In an optical fibre, core and cladding were made with materials of refractive indices 1.5 and 1.414 respectively. To observe total internal reflection, what will be the range of incident angle with the axis of optical fibre?

Options

A. `0^(@) - 60^(@)`
B. `0^(@)- 48^(@)`
C. `0^(@) - 30^(@)`
D. `0^(@)- 82^(@)`

Answer

C

Explanation

`mu_(1)=` refractive index of core `=1.5=3//2`
`mu_(2)=` refractive index of cladding `=1.414=sqrt(2)`
Apply Snell.s law at the interface of core and cladding
`(sin(90^(@)-theta_(0)))/(sin90^(@))=""_(1)mu_(2)`
`cos theta_(0)=(mu_(2))/(mu_(1))`
From Snell.s law at point P, interface of core and air
`:. (sin theta_(c))/(sintheta_(0))=""_(a) mu_(1)`
`sin theta_(c)= mu_(1) sin theta_(0)`
`= mu_(1) sqrt(1-cos^(2) theta_(0))= mu_(1) (sqrt(mu_(1)^(2)-mu_(2)^(2)))/(mu_(1))`
`sin theta_(c)= sqrt(mu_(1)^(2)-mu_(2)^(2))`
`=sqrt(9/4 -2)`
`sin theta_(c)= 1/2`
` theta_(c)= 30^(@)`