Optics Ques 203
- If $\varepsilon _0$ and $\mu _0$ are, respectively, the electric permittivity and magnetic permeability of free space, $\varepsilon$ and $\mu$ the corresponding quantities in a medium, the index of refraction of the medium in terms of the above parameters is…….
$(1992,1 M)$
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Answer:
Correct Answer: 203.$( \sqrt{\frac{\varepsilon \mu}{\varepsilon _0 \mu _0}})$
Solution:
Formula:
- Speed of light in vacuum, $c=\frac{1}{\sqrt{\varepsilon _0 \mu _0}}$ and
speed of light in medium, $v=\frac{1}{\sqrt{\varepsilon \mu}}$
Therefore, refractive index of the medium is
$ \mu=\frac{c}{v}=\frac{1 / \sqrt{\varepsilon _0 \mu _0}}{1 / \sqrt{\varepsilon \mu}}=\sqrt{\frac{\varepsilon \mu}{\varepsilon _0 \mu _0}} $
BETA
