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?

Respuesta :

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:
[tex]n= \frac{c}{v_a} [/tex]

For material a, the relationship becomes
[tex]n_a = \frac{c}{v_a} [/tex]
where [tex]v_a[/tex] is the speed of light in material a.

For material b,
[tex]n_b = \frac{c}{v_b} [/tex]
where [tex]v_b[/tex] is the speed of light in material b.

The ratio between the two refractive indices is
[tex] \frac{n_a}{n_b} = \frac{c}{v_a} \frac{v_b}{c}= \frac{v_b}{v_a} [/tex] (1)
And we know that the speed of light in material a is 1.4 times the speed of light in material b, so
[tex]v_a = 1.4 v_b[/tex]
and if we substitute this into (1), we find
[tex] \frac{n_a}{n_b}= \frac{v_b}{1.4 v_b}= \frac{1}{1.4}=0.71 [/tex]