Animated Solution for Physics - Optics: A thin lens of refractive index 1.5 has a focal length of 15 cm in air. When the lens is placed in a medium of refractive index 4/3, its focal length will become ...... cm.

Have you ever opened your eyes underwater in a swimming pool? Everything looks blurry, doesn't it? Your eye's natural lens, which works perfectly in the air, suddenly loses its ability to focus light sharply when submerged in water. This everyday phenomenon is exactly what we are going to explore in this beautiful physics problem.

When light travels from one medium to another, it bends. This bending, known as refraction, is the fundamental principle behind how lenses work. A convex lens takes parallel rays of light and bends them inward, bringing them together at a specific point called the focal point. The distance from the optical center of the lens to this point is the focal length.

But here is the crucial secret: the focal length is not just a property of the lens itself. It is a property of the relationship between the lens and the environment it sits in. The lens bends light because it is optically denser than the air around it. If we change the environment, we change the bending power.

To quantify this, physicists use a brilliant equation known as the Lens Maker's Formula:

This formula tells us that the reciprocal of the focal length depends on two main factors. The first is the relative refractive index of the lens with respect to its surroundings (${\mu }_l/{\mu }_s$). The second is the physical curvature of its surfaces ($R_1$ and $R_2$). Since we are using the exact same lens in both air and the liquid medium, the radii of curvature will remain completely constant.

Let's start with the familiar scenario. Our lens is sitting in the air. The refractive index of the glass lens is given as ${\mu }_l=1.5$, and for air, we take ${\mu }_a=1$. Substituting these values, along with the given focal length of $15\text{ cm}$, we get our first equation:

Notice how the curvature term is isolated. We don't need to know the exact values of $R_1$ and $R_2$, because they will act as a constant multiplier that we can eliminate later.

Now, we take our lens and plunge it into a liquid medium. The surrounding refractive index is now ${\mu }_m=4/3$. We substitute this into the denominator of the relative refractive index term to find the new focal length, $f_{\text{medium}}$:

Simplifying the fraction $1.5/(4/3)$ gives us $4.5/4$, which is exactly $9/8$. Subtracting $1$ leaves us with $1/8$. This gives us our second equation:

In physics, we often find that we don't need to know every single detail to solve a problem. To eliminate the unknown curvature term, we simply divide the first equation by the second. The $\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ terms cancel out beautifully.

This is a profound result. It tells us that the focal length in this specific medium is exactly four times the focal length in air!

Finally, we just multiply $4$ by the original focal length of $15\text{ cm}$:

So, the new focal length is $60\text{ cm}$. The light rays now converge much further away because the lens has less bending power in the denser medium. The difference in optical density between the glass ($1.5$) and the liquid ($1.33$) is much smaller than between the glass ($1.5$) and air ($1.0$).

Before we conclude, let's push the boundaries of our understanding. What if we placed this glass lens in a liquid that is even denser than the glass itself? Say, a liquid with a refractive index of $1.6$?

The relative refractive index (${\mu }_l/{\mu }_s$) would become $1.5/1.6$, which is less than $1$. When we subtract $1$ in the Lens Maker's Formula, the entire term becomes negative! This means the focal length becomes negative. Our converging convex lens would suddenly start diverging light like a concave lens. Physics is full of such fascinating surprises!