Consider an equilateral prism of refractive index 1.5 and a parallelepiped block of refractive index 2.0 arranged as shown in the figure such that their adjacent faces are parallel. A light ray enters the prism from air at an angle of incidence such that the ray travels through the prism parallel to its base. What is the angle of emergence $\theta$?
Show Hint
For parallel slabs/interfaces, the intermediate refractive index values do not affect the final angle of emergence.
You can directly link the initial medium (prism) to the final medium (air) using $n_{\text{prism}} \sin(r_2) = n_{\text{final}} \sin(\theta)$.
This shortcut avoids any calculations regarding the parallelepiped block entirely!
Step 1: Understanding the Question:
This problem involves a multi-interface ray-tracing scenario where light passes through a symmetric equilateral prism, a gap of air, and a parallel-sided block. We are asked to determine the final angle of emergence. Step 2: Key Formulas and Approach:
1. Snell's Law at any refracting interface:
\[ n_i \sin \theta_i = n_r \sin \theta_r \]
2. Geometry of an equilateral prism ($A = 60^\circ$):
When a ray travels parallel to the base, the refraction is symmetric:
\[ r_1 = r_2 = \frac{A}{2} = 30^\circ \]
3. For any series of parallel interfaces, the relation $n \sin \theta = \text{constant}$ holds true across all parallel media. Step 3: Detailed Explanation:
The equilateral prism has an angle of $A = 60^\circ$ and a refractive index $n_{\text{prism}} = 1.5$.
Since the ray inside the prism travels parallel to the base, the path is symmetric. The angle of refraction at the first face ($r_1$) and the angle of incidence on the second face ($r_2$) are equal:
\[ r_1 = r_2 = \frac{A}{2} = 30^\circ \]
The ray exits the second face of the prism into a thin parallel air gap ($n_{\text{air}} = 1.0$) at an angle of refraction $\theta_{\text{air}}$. By Snell's Law:
\[ n_{\text{prism}} \sin(r_2) = n_{\text{air}} \sin(\theta_{\text{air}}) \]
\[ 1.5 \sin(30^\circ) = 1.0 \sin(\theta_{\text{air}}) \]
\[ \sin(\theta_{\text{air}}) = 1.5 \times \frac{1}{2} = 0.75 = \frac{3}{4} \]
Since the adjacent faces of the prism and the block are parallel, the angle of incidence on the first face of the parallelepiped block is also $\theta_{\text{air}}$.
For a parallel-sided block of refractive index $n_{\text{block}} = 2.0$, the angle of emergence $\theta$ into air on the opposite side is related to the initial angle of incidence by:
\[ n_{\text{air}} \sin(\theta_{\text{air}}) = n_{\text{block}} \sin(\theta_{\text{block}}) = n_{\text{air}} \sin(\theta) \]
This simplifies to:
\[ \sin(\theta) = \sin(\theta_{\text{air}}) = \frac{3}{4} \]
Solving for the angle of emergence $\theta$:
\[ \theta = \sin^{-1}\left(\frac{3}{4}\right) \]
Step 4: Final Answer:
The angle of emergence is $\sin^{-1}(3/4)$, which corresponds to Option (A).
