An optical fiber has an acceptance angle of 30° and core refractive index of [value not printed in the source paper] calculate refractive index of cladding?

Note: The core index is missing from the printed paper. As the paper allows, we assume \(n_1=1.5\) (a typical glass core) and take the fibre to be in air (\(n_0=1\)).

Given: \(\theta_a=30^\circ\), \(n_1=1.5\), \(n_0=1\).

Formula: \(NA=n_0\sin\theta_a=\sqrt{n_1^2-n_2^2}\)

\(NA=\sin30^\circ=0.5\)

\(n_2^2=n_1^2-NA^2=2.25-0.25=2.00\Rightarrow n_2=\sqrt2\)

Refractive index of cladding \(n_2\approx1.414\) (for \(n_1=1.5\)). In general, \(n_2=\sqrt{n_1^2-0.25}\).

Check: \(\sqrt{1.5^2-1.414^2}=\sqrt{2.25-2.0}=0.5=\sin30^\circ\). The result is consistent.