Animated Solution for Physics - Optics: A convex lens $A$ of focal length $20\mathrm{cm}$ and a concave lens $B$ of focal length $5\mathrm{cm}$ are kept along the same axis with a distance $d$ between them. If a parallel beam of light falling on $A$ leaves $B$ as parallel beam, then $d$ is equal to ...... cm.

Imagine a parallel beam of light traveling through the vastness of space. It first encounters a convex lens, $A$, which possesses a focal length of $20\mathrm{cm}$. Shortly after, the light is intercepted by a concave lens, $B$, with a focal length of $5\mathrm{cm}$. These two lenses are placed coaxially, separated by an unknown distance $d$.

Our objective is to determine this exact separation distance $d$, given a very specific and beautiful optical condition: the light that enters the system as a parallel beam must also exit the system as a perfectly parallel beam.

Let's trace the journey of the light step by step. When a parallel beam of light strikes a convex lens, the laws of optics dictate that it must converge. Specifically, the lens bends the rays so that they aim to meet at its second principal focus.

Since the focal length of lens $A$ is $20\mathrm{cm}$, these rays are directed towards a point exactly $20\mathrm{cm}$ behind lens $A$. Let's call this convergence point $I_1$. If lens $B$ were not there, the light would simply focus at $I_1$ and then diverge again.

However, before the rays can reach $I_1$, they are intercepted by the concave lens $B$. The problem states a crucial condition: after passing through lens $B$, the rays become parallel once again.

Now, we must ask ourselves a fundamental question: When does a concave lens render converging rays parallel?

By the principle of reversibility of light, we know that parallel rays incident on a concave lens will diverge, appearing to originate from its first principal focus. Reversing this logic, if incident rays are converging exactly towards the first principal focus of a concave lens, the lens will bend them just enough to make them emerge parallel to the principal axis.

This is the master key to the problem! The point $I_1$, which is the focus of the convex lens $A$, must simultaneously be the focus of the concave lens $B$. They share the exact same focal point in space.

Since the focal length of lens $B$ is $5\mathrm{cm}$, the distance from lens $B$ to this shared focal point $I_1$ must be exactly $5\mathrm{cm}$.

Now, let's look at the simple geometry of our setup along the principal axis. The total distance from lens $A$ to the focus $I_1$ is $20\mathrm{cm}$. The distance from lens $B$ to the same focus $I_1$ is $5\mathrm{cm}$. The distance $d$ between the two lenses is simply the difference between these two values.

Substituting the given values:

Therefore, the two lenses must be separated by exactly $15\mathrm{cm}$ to achieve this optical effect. It is a beautiful interplay of converging and diverging powers, perfectly balanced by their spatial separation!