The refractive indices of the core and cladding of a step index fibre are 1.50 and 1.46 respectively. Calculate (1) the numerical aperture, (2) the acceptance angle, (3) the critical angle at the core–cladding interface and (4) the fractional refract…

Answer: (1) NA \(=0.344\); (2) \(\theta_a=20.13^\circ\); (3) \(\theta_c=76.74^\circ\); (4) \(\Delta=0.0267\).

Given \(\mu_1=1.50\), \(\mu_2=1.46\), fibre in air.

(1) \(NA=\sqrt{\mu_1^2-\mu_2^2}=\sqrt{2.25-2.1316}=\sqrt{0.1184}=0.344\)

(2) \(\theta_a=\sin^{-1}(NA)=\sin^{-1}(0.344)=20.13^\circ\) (full acceptance cone \(=40.26^\circ\)).

(3) \(\theta_c=\sin^{-1}\dfrac{\mu_2}{\mu_1}=\sin^{-1}\dfrac{1.46}{1.50}=\sin^{-1}(0.9733)=76.74^\circ\)

(4) \(\Delta=\dfrac{\mu_1-\mu_2}{\mu_1}=\dfrac{0.04}{1.50}=0.0267\) (2.67%).

Check: \(\mu_1\sqrt{2\Delta}=1.50\sqrt{0.0533}=0.346\approx NA\). ✓