The frequence of a light wave in a material in `2xx10^(14)` Hz and wavelength is `5000.Å.` The refractive index of material will be

Options

A. `1.40`
B. `1.50`
C. `3.00`
D. `1.33`

Answer

C

Explanation

Velocity of light waves in material is
`v=nlamda` ..(i)
Refractive index of material is
`mu=(c)/(v)` ..(ii)
Where `c` is speed of light in vacuum or air.
or `mu=(c)/(nlamda)` ..(iii)
Given, `n=2xx10^(14)Hz`
`lamda=5000 Å = 5000 xx 10^(-10) m`,
`c = 3 xx 10^(8) m//s`
Hence, from Eq (iii) we get
`mu=(3xx10^(8))/(2xx10^(14)xx5000xx10^(-10))=3.00`