DP Physics (last assessment 2024)

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

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

a.i.

[Maximum mark: 2]

18N.3.SL.TZ0.a.i

An optic fibre consists of a glass core of refractive index 1.52 surrounded by cladding of refractive index n. The critical angle at the glass–cladding boundary is 84°.

Calculate n.

[2]

Markscheme

« ${\text{sin}}\,{\theta _{\text{c}}} = \frac{{{n_1}}}{{{n_2}}}$ » n₁ = 1.52 × sin 84.0° ✔

n₁ = 1.51 ✔

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.a.i

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