An optical fiber 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 fiber. The highest `alpha` for which ray can be travelled through fiber is

Options

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

Answer

B

Explanation

For TIR,
`i gt i_(c )`
`rArr " " sini gt sin i_(c ) `
`rArr " " sini gt (n_(2))/(n_(1))" "`….(i)
Snell.s law, sin `alpha = n_(1) ` sin r
`rArr " " n_(1) = (sin alpha)/( sin r) " "`… (ii)
`r + i = 90^(@)`
sin r = sin(90 - i)
From eq. (ii) `rArr ` sin `alpha = n_(1) ` sin(90 - i)
`rArr " " sin alpha = n_(1) cos `i

Now, sin i` = sqrt(1 - cos^(2) i) = sqrt( 1 - ( (sin alpha)/(n_(1)))^(2) )`
From (i) `rArr " " sin i gt (n_(2))/(n_(1))`
`rArr " "1 - ((sin alpha)/(n_(1)))^(2) gt ((n_(2))/(n_(1)))^(2) rArr ( (sin^(2) alpha)/(n_(1)^(2))) lt 1 - ( (n_(2))/(n_(1)) )^(2) `
`rArr sin^(2) alpha gt n_(1)^(2) - n_(2)^(2)`
`rArr sin alpha lt sqrt(n_(1)^(2) - n_(2)^(2))`
`alpha_(max) = sin^(-1) ( sqrt(n_(1)^(2) - n_(2)^(2)) ) `
Note : Use direct formula for objective question.