| Date | May 2013 | Marks available | 2 | Reference code | 13M.3.SL.TZ1.19 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 1 |
| Command term | Sketch | Question number | 19 | Adapted from | N/A |
Question
This question is about an optic fibre.
Monochromatic light enters an optic fibre, from air, along direction A that is at an angle θ to the axis of the fibre.
The refractive index of the core is 1.62 and the refractive index of the cladding is 1.52. The critical angle at the core-cladding boundary is 70º .
Calculate the greatest angle of incidence θ that can be used with this fibre.
Sketch the path of the light in the core on the diagram above.
Markscheme
refracted angle inside core = 20º;
sin θ= sin 20º x 1.62;
34º;
refraction angle on entering core sensible and smaller than incidence angle; equal angles of reflection at cladding; (judge by eye)
Examiners report
There were very few correct answers although this is fairly basic geometrical optics. Some attempted to use the equation n1 sinθ1 = n2 sin θ2 for refraction at the core/cladding boundary.
Very few answers showed refraction on entry, but many showed total internal reflection correctly.
