Question

Light from a source located in a medium (refractive index $ = \mu_0 $ ) enters an optical fibre with core refractive index $ \mu_1 $ and clad refractive index $ \mu_2 $ , as shown in the figure. The maximum value of incident angle $ \theta $ which undergo total internal reflection in the fibre is

• A. $ \theta=\cos^{-1}\left(\frac{\sqrt{\mu^{2}_{1}-\mu^{2}_{2}}}{\mu_{0}}\right) $
• B. $ \theta=\tan^{-1}\left(\frac{\mu_{1}-\mu_{2}}{\mu_{0}}\right) $
• C. $ \theta=\sin^{-1}\left(\frac{\sqrt{\mu^{2}_{1}-\mu^{2}_{2}}}{\mu_{0}}\right) $
• D. None of the above

Answer 1

Answer: (C)

$\mu_0 \,\sin\,\theta = \mu_1\,\sin \,\theta_1$
$\sin\,\theta = (\frac{\mu_1}{\mu_0}) \sin \,\theta_1$
If $\theta_c$ is critical angle, then $\sin \,\theta_c = \frac{\mu_2}{\mu_1}$
$\sin \,\theta = \frac{\mu_1}{\mu_2} , \sin \,\theta_1 = (\frac{\mu_1}{\mu_0}) \sin (90^{\circ} - \theta_c)$
$ = \frac{\mu_1}{\mu_2} \cos \,\theta_c = (\frac{\mu_1}{\mu_2}) \sqrt{1- (\frac{\mu_2}{\mu_1})^2}$
$\sin\,\theta = \frac{\mu_1^2 - \mu_2^2}{\mu_0}$