When a beam of white light strikes a prism, the blue light bends more when refracted than the red light. A material has a higher index of refraction for light toward the blue end of the visible spectrum. Therefore the $textbf{blue light}$ is refracted the most while the $textbf{red light}$ is the least.

When a beam of white light strikes a prism, the blue light bends more when refracted than the red light. A material has a higher index of refraction for light toward the blue end of the visible spectrum. Therefore the $textbf{blue light}$ is refracted the most while the $textbf{red light}$ is the least.

The angle of refraction in this case is 90$text{textdegree}$. Since if the incident angle is just bellow the critical angle, the light emerges almost parallel to the interface. Which is at 90$text{textdegree}$ from normal. Hence the angle of refraction is 90$text{textdegree}$.

90$text{textdegree}$.

From the formula for the critical angle we have

$$
alpha_c=arcsinleft(frac{n_1}{n_2}right) = arcsinleft(frac{1.33}{1.65}right) = 53.7^circ.
$$

The formula for the critical angle here reads

$$
alpha_c=arcsinleft(frac{n_{air}}{n_{glass}}right)Rightarrow39.1^circ=arcsinleft(frac{1}{n_{glass}}right).
$$
Taking sine of both sides we get

$$
sin39.1^circ = frac{1}{n_{glass}}Rightarrow n_{glass} =frac{1}{sin39.1^circ} =1.59
$$

Here there is no refraction of light on cathetus since the light goes in perpendicularly to cathetus. Then the angle of incidence on hypotenuse is $30^circ$ so the application of Snell’s law gives

$$
n_{red}sin30^circ=n_{air}sinalpha_{red};quad n_{violet}sin30^circ =n_{air}sinalpha_{violet}.
$$
This then gives by putting $n_{air} = 1$

$$
sinalpha_{red}=n_{red}sin30^circ;quad sinalpha_{violet}=n_{violet}sin30^circ.
$$

Taking sine of both sides of both equations we finally get

$$
alpha_{red}=arcsin(n_{red}sin30^circ)=45.3^circ;quadalpha_{violet}=arcsin(n_{violet}sin30^circ)=48.8^circ;
$$
yielding

$$
Deltaalpha = alpha_{violet}-alpha_{red} =3.5^circ.
$$