Let $f_{1}$, $f_{2}$ and $f_{3}$ be three focal lengths of the lenses.

Combined Focal length: $f=12cm$

So, $f1 =f_{1}1 +f_{2}1 +f_{3}1 $

When the third lens is removed, $f_{′′}1 =f_{1}1 +f_{2}1 $, where $f_{′′}=760 $

This means, $121 =760 1 +f_{3}1 $

$⇒f_{3}=−30cm$

Since $f_{3}$ is negative, it is a diverging lens having focal length $30cm$.

An object is placed at O on the common principal axis. The lens A produces an image at $I_{1}$ and this image acts as the object for the second lens B. The final image is produced at $I$ as shown in figure.

PO = u, object distance for the first lens (A),

PI = v, final image distance and

$PI_{1}=v_{1}$, image distance for the first lens (A) and also object distance for second lens (B).

For the image $I_{1}$ produced by the first lens A,

$v_{1}1 −u1 =f_{1}1 $ .... (1)

For the final image I, produced by the second lens B,

$v1 −v_{1}1 =f_{2}1 $ ... (2)

Adding equations (1) and (2),

$v1 −u1 =f_{1}1 +f_{2}1 $ ... (3)

If the combination is replaced by a single lens of focal length F such that it forms the image of O at the same position I, then

$v1 −u1 =F1 $ ... (4)

From equations (3) and (4),

$F1 =f_{1}1 +f_{2}1 $...... (5)

This F is the focal length of the equivalent lens for the combination.

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