1. System Configuration

We have two converging lenses with focal lengths $f_1$ and $f_2$ separated by a distance $d$. Let the object be placed at a distance $u$ from the first lens. We require the size of the final image (magnification) to be independent of $u$.

2. Total Magnification Calculation

The total magnification $m$ is the product of individual magnifications: $m = m_1 \times m_2$.

For Lens 1: $$ m_1 = \frac{f_1}{f_1 + u} $$ Image position $v_1$ is given by $v_1 = \frac{u f_1}{u + f_1}$.

For Lens 2: The object for the second lens is the image formed by the first. The distance of this object from the second lens is $u_2 = -(d – v_1)$ (assuming standard sign conventions where light travels left to right). More simply, in terms of separation $d$, the magnification $m_2$ is: $$ m_2 = \frac{f_2}{f_2 + (v_1 – d)} $$ (Note: $v_1 – d$ is the effective object coordinate for lens 2 relative to its own optical center).

3. Analyzing the Dependency

Substitute $m_1$ and $m_2$ into the total magnification equation:

Substitute $v_1(f_1 + u) = u f_1$ (derived from $v_1 = \frac{u f_1}{u + f_1}$):

For the image size ($m$) to be independent of the object distance ($u$), the term involving $u$ in the denominator must be zero.

Physical Interpretation: This corresponds to a telescopic (afocal) system where the secondary focus of the first lens coincides with the primary focus of the second lens.