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(B) For the light ray to undergo total internal reflection at the core-cladding interface,the angle of incidence $i$ must be greater than the critical

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$\sin^{-1}\sqrt{\mu_1^2 - \mu_2^2}$

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(B) For the light ray to undergo total internal reflection at the core-cladding interface,the angle of incidence $i$ must be greater than the critical angle $c$.$i > c \Rightarrow \sin i > \sin c \Rightarrow \sin i > \frac{\mu_2}{\mu_1}$ ... $(1)$From Snell's law at the air-core interface,where the refractive index of air is $1$:$1 \cdot \sin \alpha = \mu_1 \cdot \sin r$ ... $(2)$In the right-angled triangle $\Delta OBA$,the angle of refraction $r$ and the angle of incidence $i$ at the core-cladding interface are related by:$r + i = 90^\circ \Rightarrow r = 90^\circ - i$Substituting this into equation $(2)$:$\sin \alpha = \mu_1 \sin(90^\circ - i) = \mu_1 \cos i$$\cos i = \frac{\sin \alpha}{\mu_1}$Using the identity $\sin i = \sqrt{1 - \cos^2 i}$:$\sin i = \sqrt{1 - \left(\frac{\sin \alpha}{\mu_1}ight)^2}$ ... $(3)$Substituting $(3)$ into $(1)$:$\sqrt{1 - \frac{\sin^2 \alpha}{\mu_1^2}} > \frac{\mu_2}{\mu_1}$$1 - \frac{\sin^2 \alpha}{\mu_1^2} > \frac{\mu_2^2}{\mu_1^2}$$1 - \frac{\mu_2^2}{\mu_1^2} > \frac{\sin^2 \alpha}{\mu_1^2}$$\mu_1^2 - \mu_2^2 > \sin^2 \alpha$$\sin \alpha Thus,the maximum angle $\alpha_{	ext{max}}$ is given by:$\alpha_{	ext{max}} = \sin^{-1}\sqrt{\mu_1^2 - \mu_2^2}$

An optical fibre consists of a core of refractive index $\mu_1$ surrounded by a cladding of refractive index $\mu_2 < \mu_1$. $A$ beam of light enters from air at an angle $\alpha$ with the axis of the fibre. The highest angle $\alpha$ for which the ray can travel through the fibre is:

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