Explanation

(a) Critical angle of core-cladding surface is
$\sin{C = \frac{\mu_{2}}{\mu_{1}} = \frac{1.49}{150} = 0.09933 = \sin{83{^\circ}21^{'}}}$
Or $C = 83{^\circ}21^{'}$
For optical fiber,
$\frac{\sin\theta_{a}}{\sin{(90 - \theta_{i})}} = \frac{\mu_{i}}{\mu_{0}}\ \ or\ \ \ \frac{\sin\theta_{a}}{\cos\theta_{i}} = \frac{\mu_{1}}{\mu_{0}}$
Or $\cos{\theta_{i} = \frac{\mu_{0}}{\mu_{1}} = \sin{\theta_{a} = \frac{1}{1.5} = \sin{10{^\circ}}}}$
$= \frac{2}{3} \times 0.1736 = 0.1157$
$\theta_{i} = 83{^\circ}23^{'}$
As the angle of incidence at the core and cladding surface is just equal to critical angle, so the ray will not suffer total internal reflection. Hence, beam can not propagate along the fibre