Two parallel light rays 1 and 2 are incident from air on a system consisting of media P, Q, and air, as shown in the figure below. The incident angle is 45°. Ray 1 passes through medium P, air and medium Q and ray passes through media P and Q before leaving the system. After passing through the system, the angular deviation (in radians) between the two rays is The dimensions of the media and their refractive indices (\(n_a, n_P\) and \(n_Q\)) are shown in the figure
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Remember this fundamental rule: A ray of light passing through any number of parallel-sided transparent slabs will emerge parallel to its incident direction if the initial and final media are the same. This can save you from performing tedious calculations using Snell's law at each interface. The different path only introduces a lateral shift.
Step 1: Understanding the Concept:
The problem asks for the angular deviation between two initially parallel light rays after they pass through a system of parallel-sided media. A key principle of optics is that a light ray passing through one or more parallel-sided slabs and emerging into the original medium will be parallel to its incident direction. Step 2: Key Formula or Approach:
The principle is based on Snell's Law applied at each interface. For a ray entering a series of parallel slabs from a medium with index \(n_i\) and exiting into a final medium with index \(n_f\), the relationship between the initial angle of incidence \(\theta_i\) and the final angle of refraction \(\theta_f\) is given by \(n_i \sin\theta_i = n_f \sin\theta_f\). Step 3: Detailed Explanation:
Both rays, ① and ②, start in air (\(n_i = n_a = 1\)) and finally emerge into air (\(n_f = n_a = 1\)). The system consists of various media (P, Q, air) arranged as parallel horizontal slabs.
For any ray passing through such a system, we can apply the generalized form of Snell's Law. Let \(\theta_i\) be the initial angle of incidence in the first medium and \(\theta_f\) be the final angle of emergence in the last medium. The relationship is:
\[ n_{\text{initial}} \sin\theta_{\text{initial}} = n_{\text{final}} \sin\theta_{\text{final}} \]
In this problem, for both rays, the initial medium is air and the final medium is also air. So, \(n_{\text{initial}} = n_{\text{final}} = n_a = 1\).
Therefore, for both ray ① and ray ②:
\[ 1 \cdot \sin\theta_{\text{incident}} = 1 \cdot \sin\theta_{\text{emergent}} \]
\[ \sin\theta_{\text{incident}} = \sin\theta_{\text{emergent}} \implies \theta_{\text{incident}} = \theta_{\text{emergent}} \]
This means that the emergent ray is parallel to the incident ray for both ray ① and ray ②. The paths of the rays will be laterally shifted, but their final direction of propagation will be the same as their initial direction.
The initial rays ① and ② are parallel to each other.
The emergent ray for ① is parallel to the incident ray ①.
The emergent ray for ② is parallel to the incident ray ②.
Since incident rays ① and ② are parallel, the emergent rays for ① and ② must also be parallel to each other.
When two rays are parallel, the angle between them is zero. Therefore, the angular deviation between the two rays is 0. Step 4: Final Answer:
The emergent rays are parallel to their respective incident rays. Since the incident rays are parallel to each other, the emergent rays will also be parallel to each other. The angular deviation between them is therefore 0. Option (A) is correct.
