If `epsilon_(0) and mu_(0)` are the electric permittivity and magnetic permeability of free space and `epsilon and mu` are the corresponding quantities in the medium, the index of refraction of the medium in terms of above parameter is :

Options

A. `((epsilonmu)/(epsilon_(0)mu_(0)))^(1//2)`
B. `((epsilonmu)/(epsilonmu))^(1//2)`
C. `((epsilon_(0)mu_(0))/(epsilonmu))`
D. `((epsilon_(0)mu_(0))/(epsilonmu))^(1//2)`

Answer

B

Explanation

Velocity of light in vacuum
`" "c=1/sqrt(mu_(0)epsilon_(0))`
Velocity of light in medium
`" "v=1/sqrt(muepsilon)`
`therefore" " mu=c/v=((muepsilon)/(mu_(0)epsilon_(0)))^(1//2)`