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A small coin is resting on the bottom of a beaker filled with liquid. A ray of light from the coin travels upto the surface of the liquid and moves along its surface. How fast is the light travelling in the liquid?
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The correct answer is:
$1.8 \times 10^8 \mathrm{~m} / \mathrm{s}$
$\because \quad \frac{1}{\mu}=\sin \theta_c=\frac{3}{5}$
$\begin{aligned}
& \Rightarrow \quad \mu=\frac{5}{3} \\
& \text { And } v=\frac{c}{\mu}=\frac{3 \times 10^8}{5 / 3} \\
& =\frac{9}{5} \times 10^8=1.8 \times 10^8 \mathrm{~ms}^{-1}
\end{aligned}$
$\begin{aligned}
& \Rightarrow \quad \mu=\frac{5}{3} \\
& \text { And } v=\frac{c}{\mu}=\frac{3 \times 10^8}{5 / 3} \\
& =\frac{9}{5} \times 10^8=1.8 \times 10^8 \mathrm{~ms}^{-1}
\end{aligned}$
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