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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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Answered 2 years ago
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Given\textbf{Given}

n=2.35n=2.35

c=3×108 m sc=3\times10^8\frac{\text{ m}}{\text{ s}}

Approach\textbf{Approach}

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

Solution\textbf{Solution}

The definition of index of refraction is

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

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

v=cnv=3×108 m s2.35v=1.28×108m s\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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