aperture much wider than the wavelength gives a broad central region that travels nearly straight with limited edge spreading; narrowing the aperture increases the angular spread.
Wavefront–ray representation
Before a straight aperture, incident wavefronts are parallel. Beyond a narrow aperture, draw curved outgoing wavefronts; rays stay perpendicular to them and fan outward. Around a body, wavefronts curve into the geometrical shadow.
Do not confuse mechanisms
Refraction is a direction change caused by a speed change at a boundary. Diffraction is spreading caused by an edge or aperture, even when the medium is unchanged.
Questions ask you to state the criterion or choose an aperture/wavelength change that resolves two sources. The evidence rewards the exact central-maximum/first-minimum relationship and the violet-light choice.
State / Which change
State the Rayleigh criterion using the central maximum and first minimum, then identify the change that improves angular resolution. A shorter wavelength or larger aperture reduces the minimum resolvable separation.
Claiming that longer-wavelength red light improves resolution or misquoting the two-pattern condition.
Representative question
State the Rayleigh criterion for resolution.
central maximum of one diffraction pattern lies over the central/first minimum of the other diffraction pattern
C.3 Wave phenomena is secure when you can connect the physical picture to the equation and its limits.