Animated Solution for Physics - Optics: What is the relation between the refractive indices ${\mu }_1$ and ${\mu }_2$ ? If the behaviour of light rays is as shown in the figure.

This is a classic conceptual problem that tests your understanding of how lenses behave when placed in different optical media. Let's break down the physics behind the two scenarios presented in the figure.

In the first figure, we observe a biconvex lens of refractive index $\mu$ submerged in a medium of refractive index ${\mu }_1$. Notice the path of the parallel light rays entering the lens. They pass straight through without any deviation!

For light to pass through an optical boundary without bending, the refractive index on both sides of the boundary must be identical. The lens is effectively invisible to the light rays. This immediately tells us that the refractive index of the lens is equal to the refractive index of the surrounding medium.

Therefore, we can establish our first relation:

Now, let's turn our attention to the second figure. The same biconvex lens is now placed in a different medium with a refractive index ${\mu }_2$. This time, the parallel rays entering the lens diverge away from the principal axis.

Wait a minute! A biconvex lens is typically a converging lens. Why is it acting like a diverging (concave) lens here? This anomaly occurs when the surrounding medium is optically denser than the material of the lens itself.

We can verify this mathematically using the Lens Maker's Formula:

For a biconvex lens, the geometric term $\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ is always positive. If the lens is diverging light, its focal length $f$ must be negative. This implies that the term $\left(\frac{{\mu }_{\text{lens}}}{{\mu }_{\text{medium}}}-1\right)$ must be negative, which means ${\mu }_{\text{medium}}>{\mu }_{\text{lens}}$.

Applying this to our specific case, the surrounding medium ${\mu }_2$ must be greater than the lens's refractive index $\mu$.

We now have two critical pieces of information:

1. ${\mu }_1=\mu$ 2. ${\mu }_2>\mu$

By combining these two relations, it becomes crystal clear that ${\mu }_1$ must be strictly less than ${\mu }_2$.