Answer to Question #155327 in Optics for akash

Question

An optical fibre is made of glass with a refractive index of 1.55 and is clad

with another glass with a refractive index of 1.51. The fibre has a core

diameter of 50µm and is used at a light wavelength of 0.8 µm.

(a) What NA does the fibre have?

(b) What is the acceptance angle?

Answer 1

(a) By the definition of the numerical apperture, we have:

$$NA=\sqrt{(n_1^2-n_2^2)}=\sqrt{(1.55^2-1.51^2)}=0.35.$$

(b) By the definition of the acceptance angle. we have:

$$\theta_a=sin^{-1}(NA)=sin^{-1}(0.35)=20.5^{\circ}.$$

Answer:

(a) $NA=0.35.$

(b) $\theta_a=20.5^{\circ}.$