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An optical fiber $(μ = 1.72)$ is surrounded by a glass coating $(μ = 1.5)$ .The critical angle for total internal reflection at the fiber-glass interface is

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\sin^{- 1} (\frac{2}{15})

\sin^{- 1} (\frac{85}{75})

\sin^{- 1} (\frac{75}{86})

\sin^{- 1} (\frac{2}{17})

answer is B.

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Detailed Solution

μ_{f} = 1.72 - denser

$μ_{g} = 1.50 - rarer$

$_{g} μ_{f} = \frac{1}{\sin C} = \frac{μ_{f}}{μ_{g}}$

$\sin C = \frac{μ_{g}}{μ_{f}} = \frac{1.50}{1.72}$

$\sin C = \frac{75}{86}$

$C = \sin^{- 1} (\frac{75}{86})$

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