Answer to Question #182245 in Optics for Abhishek Haste

Question

A step index fiber is made with a core of refractive index 1.36, a diameter of 49pm and a fractional refractive index change of 0.025. it is operated at a wavelength of 1.3. Then the numerical aperture of the cable is

Answer 1

The refractive index is inversely proportional to the wavelength.

Let the refractive index of cladding be n' .

Hence, we know

$\Delta n = \dfrac{n-n'}{n}$

$0.025 = \dfrac{1.36-n'}{1.36}$

$n' = 1.326$ = Refractive index of cladding.

We know,

Numerical aperture, $N_a = \sqrt{(n_1^2-n'^2)}$

$n_1 = 1.36$

$n' = 1.326$

$N_a = \sqrt{{1.36^2-1.326^2}}$

$N_a = 0.302$