Again since the light hits first cathetus of a prism perpendicularly it does not refract there, but only refracts on hypotenuse. From the figure we see that the angle of incidence is $30^circ$ do from the Snell’s law we calculate

$$
n_{glass}sin30^circ=n_{air}sin57^circ
$$
and this gives

$$
n_{glass}=frac{sin57^circ}{sin30^circ} =1.68.
$$

Using Snell’s law we have

$$
n_{air}sin45^circ=n_{liquid}sinalpha_{reff}.
$$

Putting $n_{air} = 1$ we obtain

$$
sinalpha_{reff} = frac{sin45^circ}{n_{liquid}}Rightarrowalpha_{reff} = arcsinleft(frac{sin45^circ}{n_{liquid}}right).
$$

Now we just put in values of the index for both wavelengths which gives

$$
alpha_{violet} =arcsinleft(frac{sin45^circ}{1.332}right) = 32.06^circ;
$$

$$
alpha_{red} =arcsinleft(frac{sin45^circ}{1.330}right) = 32.12^circ.
$$

This gives

$$
alpha_{violet} – alpha_{red} = 0.06^circ=3.6′.
$$