Answer to Question #109868 in Optics for Noshad

Question

The optical fiber has a ratio of core to cladding is 1.02 and refractive index of core is 1.50. Determine the i) critical angle, ii) Numerical aperture, iii) acceptance angle and iv) fractional index change.

Answer 1

As per the given question,

The ratio in the refractive index of the core to cladding =1.02

The refractive index of the core is $(\mu_1)=1.5$

Let the refractive index of the cladding is $(\mu_2)$

hence,

$\dfrac{\mu_1}{\mu_2}=1.02$

$\mu_2=\dfrac{\mu_1}{1.02}=\dfrac{1.5}{1.02}=1.47$

a) Critical angle $C=\sin^{-1}({ \frac{\mu2}{\mu_1})}$

$C=\sin^{-1}(\frac{1.47}{1.5})=\sin^{-1}(0.98)=78.52^\circ$

b) Numerical aperture $(NA)=\sqrt{\mu_1^2-\mu^2}=\sqrt{1.5^2-1.47^2}$

$=\sqrt{0.0891}$

$=0.2984$

c)Acceptance angle $(\theta_o)=\sin^{-1}(NA)$

$=\sin^{-1}(0.2984)$

$=17.36^\circ$

d) Fractional index change

$\Delta=\dfrac{\mu_1-\mu_1}{\mu_1}$

$\Delta=\dfrac{1.5-1.47}{1.5}=0.02$