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(B) The acceptance angle $	heta_a$ is defined as the maximum angle that a light ray can make with the axis of the optical fibre and still be guided t

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$\sin^{-1}\sqrt{n_1^2 - n_2^2}$

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(B) The acceptance angle $	heta_a$ is defined as the maximum angle that a light ray can make with the axis of the optical fibre and still be guided through the core by total internal reflection.Applying Snell's Law at the air-core interface: $1 \cdot \sin 	heta_a = n_1 \cdot \sin 	heta_r$,where $	heta_r$ is the angle of refraction.At the core-cladding interface,for total internal reflection to occur,the angle of incidence $	heta_i$ must be at least the critical angle $	heta_c$,where $\sin 	heta_c = n_2/n_1$.Since $	heta_r + 	heta_i = 90^\circ$,we have $\sin 	heta_r = \cos 	heta_i = \sqrt{1 - \sin^2 	heta_i} = \sqrt{1 - (n_2/n_1)^2} = \frac{\sqrt{n_1^2 - n_2^2}}{n_1}$.Substituting this into the first equation: $\sin 	heta_a = n_1 \cdot \frac{\sqrt{n_1^2 - n_2^2}}{n_1} = \sqrt{n_1^2 - n_2^2}$.Therefore,$	heta_a = \sin^{-1}\sqrt{n_1^2 - n_2^2}$.

What should be the maximum acceptance angle at the air-core interface of an optical fibre if $n_1$ and $n_2$ are the refractive indices of the core and the cladding,respectively?

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