A step-index optical fibre has a core of refractive index 1.52 and a cladding of refractive index 1.48.
(a) Calculate the critical angle at the core–cladding boundary. [2]
(b) Calculate the speed of light in the core. [1]
(c) Give two reasons why the fibre has a cladding. [2]
(d) Explain how modal dispersion arises and why using a narrower core reduces it. [2]
- (a) $\sin c = \dfrac{n_2}{n_1} = \dfrac{1.48}{1.52}$(1)
- (a) $c = 76.8°$ allow 77°(1)
- (b) $\dfrac{3.00\times10^{8}}{1.52} = 1.97\times10^{8}$ m s$^{-1}$(1)
- (c) any two from: protects the core from damage / scratches (which could let light escape); prevents signal crossing to adjacent fibres (keeps data secure); has a lower refractive index than the core so total internal reflection occurs at the boundary allow: increases the critical angle, reducing modal dispersion — first reason(1)
- (c) second, different reason from the list above(1)
- (d) rays entering at different angles take different paths of different lengths, so arrive at different times (pulse broadening)(1)
- (d) a narrow core allows only (nearly) one path / fewer paths, so less difference in arrival time(1)
- Answer: (a) 76.8° (b) $1.97\times10^{8}$ m s$^{-1}$ (c) protection; prevents crossover (d) different path lengths; narrow core → fewer paths
- Examiner tip: For the critical angle at a core–cladding boundary, sin c = n₂/n₁ — not 1/n₁, which only applies to an air boundary.