MYP Physics · topic MYP-11.03
Total internal reflection
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
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. 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.