CMP: The following figure shows a simple version of a zoom

The converging lens has focal length $f_{1’}$ and the diverging lens has focal length $f_2 = –|f_2|$. The two lenses are separated by a variable distance d that is always less than $f_{1’}$ also the magnitude of the focal length of the diverging lens satisfies the inequality $|f_2| > (f_1 –d)$. If the rays that emerge from the diverging lens and reach the final image point are extended backward to the left to the diverging lens, they will eventually expand to the original radius $r_0$ at the same point Q. To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius $r_0$ entering the converging lens.

To the right of the diverging lens the final image I’ is formed at a distance $s’_2$.

$\frac{(f_1-d)|f_2|}{|f_2|-f_1+d}$

The image at focal point of the first lens, a distance $f_1$ to the right of first lens, serves as the object for the second lens. The image is at distance $f_1$–d to the second lens so

$s_2=–(f_1 –d) = d – f_1$ so $s’_2 =\frac{s_2f_2}{s_2-f_2}=\frac{(f_1-d)f_2}{|f_2|-f_1+d}$