A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is \( \sqrt{2} \), and \( \sqrt{3} \).
Show Hint
To calculate the refraction angles inside a prism, use Snell’s Law at each interface. The refractive index determines how much the ray bends when passing through different materials.
We are given a triangular prism with an isosceles right-angled base ABC, and a ray of light \( MN \) is incident normally on face AB of the prism. Since the ray is incident normally (at \( 0^\circ \)) on face AB, it will travel straight through without any refraction at the first interface. This means the angle of incidence at face AB is \( 0^\circ \), and the light ray does not bend as it enters the prism.
Upon entering the prism, the light ray will travel along the interior of the prism. At the second interface, where the ray meets side BC, refraction occurs. The amount of bending depends on the refractive index of the prism material and the angle at which the ray strikes the second face of the prism.
For an isosceles right-angled prism, the angle between the two faces (AB and BC) is \( 45^\circ \). As the ray exits the prism, it will be refracted depending on the refractive index of the prism.
### Case 1: Refractive index \( n = \sqrt{2} \)
When the refractive index of the prism is \( n = \sqrt{2} \), the light ray will bend at the second interface. Using Snell's Law, we can calculate the angle of refraction at face BC:
\[
n_1 \sin(\theta_1) = n_2 \sin(\theta_2)
\]
where \( n_1 = 1 \) (refractive index of air), \( n_2 = \sqrt{2} \) (refractive index of the prism), and \( \theta_1 = 45^\circ \).
\[
\sin(\theta_2) = \frac{\sin(45^\circ)}{\sqrt{2}} = \frac{1/\sqrt{2}}{\sqrt{2}} = \frac{1}{2}
\]
Thus, \( \theta_2 = 30^\circ \). This means that the light ray will refract as it exits the prism, and the angle of refraction at face BC will be \( 30^\circ \).
### Case 2: Refractive index \( n = \sqrt{3} \)
When the refractive index of the prism is \( n = \sqrt{3} \), we follow the same steps to calculate the angle of refraction. Using Snell's Law again:
\[
\sin(\theta_2) = \frac{\sin(45^\circ)}{\sqrt{3}} = \frac{1/\sqrt{2}}{\sqrt{3}} = \frac{1}{\sqrt{6}}
\]
Thus, \( \theta_2 = 24.57^\circ \). In this case, the light ray will refract even more sharply as it exits the prism, and the angle of refraction at face BC will be \( 24.57^\circ \).
### Conclusion
For both cases, the ray passes through the prism with the direction of propagation changing due to the refractive indices of the material. The higher the refractive index of the material, the more the light will bend when exiting the prism. The ray will eventually exit the prism at an angle relative to the base of the triangle, depending on the refractive index.
