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Question

# Find the speed of light in antimony trioxide if it has an index of refraction of 2.35.

Solution

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$\textbf{Given}$

$n=2.35$

$c=3\times10^8\frac{\text{ m}}{\text{ s}}$

$\textbf{Approach}$

In this problem, we will use a definition of the index of refraction.

$\textbf{Solution}$

The definition of index of refraction is

\begin{align} {n}=\frac{c}{v} \end{align}

We need to calculate the speed of light in medium, so we write

\begin{align} &{v}=\frac{c}{n} \\ &{v}=\frac{3\times10^8\frac{\text{ m}}{\text{ s}}}{2.35} \\ &\boxed{{v}=1.28\times10^8\frac{ m}{\text{ s}}} \end{align}

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