Question

If the
refractive index of benzere is 2.419,
what is the speed of light in benzene?

Answer 1

`v=1.24* 10^8\ m/s`

Step-by-step explanation:

Given that,

The refractive index of benzene is 2.419

We need to find the speed of light in benzene. The ratio of speed of light in vacuum to the speed of light in the medium equals the refractive index. So,

`n=(c)/(v)\\\\v=(c)/(n)\\\\v=(3* 10^8)/(2.419)\\\\v=1.24* 10^8\ m/s`

So, the speed of light in bezene is
`1.24* 10^8\ m/s`.