Animated Solution for Physics - Optics: A convex lens of focal length $20\text{ cm}$ produces images of the same magnification $2$ when an object is kept at two distances $x_1$ and $x_2$ ($x_1>x_2$) from the lens. The ratio of $x_1$ and $x_2$ is

Imagine you are given a convex lens and told it produces an image twice the size of the object. Sounds simple, right? But there is a hidden trap here!

The problem states the magnification is $2$, but it deliberately omits whether the image is real or virtual. A convex lens is a versatile optical tool. It can form a real, inverted magnified image when the object is placed between $f$ and $2f$. It can also form a virtual, erect magnified image when the object is placed between the optical center and the focus.

To solve this problem, we must explore both possibilities using our standard optical equations.

We will rely on two fundamental formulas. First, the magnification formula:

Second, the thin lens formula:

Let's apply these to our two distinct scenarios.

The Real and Inverted Image

Let's assume the lens forms a real image. By sign convention, a real image formed by a single lens is always inverted, which means the magnification must be negative.

So, we set $m=-2$. Using the magnification formula:

Now, we substitute this relation and the given focal length $f=+20\text{ cm}$ into the lens formula:

To solve for $u$, we find a common denominator:

Cross-multiplying gives us:

Since distance is a positive scalar, our first object distance is $x_1=30\text{ cm}$.

The Virtual and Erect Image

Now, let's consider the second possibility: a virtual image. A virtual image is always erect, meaning the magnification is positive.

So, we set $m=+2$. Using the magnification formula:

We substitute this new relation into the lens formula:

Again, finding a common denominator:

Cross-multiplying yields:

Thus, our second object distance is $x_2=10\text{ cm}$. Notice that $x_1>x_2$ ($30>10$), which perfectly aligns with the condition given in the problem!

We have successfully navigated the trap and found both object distances. The final step is to calculate their ratio.

Substituting our values:

This problem is a brilliant exercise in remembering that a single magnitude of magnification can correspond to two entirely different physical setups in optics!