An optical fibre consists of core of `mu_(1)` surrounded by a cladding of `mu_(2) lt mu_(1)`. A beam of light enters from air at an angle `alpha` with axis of fibre. The highest `alpha` for which ray can be travelled through fibre is

Options

A. `cos^(-1) sqrt(mu_(2)^(2)-mu_(1)^(2))`
B. `sin^(-1) sqrt(mu_(1)^(2)-mu_(2)^(2))`
C. `tan^(-1)sqrt(mu_(1)^(2)-mu_(2)^(2))`
D. `sec^(-1)sqrt(mu_(1)^(2)-mu_(2)^(2))`

Answer

B

Explanation

`90 - r gt theta_(e)` where `sin theta_(c) = (mu_(2))/(mu_(1))`
So that TIR take place at core and cladding interface
`90 - theta_(c) gt r`
`Sin (90 - theta_(c)) gt sin r`....(1)
Using snell law
`sin alpha = mu_(1) sinr` ...(2)
from (1) & (2)
`cos theta_(c) gt (sin alpha)/(mu_(1))`
`sin alpha lt mu_(1) cos theta_(c)`
`sin alpha lt sqrt(mu_(1)^(2)-mu_(2)^(2))`
`:. alpha_(max) = sin^(-1) sqrt(mu_(1)^(2)-mu_(2)^(2))`