IB · C · extended response
Testing a fibre from refraction data
Wave behaviour · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
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.
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
Ready to self-mark?Reveal the detailed answer guide
(a)
Find refractive index, speed in the material and critical angle. Predict what happens at the 45° internal incidence.
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
Snell's law gives nair sin i = nmaterial sin r. The graph gradient sin i/sin r is 1.50.
- 2
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°.
- 3
The internal angle 45° exceeds θc and light travels from higher to lower index, meeting both conditions for total internal reflection.
(b)
Explain which wave quantities change on entry and evaluate whether all rays in a fibre necessarily remain trapped.
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
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.
- 2
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.
- 3
Real fibres have cladding, losses and bending limits. A tight bend can change incidence angles and allow previously guided light to escape.
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.
- 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.
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