Answer to Question #113622 in Optics for Rohith

Question

The numerical aperture of an optical fiber is 0.39. If the difference in the refractive indices of the material of its core and cladding is 0.05, calculate the refractive index of material of the core.

Answer 1

As per the given question,

The numerical aperture of the optical fiber =0.39

The difference in the refractive index of the core and cladding =0.05

refractive index of the material of the core=?

We know that numerical aperture $(NA)=n_i\sqrt{2\Delta}$

$\Rightarrow n_i=\dfrac{NA}{\sqrt{2\Delta}}=\dfrac{0.39}{\sqrt{2\times0.05}}$

$\Rightarrow n_i=\dfrac{0.39}{0.32}=1.218$