Light takes time $t_{1}$ s to travel a distance X cm in vacuum and the same light takes time $t_{2}$ to travel 10X cm in medium. The critical angle for the corresponding medium is

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\sin^{- 1} (\frac{10 t_{2}}{t_{1}})

\sin^{- 1} (\frac{10 t_{1}}{t_{2}})

\sin^{- 1} (\frac{t_{2}}{10 t_{1}})

\sin^{- 1} (\frac{t_{1}}{10 t_{2}})

answer is C.

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Detailed Solution

speed of light in vacuum

v_{r} = \frac{x}{t_{1}}

Speed of light in medium $v_{d} = \frac{10}{t_{2}} [∴ r = \frac{d}{t}]$

$_{v} μ_{m} = \frac{1}{\sin C}$

$\frac{μ_{m}}{μ_{v}} = \frac{1}{\sin C}$

$\sin C = \frac{μ_{r}}{μ_{m}}$

$μ α \frac{1}{v} \Rightarrow \sin C = \frac{v_{m}}{v_{r}}$

$\sin C = \frac{10}{t_{2}} \times \frac{t_{1}}{1}$

$\sin C = \frac{10 t_{1}}{t_{2}}$

$C = \sin^{- 1} (\frac{10 t_{1}}{t_{2}})$

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