Question

A silicon optical fibre with a core diameter large enough has a core refractive index of 1.50 and a cladding refractive index 1.47. Determine
(i) the critical angle at the core cladding interface,
(ii) the numerical aperture for the fibre
(iii) the acceptance angle in air for the fibre.

Answer 1

Here,${\mu }_1=1.50;{\mu }_2=1.47;$
${\mu }_0=1$
(i) Critical angle ${\theta }_c$ at the core-cladding interface is given by
${\theta }_c={sin}^{-1}\frac{1.47}{1.50}=78.5^0$
(ii) Numerical aperture,$NA={({{\mu }_1}^2-{{\mu }_2}^2)}^{\frac{1}{2}}$
$={[{\left(1.50\right)}^2-{\left(1.47\right)}^2]}^{\frac{1}{2}}={(2.25-2.16)}^{1/2}=0.30$
(iii) Acceptance angle ${\theta }_a={sin}^{-1}\left(NA\right)$
$={sin}^{-1}\left(0.30\right)={17.4}^0$