The prism shown in Fig. has a refractive index of 1.66, and the angles Aare . Two light rays m and nare parallel as they enter the prism. What is the angle between them after they emerge?

The refractive index of the prism is$n_a=1.66$ . We draw the two rays through the prism as shown below. The angle between both rays when they emerge is shown in the figure below.

First, let us find the angle${\theta }_b$ , by using Snell's law. The refractive index of an optical material is expressed by n and represents the speed of light in the vacuum divided by the speed of light in the material. Snell's law is given by an equation in the form

$n_a\sin {\theta }_a=n_b\sin {\theta }_b$

Where${\theta }_a=25°$ and$n_b=1$ because it is the refractive index of air. Solve equation (1) for${\theta }_b$ , and plug the values for$n_a$ , ${\theta }_a$and$n_b$

$\sin {\theta }_b=\frac{n_a\sin {\theta }_a}{n_b}=\frac{\left(1.66\right)\sin 25°}{1}=0.7$

Hence, the angle is

${\theta }_b=44.55°$

From the figure below, the anglecould be calculated by

$\beta =90°-44.55°=45.55°$

From the triangle the angle c is calculated by

$c=90°-\left(A+\beta \right)=90°-\left(25°+45.55°\right)=19.55°$

Now, the angle between the two rays is calculated by

$a=2c=2\left(19.55°\right)=39.1°$