MYP Physics · topic MYP-12.04
Polarisation as transverse-wave evidence
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.
Finding a hidden polariser axis
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. 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.