Animated Solution for Physics - Optics: The same size images are formed by a convex lens when the object is placed at 20 cm or at 10 cm from the lens. The focal length of convex lens is ………… cm.

The problem of "same size images" in optics is a classic trap that has caught countless students off guard. Let's break down the physics and the math behind this fascinating phenomenon.

When a problem states that a convex lens forms images of the same size for two different object distances, it is crucial to interpret this correctly.

Many students mistakenly assume that the image is the same size as the object (which would mean $|m|=1$). However, if you try to solve the equations with that assumption, you will quickly run into a mathematical contradiction! The problem actually means that the two images are equal in size to each other.

Furthermore, a convex lens cannot form two real images of the same size for different object distances. As the object moves away from the focus, the real image strictly diminishes in size. Therefore, to get the same size, one image must be virtual and erect (when the object is close to the lens), and the other must be real and inverted (when the object is further away).

This physical reality gives us our master constraint: the magnifications must be equal in magnitude but opposite in sign.

To avoid calculating the image distance $v$ for both cases, we can use a powerful shortcut. We know the standard magnification formula:

By substituting $v$ from the thin lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, we can express magnification purely in terms of focal length $f$ and object distance $u$:

This equation is a massive time-saver in competitive exams!

Now, let's apply our master equation to the two cases given in the problem.

For the first case, the object is at $10\text{ cm}$. Using the standard sign convention, $u_1=-10\text{ cm}$.

For the second case, the object is at $20\text{ cm}$, so $u_2=-20\text{ cm}$.

We established earlier that $m_1=-m_2$. Let's substitute our expressions into this constraint:

Notice how beautifully the $f$ in the numerator cancels out on both sides! This leaves us with a simple linear equation:

Cross-multiplying carefully to avoid sign errors, we get:

Expanding the negative sign:

Bringing the $f$ terms to one side:

And there we have it! The focal length of the convex lens is exactly $15\text{ cm}$. By understanding the physical nature of the images rather than blindly plugging in numbers, we navigated the trap perfectly.