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The refractive index of the core of an optical fibre is $μ_{2}$ and that of the cladding is $μ_{1}$ . The angle of incidence on the face of the core so that the light ray just undergoes total internal reflection at the cladding is

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$\sin^{- 1} (\frac{μ_{1}}{μ_{2}})$

$\sin^{- 1} \sqrt{μ_{2}^{2} - μ_{1}^{2}}$

$\sin^{- 1} \sqrt{μ_{2} - μ_{1}}$

$\sin^{- 1} \sqrt{μ_{1}^{2} + μ_{2}^{2}}$

answer is B.

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Detailed Solution

$\sin r = \frac{μ_{1}}{μ_{2}}$

$\sin i = μ_{2} \cos r$

$sini = μ_{2} \sqrt{1 - \frac{μ_{1}^{2}}{μ_{2}^{2}}} = \sqrt{μ_{2}^{2} - μ_{1}^{2}};$

$i = \sin^{- 1} \sqrt{μ_{2}^{2} - μ_{1}^{2}}$

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