Question

A silica optical fiber has a core refractive index of 1.50 and a cladding refractive index of 1.42. Calculate the critical angle at the core-cladding interface.​

Answer 1

Explanation:

Here,μ

1

=1.50;μ

2

=1.47;

μ

0

=1

(i) Critical angle θ

c

at the core-cladding interface is given by

θ

c

=sin

−1

1.50

1.47

=78.5

0

(ii) Numerical aperture,NA=(μ

1

2

−μ

2

2

)

2

1

=[(1.50)

2

−(1.47)

2

]

2

1

=(2.25−2.16)

1/2

=0.30

(iii) Acceptance angle θ

a

=sin

−1

(NA)

=sin

−1

(0.30)=17.4

0