Refraction of Light Ray Through Lens Shaped Media

To determine the correct matching between Column I and Column II, we analyze the refraction of the light ray at each interface using Snell's Law.

The light ray travels from right to left:

1. First interface: from medium with refractive index $\mu_1$ to the lens of refractive index $\mu_2$.
2. Second interface: from the lens of refractive index $\mu_2$ to medium with refractive index $\mu_3$.

Recall the fundamental rules of refraction:

• If a ray travels from a rarer medium to a denser medium ($\mu_{\text{incident}} < \mu_{\text{refracted}}$), it bends towards the normal.
• If a ray travels from a denser medium to a rarer medium ($\mu_{\text{incident}} > \mu_{\text{refracted}}$), it bends away from the normal.
• If there is no deviation at an interface (the ray passes undeviated), the refractive indices of the two adjacent media must be equal ($\mu_{\text{incident}} = \mu_{\text{refracted}}$).

Analysis of the Ray Diagrams in Column II:

1. Diagram (p):

   • The lens is convex. The incident ray is horizontal in medium $\mu_1$ and strikes the upper portion of the first curved surface (which is convex towards the right, so its center of curvature lies to the left, inside the lens).
   • The normal to this first surface at the point of incidence points downward and to the left.
   • The ray bends downwards into the lens, which is towards the normal. Thus, $\mu_1 < \mu_2$.
   • At the second interface (between $\mu_2$ and $\mu_3$), the ray passes straight through without any deviation. Thus, $\mu_2 = \mu_3$.
   • Therefore, diagram (p) corresponds to:
     $\mu_1 < \mu_2 \quad \text{and} \quad \mu_2 = \mu_3$

2. Diagram (q):

   • The lens is convex. The incident ray is horizontal in medium $\mu_1$.
   • At the first interface, the normal points downward and to the left. The ray bends upwards, which is away from the normal. Thus, $\mu_1 > \mu_2$.
   • At the second interface, the ray continues to bend further away from the normal as it emerges into medium $\mu_3$. Thus, $\mu_2 > \mu_3$.
   • Therefore, diagram (q) corresponds to:
     $\mu_1 > \mu_2 \quad \text{and} \quad \mu_2 > \mu_3$

3. Diagram (r):

   • The lens is concave. The first surface has its center of curvature to the right (outside the lens).
   • For the upper half, the normal to the first surface points upward and to the left (towards the inside of the lens).
   • The ray bends upwards, which is towards the normal. Thus, $\mu_1 < \mu_2$.
   • At the second interface, the ray passes straight without any deviation into medium $\mu_3$. Thus, $\mu_2 = \mu_3$.
   • Therefore, diagram (r) corresponds to:
     $\mu_1 < \mu_2 \quad \text{and} \quad \mu_2 = \mu_3$

4. Diagram (s):

   • The lens is concave.
   • At the first interface, the normal points upward and to the left. The ray bends downwards, which is away from the normal. Thus, $\mu_1 > \mu_2$.
   • At the second interface, the ray bends further downward, away from the normal. Thus, $\mu_2 > \mu_3$.
   • Therefore, diagram (s) corresponds to:
     $\mu_1 > \mu_2 \quad \text{and} \quad \mu_2 > \mu_3$

5. Diagram (t):

   • The lens is concave.
   • At the first interface, the ray bends downwards, which is away from the normal. Thus, $\mu_1 > \mu_2$.
   • At the second interface, the ray passes undeviated into medium $\mu_3$. Thus, $\mu_2 = \mu_3$.
   • Therefore, diagram (t) corresponds to:
     $\mu_1 > \mu_2 \quad \text{and} \quad \mu_2 = \mu_3$

Matching the Columns:

• A) $\mu_1 < \mu_2$: matches with p, r
• B) $\mu_1 > \mu_2$: matches with q, s, t
• C) $\mu_2 = \mu_3$: matches with p, r, t
• D) $\mu_2 > \mu_3$: matches with q, s