In an optical fibre, core and cladding were made with materials of refractive indices 1.5 and 1.414…
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?
$0^{\circ}-60^{\circ}$
$0-48^{\circ}$
$0^{\circ}-30^{\circ}$
$0^{\circ}-82^{\circ}$
Solution
The range of incident angle
$\sin \theta=\frac{\sqrt{\mu_1^2-\mu_2^2}}{\mu_0}$
$\theta=\sin ^{-1} \sqrt{\left(\mu_1\right)^2-\left(\mu_2\right)^2} \quad\left[\right.$ for air $\left.\mu_0=1\right]$
$\theta=\sin ^{-1} \sqrt{(1.5)^2-(1.414)^2}$
$=\sin ^{-1}(0.5006)$
$\theta=30^{\circ}$
Hence, option (c) is correct.