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(B) The acceptance angle $	heta_a$ is defined as the maximum angle at which light can enter the optical fiber and still be guided through the core by

Vedclass · Accepted Answer

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

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(B) The acceptance angle $	heta_a$ is defined as the maximum angle at which light can enter the optical fiber and still be guided through the core by total internal reflection.Using Snell's Law at the air-core interface: $n_0 \sin 	heta_a = n_1 \sin 	heta_r$,where $n_0 = 1$ (for air).At the core-cladding interface,the critical angle condition for total internal reflection is $\sin 	heta_c = n_2/n_1$.Since $	heta_r = 90^\circ - 	heta_c$,we have $\sin 	heta_r = \cos 	heta_c = \sqrt{1 - \sin^2 	heta_c} = \sqrt{1 - (n_2/n_1)^2} = \frac{\sqrt{n_1^2 - n_2^2}}{n_1}$.Substituting this into the first equation: $\sin 	heta_a = n_1 	imes \frac{\sqrt{n_1^2 - n_2^2}}{n_1} = \sqrt{n_1^2 - n_2^2}$.Therefore,$	heta_a = \sin^{-1}(\sqrt{n_1^2 - n_2^2})$.

What is the maximum acceptance angle at the entrance surface of an optical fiber,if $n_1$ and $n_2$ are the refractive indices of the core and cladding,respectively?

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