Question

A beam of white light is incident on a prism at an angle of incidence of 71.2 degrees. The speed of red light in the prism material is 1.984 x 10^8 meters per second, and the speed of violet light in the prism material is 1.951 x 10^8 meters per second.a. At what angle does the red light enter the prism?b. At what angle does the violet light enter the prism?

Answer 1

ANSWER

$\begin{gathered} (a)\text{ }38.8\degree \\ (b)\text{ }38\degree \end{gathered}$

EXPLANATION

Parameters given:

Incident angle of white light, θ1 = 71.2 degrees

Speed of red light in the prism, vr = 1.984 * 10^8 m/s

Speed of violet light in the prism, vv = 1.951 * 10^8 m/s

Speed of light in air, v = 3 * 10^8 m/s

(a) To find the angle at which the red light enters the prism, apply the relationship given by Snell's law:

${\frac{v}{v_r}}=\frac{\sin\theta_1}{\sin\theta_r}$

where v = speed of light in air

vr = speed of red light in the prism

θr = angle of refraction (angle that the light enters the prism)

Hence, solving for θr, we have that the angle at which the red light enters the prism is:

$\begin{gathered} \frac{3*10^8}{1.984*10^8}=\frac{\sin71.2}{\sin\theta_r} \\ \\ \sin\theta_r=\frac{1.984*\sin71.2}{3}=0.6261 \\ \\ \theta_r=\sin^{-1}(0.6261) \\ \theta_r=38.8\degree \end{gathered}$

That is the answer.

(b)To find the angle at which the violet light enters the prism, apply the same formula above for violet light:

$\frac{v}{v_v}=\frac{\sin\theta_1}{\sin\theta_v}$

Hence, solving for θv, we have that the angle at which the violet light enters the prism is:

$\begin{gathered} \frac{3*10^8}{1.951*10^8}=\frac{\sin71.2}{\sin\theta_v} \\ \\ \sin\theta_v=\frac{1.951*\sin71.2}{3}=0.6156 \\ \\ \theta_v=\sin^{-1}(0.6156) \\ \theta_v=38\degree \end{gathered}$

That is the answer.