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A tank is filled with water to a height of 12.5 cm. The apparent depth of needle lying at the bottom of the tank is measured by a microscope as 9.4 cm. The speed of light in water is

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

$1.5 \times 10^{8} {ms}^{- 1}$

$2.256 \times 10^{8} {ms}^{- 1}$

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

answer is C.

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Detailed Solution

$μ = \frac{C_{0}}{C_{m}} = \frac{d_{R}}{d_{A}} \Rightarrow \frac{3 \times 10^{8}}{c_{m}} = \frac{12.5 cm}{9.4 cm}$

$C_{m} = \frac{3 \times 9.4}{12.5} \times 10^{8} {ms}^{- 1}$

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