Animated Solution for Physics - Optics: Find the distance of the image from object $O$, formed by the combination of lenses in the figure.

A Triple Lens Relay

Imagine light as a runner in a relay race, passing through a beautifully orchestrated sequence of three lenses: a convex lens ($L_1$), a concave lens ($L_2$), and another convex lens ($L_3$). Our objective is to track the journey of light originating from an object $O$ placed $30\text{ cm}$ in front of the first lens, and ultimately find the exact location of the final image.

To solve this, we must break the problem down into three distinct legs. The image formed by the first lens will act as the object for the second lens, and the image from the second will act as the object for the third.

The First Convex Lens

The light begins its journey by striking the first convex lens, $L_1$. We apply the standard thin lens formula:

Given the object is on the left, we use the Cartesian sign convention, making $u_1=-30\text{ cm}$. The focal length of the convex lens is $f_1=+10\text{ cm}$. Substituting these values:

This tells us that $L_1$ is attempting to form a real image $I_1$ exactly $15\text{ cm}$ to its right.

The Concave Interceptor

But wait! Before the light rays can converge to form $I_1$, they are intercepted by the concave lens $L_2$, which is situated just $5\text{ cm}$ behind $L_1$.

Because the rays are converging towards a point behind $L_2$, the image $I_1$ acts as a virtual object for $L_2$. The distance from $L_2$ to this virtual object is $15\text{ cm}-5\text{ cm}=10\text{ cm}$. Since it lies to the right of $L_2$, the object distance is positive: $u_2=+10\text{ cm}$.

Now, we apply the lens formula for $L_2$, which has a focal length $f_2=-10\text{ cm}$:

This is a magical moment in optics! The concave lens perfectly neutralizes the convergence of the incoming rays, sending them out perfectly parallel to each other.

The Final Convergence

Finally, this bundle of parallel rays strikes the third lens, $L_3$. Because the incident rays are parallel, the object distance for $L_3$ is effectively infinity ($u_3=\infty$).

For any convex lens, parallel incident rays are forced to converge exactly at its focal plane. Since the focal length of $L_3$ is $f_3=+30\text{ cm}$, the final image $I_3$ is formed $30\text{ cm}$ to the right of $L_3$:

The question specifically asks for the total distance from the original object $O$ to the final image $I_3$. We simply sum up all the segments of our relay race:

- Object to $L_1$: $30\text{ cm}$ - $L_1$ to $L_2$: $5\text{ cm}$ - $L_2$ to $L_3$: $10\text{ cm}$ - $L_3$ to Final Image: $30\text{ cm}$

Total Distance $=30+5+10+30=75\text{ cm}$.

By systematically treating each lens as an independent stage, even the most complex optical systems become beautifully simple to unravel.