Question

A right-angled prism ABC (refractive index \( \sqrt{2} \)) is kept on a plane mirror as shown in the figure. A ray of light is incident normally on the face AC. Trace the path of the ray as it passes through the prism.

Answer 1

Path of the Ray through a Prism

Given:

• Refractive index of the prism: \( n = \sqrt{2} \)
• Angle of the prism at \( A = 90^\circ \)
• Angle of incidence on face AC: \( 0^\circ \) (since the light is incident normally)
• Angle \( \angle BAC = 60^\circ \)

The path of the ray follows these steps:

1. The ray is incident normally on the face AC of the prism, so the angle of incidence is \( 0^\circ \).
2. The ray passes through the prism without deviation until it reaches the second face AB.
3. The ray refracts at the interface between the two media at the face AB.

To find the angle of refraction at face AB, we apply Snell’s Law at the boundary:

\[ n_{\text{prism}} \sin \theta_1 = n_{\text{air}} \sin \theta_2 \]

Where:

• \( \theta_1 \) is the angle inside the prism
• \( \theta_2 \) is the angle of refraction in air

Since the light is passing from a medium with refractive index \( \sqrt{2} \) (the prism) into air with refractive index 1, we can apply Snell’s law to find the angle of refraction:

\[ \sqrt{2} \sin \theta_1 = 1 \cdot \sin \theta_2 \]

Thus, using this relation, you can find the angle \( \theta_2 \), the angle of refraction in air.