A optical fiber is made up of a core material with refractive index 1.68 and a cladding material of refractive index 1.44. What is the acceptance angle of the fiber kept in air medium? What is the answer if there is no cladding?

Explanation

Give: `n_(1) = 1.68,n_(2) = 144,n_(1) = 1`
Acceptance angle, `i_(a) = sin^(-1) (sqrt(n_(1)^(2)-n_(2)^(2)))`
`i_(a) = sin^(-1) (sqrt((1.68)^(2)-(1.44)^(2)))=sin^(-1)(0.865)`
If there is no cladding then, `n_(1)=1`
Acceptance angle, `i_(a) = sin^(-1)(sqrt(n_(1)^(2)-1))`
`i_(a)=sin^(-1)(sqrt((1.68)-1))=sin^(-1)(1.35)`
`sin^(-1)` (more than 1) is not possible But, this includes the range `0^(@) "to" 90^(@)`. Hence, all the rays entering the core from flat surface will undergo total internal reflection.
Note: If there is no cladding then is a condition on the refractive index `(n_(1))` of the core, `i_(a)=sin^(-1)(sqrt(n_(1)^(2)-1))`
Hence, as per mathematical rule, `(n_(1)^(2)-1)le1or(n_(1)^(2))le2orn_(1)lesqrt(2)`
Hence, in air (no cladding) the refractive index `n_(1)` of the should be, `n_(1) le 1.414`