Refractive Index from Permittivity and Permeability – Rankers Physics
Topic: Magnetic Effects of Current
Subtopic: Magnetic Properties of Matter

Refractive Index from Permittivity and Permeability

\(\varepsilon_0\) and \(\mu_0\) are the electric permittivity and magnetic permeability of free space respectively. If the corresponding quantities of a medium are \(2\varepsilon_0\) and \(1.5\mu_0\) respectively, the refractive index of the medium will nearly be
3
2
\(\sqrt{2}\)
\(\sqrt{3}\)

Solution:

The speed of light in free space is \(c = \frac{1}{\sqrt{\varepsilon_0\mu_0}}\) and in a medium is \(v = \frac{1}{\sqrt{\varepsilon\mu}}\). The refractive index \(n = \frac{c}{v} = \sqrt{\frac{\varepsilon\mu}{\varepsilon_0\mu_0}} = \sqrt{2 \times 1.5} = \sqrt{3}\).

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