Question

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?

• A. $0^{\circ}-60^{\circ}$
• B. $0-48^{\circ}$
• C. $0^{\circ}-30^{\circ}$
• D. $0^{\circ}-82^{\circ}$

Answer 1

Answer: (C)

The range of incident angle
$\sin \theta =\frac{\sqrt{\mu_{1}^{2}-\mu_{2}^{2}}}{\mu_{0}}$
$\theta=\sin ^{-1} \sqrt{\left(\mu_{1}\right)^{2}-\left(\mu_{2}\right)^{2}}$ [for air $\mu_{0}=1$]
$\theta=\sin ^{-1} \sqrt{(1.5)^{2}-(1.414)^{2}}$
$=\sin ^{-1}(0.5006)$
$\theta =30^{\circ}$