IB Diploma Physics · topic C.3
Wave phenomena
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
Testing a fibre from refraction data
A transparent material is surrounded by air of refractive index 1.00. Refraction measurements for light entering from air are supplied as sine values. Take c = 3.0×10⁸ m s⁻¹. A ray inside the material subsequently reaches a material–air boundary at 45° to the normal. Supplied data (sin(refracted angle); sin(incident angle)): (0, 0), (0.2, 0.3), (0.4, 0.6), (0.6, 0.9). (a) Find refractive index, speed in the material and critical angle. Predict what happens at the 45° internal incidence. [3 marks] (b) Explain which wave quantities change on entry and evaluate whether all rays in a fibre necessarily remain trapped. [3 marks]
Air-to-material refraction
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): sin(incident angle)
Horizontal axis (x): sin(refracted angle)
Read the exact data as a table
| sin(refracted angle) | sin(incident angle) |
|---|---|
| 0 | 0 |
| 0.2 | 0.3 |
| 0.4 | 0.6 |
| 0.6 | 0.9 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find refractive index, speed in the material and critical angle. Predict what happens at the 45° internal incidence.
Reveal the worked solution
- Snell's law gives nair sin i = nmaterial sin r. The graph gradient sin i/sin r is 1.50.
- Wave speed in the material is c/n = 2.0×10⁸ m s⁻¹. The critical angle obeys sin θc = 1/1.50, giving about 41.8°.
- The internal angle 45° exceeds θc and light travels from higher to lower index, meeting both conditions for total internal reflection.
- The boundary must transmit the same oscillation frequency. Since v = fλ and speed drops by a factor of 1.50, wavelength also drops by that factor.
- A ray arriving below the critical angle can refract into the surrounding lower-index material, so a transparent tube does not trap all entry directions.
- Real fibres have cladding, losses and bending limits. A tight bend can change incidence angles and allow previously guided light to escape.
Answer (a) n = 1.50; speed = 2.0×10⁸ m s⁻¹; critical angle ≈41.8°; total internal reflection occurs at 45°. (b) Frequency stays fixed; speed and wavelength fall on entry; only rays meeting the total-internal-reflection condition are trapped in the ideal model.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A fringe investigation that tests two hypotheses
Two coherent slits of separation 0.30 mm illuminate a screen. For each screen distance D, the separation between the centres of the first and eleventh bright fringes is measured and divided by ten. These mean fringe-spacing data are supplied. The small-angle condition applies. Supplied model: Use fringe spacing w = λD/d under the small-angle approximation. Supplied data (D / m; Mean fringe spacing / mm): (0.5, 1), (1, 2), (1.5, 3), (2, 4). (a) Infer the light wavelength and predict fringe spacing if slit separation is doubled at D = 1.5 m. [3 marks] (b) Explain how the method reduces relative reading uncertainty, and design a check that the pattern is interference rather than a screen ruler artefact. [3 marks]
Fringe spacing versus screen distance
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Mean fringe spacing / mm
Horizontal axis (x): D / m
Read the exact data as a table
| D / m | Mean fringe spacing / mm |
|---|---|
| 0.5 | 1 |
| 1 | 2 |
| 1.5 | 3 |
| 2 | 4 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Infer the light wavelength and predict fringe spacing if slit separation is doubled at D = 1.5 m.
Reveal the worked solution
- For two-slit interference, w = λD/d. The slope w/D is 2.0 mm m⁻¹ = 0.0020.
- λ = slope×d = 0.0020(0.00030) = 6.0×10⁻⁷ m.
- Doubling d halves w at fixed λ and D. At 1.5 m, the original 3.0 mm spacing becomes 1.5 mm.
- First-to-eleventh fringe centres contain ten intervals, not eleven. A 30 mm span is less sensitive to a fixed endpoint error than one 3 mm interval.
- Keep wavelength and slit separation fixed while changing D; proportional spacing supports the wave model. Then change d while keeping D fixed and test the inverse relationship.
- Record repeat endpoint readings, keep the screen perpendicular to the central ray and shield stray light. A ruler artefact would not track both controlled physical changes.
Answer (a) Wavelength = 600 nm; doubled-separation fringe spacing = 1.5 mm. (b) Measure a long span over many intervals and vary a physical input with a quantitative prediction.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A diffraction grating with an impossible order
Monochromatic light falls normally on a grating with 500 lines per millimetre. Measured sine of the diffraction angle against order is supplied below. Use c = 3.0×10⁸ m s⁻¹. Angles are measured from the central direction, not from the grating surface. Supplied model: Use d sinθ = nλ and spacing d = 1/(lines per unit length). Supplied data (Order n; sin θ): (0, 0), (1, 0.3), (2, 0.6), (3, 0.9). (a) Find grating spacing, wavelength, frequency and the largest possible order. [3 marks] (b) Explain how to distinguish an aperture's diffraction envelope from the grating's narrow interference maxima, and improve an angle measurement. [3 marks]
Diffraction orders
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): sin θ
Horizontal axis (x): Order n
Read the exact data as a table
| Order n | sin θ |
|---|---|
| 0 | 0 |
| 1 | 0.3 |
| 2 | 0.6 |
| 3 | 0.9 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find grating spacing, wavelength, frequency and the largest possible order.
Reveal the worked solution
- 500 lines mm⁻¹ is 5.0×10⁵ lines m⁻¹, so d = 1/(5.0×10⁵) = 2.0×10⁻⁶ m.
- From d sinθ = nλ, λ/d is the graph gradient, 0.30; therefore λ = 6.0×10⁻⁷ m and f = c/λ = 5.0×10¹⁴ Hz.
- The condition sinθ≤1 requires n≤d/λ = 3.33. Integer orders stop at n = 3; a fourth order would require sinθ = 1.2.
- The grating equation predicts directions of principal maxima, while finite slit width determines their broad intensity envelope.
- Measure both +n and −n peaks; half their angular separation reduces an offset in the estimated central direction.
- Repeat and use higher visible orders where practical, while avoiding weak peaks and confusion between order number and wavelength. Never view a laser directly.
Answer (a) d = 2.0 μm; λ = 600 nm; f = 5.0×10¹⁴ Hz; maximum order = 3. (b) Diffraction spreads each slit contribution; interference between the many coherent slit contributions selects the narrow orders.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Finding a hidden polariser axis
Unpolarised light of intensity 80 W m⁻² passes through an ideal first polariser and then an ideal analyser. The analyser data are supplied for angles relative to the first axis. Use I = I0 cos²θ for linearly polarised light at an analyser. Supplied data (Relative angle / °; Transmitted intensity / W m⁻²): (0, 40), (30, 30), (45, 20), (60, 10), (90, 0). (a) Explain the first intensity reduction and predict the intensity when a third polariser at 45° is inserted between crossed first and last polarisers. [3 marks] (b) Explain what the extinction supports about light and give one limitation when using this model to reduce glare. [3 marks]
Analyser transmission
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Transmitted intensity / W m⁻²
Horizontal axis (x): Relative angle / °
Read the exact data as a table
| Relative angle / ° | Transmitted intensity / W m⁻² |
|---|---|
| 0 | 40 |
| 30 | 30 |
| 45 | 20 |
| 60 | 10 |
| 90 | 0 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Explain the first intensity reduction and predict the intensity when a third polariser at 45° is inserted between crossed first and last polarisers.
Reveal the worked solution
- An ideal polariser transmits half the average intensity of unpolarised light, giving 80/2 = 40 W m⁻².
- The middle axis at 45° transmits 40cos²45° = 20 W m⁻² and defines a new output polarisation direction.
- The last axis is another 45° from the middle, so it transmits 20cos²45° = 10 W m⁻². Without the middle element, the crossed pair transmits zero in the ideal model.
- A transverse oscillation has a selectable direction perpendicular to propagation. A simple longitudinal wave has no equivalent independent transverse axis to filter.
- The evidence supports the transverse nature of light, not every microscopic detail of a classical-wave picture.
- Reflected glare may be only partially polarised, and real filters leak light and absorb other components. Measured minima therefore need not be exactly zero.
Answer (a) First output = 40 W m⁻²; with the middle polariser, final output = 10 W m⁻². (b) Polarisation supports transverse electric-field oscillations; real glare and real filters need not reach complete extinction.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.