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A small coin is resting at the bottom of a beaker filled with a liquid A ray of light travels from coin upto surface and moves along surface as shown. How fast is the light travelling in the liquid

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$3 \times 10^{8} {ms}^{- 1}$

$2 \times 10^{8} {ms}^{- 1}$

$1.8 \times 10^{8} {ms}^{- 1}$

$2.4 \times 10^{8} {ms}^{- 1}$

answer is C.

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Detailed Solution

$μ = \frac{\sin i}{\sin r} = \frac{C_{o}}{C_{m}} \Rightarrow \frac{30 / 50}{1} = \frac{3 \times 10^{8}}{C_{m}}$

$C_{m} = 3 \times 10^{8} \times \frac{3}{5} = 1.8 \times 10^{8} m s^{- 1}$

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