Obtain the equivalent focal length of a combination of thin lenses placed in contact.

(N/A) Consider two lenses $A$ and $B$ of focal lengths $f_{1}$ and $f_{2}$ placed in contact with each other. Let the object be placed at a point $O$ beyond the focus of the first lens $A$.
The first lens produces an image at $I_{1}$. Since the image $I_{1}$ is real,it serves as a virtual object for the second lens $B$,producing the final image at $I$.
Formation of the image by the first lens is presumed only to facilitate the determination of the position of the final image. In fact,the direction of rays emerging from the first lens gets modified in accordance with the angle at which they strike the second lens.
Since the lenses are thin,we assume the optical centers of the lenses to be coincident. Let this central point be denoted by $P$.
For the image formed by the first lens $A$:
$\frac{1}{v_{1}} - \frac{1}{u} = \frac{1}{f_{1}} \quad \dots (1)$
For the image formed by the second lens $B$:
$\frac{1}{v} - \frac{1}{v_{1}} = \frac{1}{f_{2}} \quad \dots (2)$
Adding equation $(1)$ and $(2)$:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \quad \dots (3)$
If the two-lens system is regarded as equivalent to a single lens of focal length $f$,we have:
$\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
The derivation is valid for any number of thin lenses in contact. If several thin lenses of focal lengths $f_{1}, f_{2}, f_{3}, \dots, f_{n}$ are in contact,the effective focal length of their combination is given by:
$\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} + \frac{1}{f_{3}} + \dots + \frac{1}{f_{n}}$