Animated Solution for Physics - Optics: What is the position and nature of image formed by lens combination shown in figure? (where, $f_1$ and $f_2$ are focal lengths)

The behavior of light as it passes through multiple lenses is the foundational principle behind almost all complex optical instruments, from the microscope in a biology lab to the telephoto lens of a wildlife photographer. In this problem, we are tasked with tracking the journey of light through a two-lens system: a converging (convex) lens followed by a diverging (concave) lens.

We are given a system with two lenses placed along the same principal axis. - Lens A is a convex lens with a focal length $f_1=+5\text{ cm}$. - Lens B is a concave lens with a focal length $f_2=-5\text{ cm}$. - The separation between the two lenses is $d=2\text{ cm}$.

An object is placed $20\text{ cm}$ to the left of Lens A. According to our standard Cartesian sign convention, the direction of incident light is taken as positive. Therefore, the object distance for the first lens is $u_1=-20\text{ cm}$. Our goal is to find the final position and nature of the image after the light has refracted through both lenses.

Lens A

To solve multi-lens problems, we take it one step at a time. First, we completely ignore Lens B and imagine where Lens A alone would form the image. We apply the thin lens formula for Lens A:

Substituting our known values:

Notice how the double negative becomes a positive. Moving it to the other side, we get:

To subtract these fractions, we find a common denominator, which is $20$:

Inverting this gives us the position of the first intermediate image, $I_1$:

Because $v_1$ is positive, this image is real and is formed to the right of Lens A.

Here is where the physics gets incredibly interesting. The light rays are converging towards the point $I_1$, which is $\frac{20}{3}\text{ cm}$ (or about $6.67\text{ cm}$) to the right of Lens A. However, Lens B is sitting just $2\text{ cm}$ away from Lens A!

This means the light rays are intercepted by Lens B before they can actually meet to form $I_1$. Because the incident rays on Lens B are converging towards $I_1$, this intermediate image $I_1$ acts as a virtual object for Lens B.

We need to find the object distance $u_2$ for Lens B. Since $I_1$ is $\frac{20}{3}\text{ cm}$ from Lens A, and Lens B is $2\text{ cm}$ to the right of Lens A, the distance from Lens B to $I_1$ is:

The positive sign is crucial here. It mathematically represents that the object is virtual and lies in the direction of the incident light.

Lens B

Now, we apply the lens formula one last time for Lens B. We use our new virtual object distance $u_2=+\frac{14}{3}\text{ cm}$ and the focal length of the concave lens $f_2=-5\text{ cm}$:

Substituting the values:

Moving the object term to the right side:

The least common multiple of $14$ and $5$ is $70$. Let's adjust the numerators:

Inverting this gives us our final image position:

The positive sign of $v_2$ tells us that the final image is formed $70\text{ cm}$ to the right of Lens B. Because it is formed on the side of the transmitted light, the refracted rays actually converge at this point, making it a real image.

By breaking the system down into isolated, sequential events and strictly adhering to the sign convention, we successfully navigated the complex path of light through a compound optical system!