KCET · Physics · Alternating Current
If \( \varepsilon_{0} \) and \( \mu_{0} \) are the permittivity and permeability of free space and \( \varepsilon \) and \( \mu \) are the corresponding quantities for a medium, then refractive index of the medium is
- A \( \sqrt{\frac{\mu_{0} \varepsilon_{0}}{\mu \varepsilon}} \)
- B \( \sqrt{\frac{\mu \varepsilon}{\mu_{0} \varepsilon_{0}}} \)
- C \( 11 \)
- D Insufficient information
Answer & Solution
Correct Answer
(B) \( \sqrt{\frac{\mu \varepsilon}{\mu_{0} \varepsilon_{0}}} \)
Step-by-step Solution
Detailed explanation
Given \(\varepsilon_{0}=\) permittivity of free space \(; \mu_{0}=\) permeabilityof free space,\(\varepsilon=\) permittivity of medium; \(\mu=\) permeability of medium
Refractive index of medium, \(n=\frac{C}{v} \rightarrow(1)\)
We know \(c=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}\)
\(\Rightarrow v=\frac{1}{\sqrt{\mu \varepsilon}}\)
Therefore, \(n=\frac{\sqrt{\mu \varepsilon}}{\sqrt{\mu_{0} \varepsilon_{0}}}=\sqrt{\frac{\mu \varepsilon}{\mu_{0} \varepsilon_{0}}}[\) from Eq. (1)]
Refractive index of medium, \(n=\frac{C}{v} \rightarrow(1)\)
We know \(c=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}\)
\(\Rightarrow v=\frac{1}{\sqrt{\mu \varepsilon}}\)
Therefore, \(n=\frac{\sqrt{\mu \varepsilon}}{\sqrt{\mu_{0} \varepsilon_{0}}}=\sqrt{\frac{\mu \varepsilon}{\mu_{0} \varepsilon_{0}}}[\) from Eq. (1)]
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