Animated Solution for Physics - Optics: A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is

Tracing Light Through a Diverging and Converging Setup

Imagine a fascinating optical setup involving two lenses working in tandem. First, we have a diverging lens, and exactly $15\text{ cm}$ away from it, a converging lens. A beam of parallel light rays comes in from the left and strikes the diverging lens. Our goal is to track these rays and find out exactly where the final image is formed and what its nature is.

The Diverging Lens (First Refraction)

What happens when parallel rays hit a diverging lens? As the name suggests, they diverge! However, if we trace these diverging rays backward, they appear to originate from a single point on the principal axis. This point is the principal focus of the diverging lens.

The focal length of this diverging lens is given as $25\text{ cm}$. Because it's a diverging lens, its focal length is taken as negative, $f_1=-25\text{ cm}$. Therefore, the virtual image, let's call it $I_1$, is formed exactly $25\text{ cm}$ to the left of the diverging lens.

The Handover (Image becomes Object)

Now comes the crucial part of any multiple-lens problem. The image formed by the first lens, $I_1$, acts as a real object for our second lens, the converging lens. We need to find the exact distance of this object from the converging lens.

The image $I_1$ is $25\text{ cm}$ to the left of the diverging lens, and the converging lens is another $15\text{ cm}$ to the right of the diverging lens. Adding these two distances together, we get the total object distance for the converging lens:

We take it as negative because the object is to the left of the converging lens, following standard sign conventions.

The Converging Lens (Final Refraction)

We know the focal length of the converging lens is $+20\text{ cm}$. Let's plug our values into the thin lens formula to find the final image position $v$:

Substituting $f=+20\text{ cm}$ and $u=-40\text{ cm}$:

The two negative signs become a positive:

Moving the $\frac{1}{40}$ to the other side, we get:

Taking the common denominator of $40$, we find:

Therefore, $v=+40\text{ cm}$.

A positive value for $v$ means the final image is real and is formed $40\text{ cm}$ to the right of the converging lens. This perfectly matches option (d). By carefully tracking the light rays step-by-step and respecting sign conventions, even complex multi-lens systems become simple to solve!