JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
In a medium the speed of light wave decreases to \(0.2\) times to its speed in free space The ratio of relative permittivity to the refractive index of the medium is \(x: 1\). The value of \(x\) is \(...........\) (Given speed of light in free space \(=3 \times 10^8 m s ^{-1}\) and for the given medium \(\mu_{ r }=1\) )
- A \(5\)
- B \(4\)
- C \(3\)
- D \(2\)
Answer & Solution
Correct Answer
(A) \(5\)
Step-by-step Solution
Detailed explanation
\(V =\frac{ C }{\mu} \Rightarrow \mu=\frac{ C }{ V }=\frac{ C }{0.2 C }\) \(\mu=5\) \(\mu=\sqrt{\epsilon_{ r } \mu_{ r }}\) \(\Rightarrow \epsilon_{ r }=\frac{\mu^2}{\mu_{ r }}\) \(\therefore \frac{\epsilon_{ r }}{\mu}=\frac{\mu}{\mu_{ r }}=5\)
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