Optical Fibres
4 marks · NormalA laser sends light into an optical fibre at an angle of incidence of 72° to the normal. The critical angle for the core–cladding boundary is 60°.
Explain whether the light will be totally internally reflected at the core–cladding boundary.
If the refractive index of the core is 1.50, calculate the maximum refractive index of the cladding for total internal reflection to occur at this boundary. (Give your answer to 2 decimal places.)
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) Yes; the angle of incidence is greater than the critical angle, so total internal reflection occurs. · must show comparison with critical angle
(b) 1.30 · must show working shown
Must show: 72° > 60°, so TIR occurs; n_core × sin c = n_cladding × sin 90°; calculations correct
Common mistakes: Using wrong formula for refractive indices.
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Compares angle of incidence with critical angle.
A1 (a) · 1 mark · States that TIR will occur because 72° > 60°.
M1 (b) · 1 mark · Uses formula: n_core × sin c = n_cladding × sin 90° (or rearranges correctly).
A1 (b) · 1 mark · Final answer for n_cladding to 2 decimal places: 1.30 (allow 1.295–1.305 for rounding).
Worked solution
a) Since the angle of incidence (72°) is greater than the critical angle (60°), total internal reflection will occur. b) For total internal reflection: n_core × sin c = n_cladding × sin 90° So: n_core × sin 60° = n_cladding × 1 n_cladding = n_core × sin 60° = 1.50 × 0.866 = 1.299 (rounded to 1.30).