Question

A beam of red light is incident at an angle of 54.3 o on an equilateral prism. If the index of refraction of red light is 1.400, at what angle does the beam emerge from the other face of the prism?

Answer 1

ANSWER

35.57°

EXPLANATION

Given:

• The incident angle, θ₁ = 54.3°

,

• The index of refraction of red light in this prism, n₃ = 1.4

Find:

• The angle at which the beam emerges from the other face of the prism, θ₂

We have the following situation,

Using Snell's law, we can find the angle α₁,

$n_1\sin\theta_1=n_3\sin\alpha_1$

Solving for α₁,

$\alpha_1=\sin^{-1}\left(\frac{n_1}{n_3}\sin\theta_1\right)=\sin^{-1}\left(\frac{1}{1.4}\sin54.3\degree\right)\approx35.45\degree$

Now, to find the angle at which the beam emerges from the other face of the prism, we have to find angle α₂, which would be the incidence angle for the second refraction.

Let's go back to the diagram of the prism,

At the top, the beam of light forms a triangle. We know that the sum of the interior angles of any triangle is 180°. We also know that angles α₁ and α₂ are complementary to the other two interior angles of that triangle, so we have,

$(90-\alpha_1)+(90-\alpha_2)+60=180$

Solving for α₂,

$\alpha_2=90+90+60-\alpha_1-180=90+90+60-35.45-180=24.55$

Now, knowing that the incidence angle at the other end of the prism is 24.55°, we can find the refraction angle using Snell's law,

$n_3\sin\alpha_2=n_2\sin\theta_2$

Solving for θ₂,

$\theta_2=\sin^{-1}\left(\frac{n_3}{n_2}\sin\alpha_2\right)=\sin^{-1}\left(\frac{1.4}{1}\sin24.55\degree\right)\approx35.57\degree$

Hence, the beam emerges from the other side of the prism at an angle of 35.57°.