Question

the absolute refractive indices of benzene and kerosene are 1.5 and 1.44. calculate the refractive indes of benzene with respect to kerosene.

Answer 1

Here's your answer mate..
Good Luck.

Answer 2

Answer: The refractive indices of benzene with respect to kerosene is $\dfrac{v_{ke}}{v_{be}}=\dfrac{25}{24}$.

Explanation:

Given that,

Refractive indices of benzene $n_{be}= 1.5$

Refractive indices of kerosene  $n_{ke}= 1.44$

We know that,

The refractive index is the ratio of the speed of light in vacuum and speed of the light in medium.

$n = \dfrac{c}{v}$

Where, c = speed of light in vacuum

v = speed of light in medium

For benzene

$n_{be}= \dfrac{c}{v_{be}}$

$1.5 = \dfrac{c}{v_{be}}.......(I)$

For kerosene

$n_{ke}= \dfrac{c}{v_{ke}}$

$1.44 = \dfrac{c}{v_{ke}}.......(II)$

Divided equation (I) by equation (II)

$\dfrac{1.5}{1.44}=\dfrac{v_{ke}}{v_{be}}$

The refractive indices of benzene with respect to kerosene is

$\dfrac{v_{ke}}{v_{be}}=\dfrac{1.5}{1.44}$

$\dfrac{v_{ke}}{v_{be}}=\dfrac{25}{24}$

Hence, The refractive indices of benzene with respect to kerosene is $\dfrac{v_{ke}}{v_{be}}=\dfrac{25}{24}$.