
Let n1 and n2 be the refractive indices of the air and optical fibre respectively. Applying Snell's law at Q, we get:
n1sinθ1=n2sinθ2
sin40∘=1.31×sinθ2
⇒sinθ2=1.310.64=0.4885≈0.5
⇒θ2=30∘
The distance covered along the length of the optical fibre after one reflection:
QN=RNcotθ2
=20×10−6×cot30∘=203×10−6m
The reflections take place throughout the length of the optical fibre.
Length of the wire=2m
Total number of reflections along its length=distancecoveredinonereflectionalongthelengthLengthofthewire
=2032×106=3100000=57735.02≈57000