Question

The speed of light in material a is 1.4 times as large as it is in material

b. what is the ratio of the refractive indices, na/nb, of these materials?

Answer 1

The refractive index of a material is the ratio between the speed of light in vacuum (c) and the speed of light in that material, v:

`n= (c)/(v_a)`

For material a, the relationship becomes

`n_a = (c)/(v_a)`
where
`v_a` is the speed of light in material a.

For material b,

`n_b = (c)/(v_b)`
where
`v_b` is the speed of light in material b.

The ratio between the two refractive indices is

`(n_a)/(n_b) = (c)/(v_a) (v_b)/(c)= (v_b)/(v_a)` (1)
And we know that the speed of light in material a is 1.4 times the speed of light in material b, so

`v_a = 1.4 v_b`
and if we substitute this into (1), we find

`(n_a)/(n_b)= (v_b)/(1.4 v_b)= (1)/(1.4)=0.71`