Explanation

(b) We know that
$\mu = \frac{velocity\ of\ light\ in\ vacuum}{velocity\ of\ light\ in\ water}$
$\frac{4}{3} = \frac{3 \times 10^{10}}{velocity\ in\ light\ in\ water}$
$\Rightarrow$ velocity of light in water =$\ 2.25 \times 10^{10}cms^{- 1}$
Time taken $= \frac{500 \times 100}{2.25 \times 10^{10}} = 2.22 \times 10^{- 6}\ s$
Equivalent optical path $= \mu \times \ $diatance travelled in water
$= \frac{4}{3} \times 500 = 666.64$ m