1. Analyzing the Geometric Configuration

We are given four dots representing the locations of sources and their images. Three dots form an equilateral triangle, and the fourth is at the centroid.

A single converging lens maps points (objects) to points (images). For a specific source $S$ and its image $I$, the line connecting them must pass through the Optical Center $O$ of the lens.

Let the vertices of the equilateral triangle be $A, B, C$ and the centroid be $G$. Due to the symmetry of the configuration (isosceles/equilateral geometry), the Optical Axis likely aligns with one of the medians of the triangle. Let’s assume the optical axis passes through vertex $B$ and the centroid $G$, intersecting the opposite side $AC$ at its midpoint $M$.

If the lens is placed such that its optical center $O$ coincides with the midpoint $M$ of side $AC$, then the points $B, G,$ and $O$ are collinear. This allows for a source at $G$ to map to an image at $B$ (or vice versa) along the principal axis.

2. Establishing the Scale

The problem states that the two point sources are $24 \text{ cm}$ apart. We need to identify which pair of points corresponds to this distance.

Let the side length of the equilateral triangle be $a$. The altitude $h$ is given by: $$ h = \frac{\sqrt{3}}{2} a $$ The distance from a vertex to the centroid is $\frac{2}{3}h = \frac{a}{\sqrt{3}}$.

From the handwritten notes, the distance constraint is applied to the distance between the centroid $G$ and the vertex $B$. $$ \text{Distance } GB = \frac{a}{\sqrt{3}} = 24 \text{ cm} $$ Solving for the side length $a$: $$ a = 24\sqrt{3} \text{ cm} $$

3. Applying the Lens Formula

We consider the pair of points on the optical axis: the Centroid $G$ and the Vertex $B$. Let $O$ be the optical center located at the midpoint of the base $AC$.

• Position of Centroid ($G$): The centroid lies at a distance of $\frac{1}{3}h$ from the base. $$ u = OG = \frac{1}{3} \left( \frac{\sqrt{3}}{2} a \right) = \frac{1}{3} \frac{\sqrt{3}}{2} (24\sqrt{3}) = 12 \text{ cm} $$
• Position of Vertex ($B$): The vertex lies at a distance of $h$ from the base. $$ v = OB = \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{2} (24\sqrt{3}) = 36 \text{ cm} $$

Assuming $G$ is the real object and $B$ is the real image (formed by a converging lens), we apply the lens formula: $$ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $$ Substituting the values: $$ \frac{1}{f} = \frac{1}{36} + \frac{1}{12} $$ $$ \frac{1}{f} = \frac{1 + 3}{36} = \frac{4}{36} = \frac{1}{9} $$