Answer
$\theta_{2,ice}-\theta_{2,water}=0.903^{0}$
Work Step by Step
Given that refractive index of ice $n_{ice}=1.309$
refractive index of water is $n_{water}=1.333$
refractive indwx of air $n_{1,air}=1.000293$
angle of incidence remains same for both ice and water and is given by
$\theta_{1,air}=60^{0}$
suppose angle of refraction in ice $\theta_{2,ice}$
and angle of refraction in water is $\theta_{2,water}$
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$
from above equation we can write
$sin\theta_{2}=\frac{n_{1}\times sin\theta_{1}}{n_{2}}$
or $\theta_{2}=sin^{-1}(\frac{n_{1}\times sin\theta_{1}}{n_{2}})$
so when light impinges from air to ice
$\theta_{2,ice}=sin^{-1}(\frac{n_{1,air}\times sin\theta_{1,air}}{n_{2,ice}})$
$\theta_{2,ice}=sin^{-1}(\frac{1.000293\times sin60^{0}}{1.309})$
$\theta_{2,ice}=sin^{-1}(\frac{1.000293\times 0.8660}{1.309})$
$\theta_{2,ice}=sin^{-1}(0.66176)$
$\theta_{2,ice}=41.434^{0}$
when light impinges from air to water
$\theta_{2,water}=sin^{-1}(\frac{n_{1,air}\times sin\theta_{1,air}}{n_{2water}})$
$\theta_{2,water}=sin^{-1}(\frac{1.000293\times sin60^{0}}{1.333})$
$\theta_{2,water}=sin^{-1}(\frac{1.000293\times 0.8660}{1.333})$
$\theta_{2,water}=sin^{-1}(0.64985)$
$\theta_{2,water}=40.531^{0}$
the difference $\theta_{2,ice}-\theta_{2,water}=41.434^{0}-40.531^{0}$
or $\theta_{2,ice}-\theta_{2,water}=0.903^{0}$
