Question

The index of refraction of the core of a typical fiber optic is n_{core} core ​  = 1.51; the cladding has n_{clad} clad ​  = 1.43. Calculate the critical angles for the total internal reflection i_{crit}i crit ​  and α_{crit}α crit ​ .

Answer 1

`i_(cr) = 71.27^(\circ)`

`\alpha_(cr) = 29.01^(\circ)`

Given:

refractive index of core,
`n_(core) = 1.51`

refractive index of clad,
`n_(clad) = 1.43`

Step-by-step explanation:

Critical angle can be defined as the incidence angle which results in the refraction angle being equal to
`90^(\circ)` at that angle of incidence.

For Total Internal Reflection to occur, the incidence angle must be greater than the critical angle.

Now, we know that the critical angle,
`\theta_(cr)` is given by:

`sin\theta_(cr) = (n_(clad))/(n_(core))`

`\theta_(cr) = sin^(- 1)((n_(clad))/(n_(core)))`

`\theta_(cr) = sin^(- 1)((1.43)/(1.51)) = sin^(- 1)(0.947) = 71.27^(\circ)`

`i_(cr) = \theta_(cr) = 71.27^(\circ)`

Now, for
`\alpha_(cr)`:

`(sin\gamma_(cr))/(sin\alpha_(cr)) = (1)/(n_(core))`

`sin\alpha_(cr)= sin(90^(\circ) - 71.27^(\circ))* 1.51`

`sin\alpha_(cr)= sin(18.734^(\circ))* 1.51`

`\alpha_(cr)= sin^(- 1)0.485 = 29.01^(\circ)`