Question

The core/cladding index difference of a single-mode optical fiber cable is 0.01. The refractive index of the material of the core is 1.5. The maximum angle of acceptance of the fiber is approximately equal to

• A. ${8.6}^{\circ }$
• B. ${2.0}^{\circ }$
• C. ${12.1}^{\circ }$
• D. ${17.5}^{\circ }$

Answer 1

The correct option is C ${12.1}^{\circ }$

$\Delta =0.01;n_1=1.5$

N.A. $=n_1\sqrt{2\Delta }$
$=1.5\sqrt{2\times 0.01}=1.5\sqrt{0.02}=0.212$

$\sin {\theta }_a=\text{N.A.}$
${\theta }_a={\sin }^{-1}\left(\text{N.A.}\right)={\sin }^{-1}\left(0.212\right)$
$\simeq {12.1}^{\circ }$