Light guidance in an optical fibre can be understood by considering a structure comprising of thin solid glass cylinder of refractive index `n_1` surrounded by a medium of lower refractive index `n_2`. The light guidance in the structure takes place due to successive total internal reflectrions at the interface of the media `n_1` and `n_2` as shown in the fugure. All rays with the angle of incidence i less than a particular value `i_m` are confined in the medium of refractive index `n_1`. The numerical aprture (NA) of the structure is defined as `sini_m`
For two structure namely `S_1` with `n_1=(sqrt45)/(4)` and `n_2=(3)/(2)`, and `S_2` with `n_1=(8)/(5)` and `n_2=(7)/(5)` and taking the refractive index of water to be `(4)/(3)` and that of air to be 1, the correct option (s) is (are) ? `

Options

A. NA of `S_1` immersed in water is the same as that of `S_2` immersed in a liquid of refractive index `(16)/(3sqrt15)`
B. NA of `S_1` immersed in liquid of refractive index `(6)/(sqrt15)` is the same as the of `S_2` immersed in water
C. NA of `S_1` placed in air is the same as that of `S_2` immersed in liquid of refractive index `(4)/(sqrt15)`
D. NA of `S_1` placed in air is the same as that of `S_2` placedin water

Answer

A, C

Explanation

Applying Snell's law at P, `n_ssini_m=n_1sin(90^circ-C)`
`n_s=refractive index of surrounding`.
Also `sinC=(n_2)/(n_1)`
Now
`NA=sini_m=(n_1)/(n_s)cosC=(n_1)/(n_s)sqrt(1-(n_2^2)/(n_1^2))`
`:'` `NA=sqrt(n_1^2-n_2^2)/(n_s)`
For `S_1` (in air)
`NA=sqrt((45)/(16)-(9)/(4))=(3)/(4)` For `S_1` (in water) `NA=(3)/(4)sqrt((45)/(16)-(9)/(4))=(9)/(16)`
For `S_1` `(in n_s=(6)/(sqrt15))
`NA=(sqrt15)/(6)sqrt((45)/(16)-(9)/(4))=(3sqrt15)/(24)`
For `S_2` (in water)
`NA=(3)/(4)sqrt((64)/(25)-(49)/(25))=(3)/(4)(sqrt15)/(5)`
`S_2` (in air)
`NA=sqrt((64)/(25)-(49)/(25))=(sqrt15)/(5)`
`S_2`(in `n_s=(4)/(sqrt15))
`NA=(sqrt15)/(4)sqrt((64)/(25)-(49)/(25))=(3)/(4)` For `S_2`( in `n_s=(16)/(3sqrt15))`
`NA=(3sqrt15)/(16)sqrt((64)/(25)-(49)/(25))=(9)/(16)` (a), (c) are correct options