Chapter 18, Problem 95A

To determine

The speed of light in antimony trioxide if it has an index of refraction of $2.35$ .

Answer to Problem 95A

  $1.27\times {10}^8\text{m/s}$

Explanation of Solution

Given:

The index of refraction of antimony trioxide, $n=2.35$

Formula used:

The index of refraction is given by,

  $n=\frac{c}{v}$

Where,

  $\begin{array}{l}c=\text{ speed of light in vacuum}=3\times {10}^8\text{m/s}\\ v=\text{ speed of light in the medium}\end{array}$

Calculation:

Using the above formula, the speed of light in antimony trioxide will be,

  $\begin{array}{c}n=\frac{c}{v}\\ 2.35=\frac{3\times {10}^8}{v}\\ v=1.27\times {10}^8\text{m/s}\end{array}$

Conclusion:

Hence, the speed of light in antimony trioxide is $1.27\times {10}^8\text{m/s}$ .