As shown in the diagram, when the incident ray is parallel to base of the prism, the emergent ray grazes along the second surface. If refractive index of the material of prism is $\sqrt{2}$, the angle $\theta$ of prism is :
Show Hint
Whenever a ray grazes the second surface of a prism with $\mu = \sqrt{2}$, the angle of incidence on that second face is always $45^\circ$.
Step 1: Understanding the Concept:
"Grazing along the second surface" means the ray is incident at the critical angle \( \theta_c \) on the internal face of the prism.
Step 2: Key Formula or Approach:
1. Critical angle: \( \sin \theta_c = \frac{1}{\mu} \)
2. Prism relation: \( r_1 + r_2 = A \) (where \( A = \theta \) is the prism angle)
Step 3: Detailed Explanation:
Given \( \mu = \sqrt{2} \), the critical angle is:
\[ \sin r_2 = \frac{1}{\sqrt{2}} \implies r_2 = 45^\circ \]
For a ray incident parallel to the base of an isosceles prism, the angle of incidence \( i \) is related to the prism angle.
If the prism angle is $\theta = 45^\circ$, and we assume a configuration where the ray enters normally to the first face ($i = 0$):
If \( i = 0 \), then \( r_1 = 0 \).
Using \( r_1 + r_2 = \theta \):
\[ 0 + 45^\circ = \theta \implies \theta = 45^\circ \]
This configuration satisfies the grazing condition exactly for \( \mu = \sqrt{2} \).
Step 4: Final Answer:
The angle of the prism is $45^\circ$.
