DP Physics (last assessment 2024)

Question 18N.3.SL.TZ0.b.i

Date	November 2018	Marks available	[Maximum mark: 2]	Reference code	18N.3.SL.TZ0.b.i
Level	SL	Paper	3	Time zone	TZ0
Command term	Show that	Question number	b.i	Adapted from	N/A

b.i.

[Maximum mark: 2]

18N.3.SL.TZ0.b.i

The diagram shows the longest and shortest paths that a ray can follow inside the fibre.

For the longest path the rays are incident at the core–cladding boundary at an angle just slightly greater than the critical angle. The optic fibre has a length of 12 km.

Show that the longest path is 66 m longer than the shortest path.

[2]

Markscheme

long path is $\frac{{12 \times {{10}^3}}}{{{\text{sin}}\,84^\circ }}$ ✔

= 12066 «m» ✔

«so 66 m longer»

Syllabus sections

Option C: Imaging

Option C: Imaging » Option C: Imaging (Core topics) » C.3 – Fibre optics

Option C: Imaging » Option C: Imaging (Core topics)

Question 18N.3.SL.TZ0.b.i

Select a Test

Syllabus sections

Confirm action