Answer
index of refraction of unknown liquid is $n_{liq}=1.632$
Work Step by Step
Given that refractive index of oil $n_{oil}=1.45$
suppose refractive index of unknown liquid is $n_{liq}$
ray of light originating in oil passes into liquid
(medium $1$ is oil and medium $2$ is liquid.)
given
angle of incidence $\theta_{1}= 64^{0}$
angle of refraction $\theta_{2}= 53^{0}$
snells law
When light travels from medium 1 to medium 2
according to Snell's law
$n_{1}sinθ_{1}=n_{2}sinθ_{2}$
Here
$n_{1}$ is refractive index of medium $1$
$θ_1$ is angle of incidence in medium $1$
$n_{2}$ is refractive index of medium $2$
$θ_2$ is angle of refraction in medium$ 2$
so in our case
$n_{oil}sinθ_{1}=n_{liq}sinθ_{2}$
so $n_{liq}=\frac{n_{oil}\times sin\theta_{1}}{sin\theta_{2}}$
putting the values of $n_{oil}=1.45$, $\theta_{1}= 64^{0}$ , $\theta_{2}= 53^{0}$
we get $n_{liq}=\frac{1.45\times sin64^{0}}{sin53^{0}}$
$n_{liq}=\frac{1.45\times 0.89879}{0.798635}=1.632$
