Question

The core/cladding index difference of a single mode step index optical fibre cable is 0.04. The refractive index of the material of the core is 1.40. The maximum angle of aaceptance of the fibre is approximately equal to

• A. ${24.6}^o$
• B. ${23.3}^o$
• C. ${16.4}^o$
• D. ${18.2}^o$

Answer 1

The correct option is B ${23.3}^o$

$\Delta =0.04$

$n_1=1.40$

$NA=n_1\sqrt{2\Delta }=1.40\sqrt{2\times 0.04}=0.39$

$sin{\theta }_a=NA$

${\theta }_a=si{n}^{-1}\left(NA\right)$

$=si{n}^{-1}\left(0.39\right)={23.3}^o$