(N/A) As the power of a lens is $P = \frac{1}{f}$,the equivalent power of a combination of lenses is $P = P_{1} + P_{2} + P_{3} + \ldots$
Note that this is an algebraic sum of individual powers.
Therefore,the sum of the terms on the right side may be positive for convex lenses and negative for concave lenses.
$A$ combination of lenses helps to obtain diverging or converging lenses of desired magnification. It also enhances the sharpness of the image.
$A$ combination of lenses is used in cameras,microscopes,telescopes,and other optical instruments.
Suppose $m_{1}$ and $m_{2}$ are the magnifications of a combination of two lenses as shown in the figure.
As shown in the figure for convex lens $L_{1}$,object distance $= OP = u$,image distance $= PI' = v'$.
Similarly,for convex lens $L_{2}$,object distance $= PI' = v'$,image distance $= PI = v$.
From the figure:
For lens $L_{1}$,magnification $m_{1} = \frac{v'}{u} \quad \ldots (1)$
For lens $L_{2}$,magnification $m_{2} = \frac{v}{v'} \quad \ldots (2)$
Now,the magnification for the lens combination is $m = \frac{v}{u}$.
Since $\frac{v}{u} = \frac{v}{v'} \times \frac{v'}{u}$,we have $m = m_{2} \times m_{1}$.
If a combination consists of more than two lenses,then $m = m_{1} \times m_{2} \times m_{3} \times \ldots \times m_{n}$.