Ray Optics

JEE Main level, numerical answer, medium

94 of 295

Question 94 of 295

In a medium the speed of light wave decreases to 0.20.2 times to its speed in free space. The ratio of relative permittivity to the refractive index of the medium is x:1x:1. The value of xx is ______.
(Given speed of light in free space =3×108ms−1=3\times {10}^{8}m{s}^{-1} and for the given medium μr=1{\mu }_{r}=1)

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Answer: 5

Speed of light in medium is related to speed of light in vacuum (c)\left(c\right) and refractive index (μ)\left(\mu \right) by v=cμv=\dfrac{c}{\mu }.
Then, we have μ=cv=c0.2c\mu =\dfrac{c}{v}=\dfrac{c}{0.2c}
⇒μ=5\Rightarrow \mu =5
Refractive index in terms of relative permeability (μr)\left({\mu }_{r}\right)and relative permittivity(ϵr)\left({\epsilon }_{r}\right) of the medium is stated as μ=ϵrμr\mu =\sqrt{{ϵ}_{r}{\mu }_{r}}
⇒ϵr=μ2μr\Rightarrow {ϵ}_{r}=\dfrac{{\mu }^{2}}{{\mu }_{r}}
∴ϵrμ=μμr=5∴\dfrac{{ϵ}_{r}}{\mu }=\dfrac{\mu }{{\mu }_{r}}=5
Hence, the value of x=5x=5.

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