For water $μ$ = 4 3/ and the velocity of light in vacuum is 3 ×10⁸ ms^–1, the time taken for light to travel a distance of 450 m in water will be(nearly)

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$1.5 μ s$

$1 μ s$

$3 μ s$

$2 μ s$

answer is C.

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Detailed Solution

$\begin{array}{l} μ = \frac{ct}{d}, μ_{w} = \frac{4}{3}; \\ V_{w} = \frac{3 \times 10^{8}}{4} \times 3 = \frac{9}{4} \times 10^{8} m / s \\ t = \frac{450 \times 4}{9 \times 10^{8}} = 2 \times 10^{- 6} = 2 μ s \end{array}$

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