Animated Solution for Physics - Optics: Two transparent media of refractive indices ${\mu }_1$ and ${\mu }_3$ have a solid lens shaped transparent material of refractive index ${\mu }_2$ between them as shown in figures in Column II. A ray traversing these media is also shown in the figures. In Column I different relationships between ${\mu }_1,{\mu }_2$ and ${\mu }_3$ are given. Match them to the ray diagram shown in Column II.

When a ray of light travels across the boundary of two transparent media, it undergoes refraction. The direction in which the ray bends is governed by Snell's Law, which states that light bends towards the normal when entering a denser medium (higher refractive index) and away from the normal when entering a rarer medium (lower refractive index). In this problem, we are tasked with analyzing the bending of light through various lens-shaped materials to deduce the relative refractive indices of the surrounding media.

Before we dive into the specific diagrams, it is vital to understand how to draw the normal at a spherical surface. The normal at any point on a spherical surface always passes through its center of curvature.

For a convex surface (bulging towards the incoming light), the center of curvature lies on the opposite side. Thus, the normal points outwards and away from the principal axis. Conversely, for a concave surface (caving inwards), the center of curvature lies on the same side as the incoming light, meaning the normal points inwards towards the principal axis.

Let's systematically break down the behavior of the light ray in each of the given diagrams:

Diagram (p): The ray enters the first convex surface and bends downwards, towards the principal axis. Since the normal at this upper convex surface points upwards and outwards, bending downwards means the ray is bending towards the normal. This implies that the light is entering a denser medium, so ${\mu }_1<{\mu }_2$. At the second convex surface, the ray passes straight through without any deviation. A lack of bending at a boundary (when not incident normally) indicates that the two media have identical refractive indices. Therefore, ${\mu }_2={\mu }_3$.

Diagram (q): Here, the incident ray is angled downwards. Upon hitting the first surface, it bends to become horizontal. This means the angle of refraction is greater than the angle of incidence, indicating that the ray has bent away from the normal. Thus, the light is entering a rarer medium, giving us ${\mu }_1>{\mu }_2$. At the second surface, the ray bends upwards, again bending away from the normal. This implies another transition into a rarer medium, so ${\mu }_2>{\mu }_3$.

Diagram (r): The incident ray is angled upwards. At the first surface, it bends to become horizontal, which means it has bent towards the normal (the angle with the normal has decreased). This indicates a transition into a denser medium, so ${\mu }_1<{\mu }_2$. At the second surface, the ray continues horizontally without any deviation. As established earlier, no bending means the refractive indices are equal, yielding ${\mu }_2={\mu }_3$.

Diagram (s): The ray is incident horizontally on a convex surface and bends upwards. Given the outward-pointing normal, bending upwards means the ray is bending away from the normal. This signifies a transition into a rarer medium, so ${\mu }_1>{\mu }_2$. At the second surface, the ray bends even further upwards, again bending away from the normal. This indicates another drop in refractive index, meaning ${\mu }_2>{\mu }_3$.

Diagram (t): The ray is incident horizontally on a concave surface and bends downwards. For a concave surface, the normal points inwards. Bending downwards in this context means the ray is bending away from the normal, implying ${\mu }_1>{\mu }_2$. At the second surface, the ray passes straight through without any further deviation, which tells us that ${\mu }_2={\mu }_3$.

By synthesizing our observations, we can confidently match the relationships in Column I with the diagrams in Column II:

(A) ${\mu }_1<{\mu }_2$ is observed in diagrams (p) and (r). (B) ${\mu }_1>{\mu }_2$ is observed in diagrams (q), (s), and (t). (C) ${\mu }_2={\mu }_3$ is observed in diagrams (p), (r), and (t). (D) ${\mu }_2>{\mu }_3$ is observed in diagrams (q) and (s).

This logical breakdown not only solves the problem but also reinforces the fundamental principles of geometric optics.