IB · C · extended response
Finding a hidden polariser axis
Wave behaviour · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
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.
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
Ready to self-mark?Reveal the detailed answer guide
(a)
Explain the first intensity reduction and predict the intensity when a third polariser at 45° is inserted between crossed first and last polarisers.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
An ideal polariser transmits half the average intensity of unpolarised light, giving 80/2 = 40 W m⁻².
- 2
The middle axis at 45° transmits 40cos²45° = 20 W m⁻² and defines a new output polarisation direction.
- 3
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.
(b)
Explain what the extinction supports about light and give one limitation when using this model to reduce glare.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
A transverse oscillation has a selectable direction perpendicular to propagation. A simple longitudinal wave has no equivalent independent transverse axis to filter.
- 2
The evidence supports the transverse nature of light, not every microscopic detail of a classical-wave picture.
- 3
Reflected glare may be only partially polarised, and real filters leak light and absorb other components. Measured minima therefore need not be exactly zero.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- 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.
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