# What is the position and nature of the image formed by lens combination shown in the figure? (f1, f2 are focal lengths )

Thin lens formula -

$\frac{1}{v}-\frac{1}{u}= \frac{1}{f}$

- wherein

$u \, and \, v$ are object and image distance from the lens.

For the first lens

$u_{1}=-20,\, \, \, f_{1}=5cm$

so, $\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f}$

$\frac{1}{v}-\frac{1}{-\left ( 20 \right )}=\frac{1}{5}$

$v_{1}=\frac{20}{3}$

So  $u_{2}$  for 2nd  lens is given by

So  $u_{2}=\frac{20}{3}-2=\frac{14}{3}$

$f_{2}=-5cm$

So  $\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$

$\frac{1}{v_{2}}-\frac{1}{\left ( 1_{\frac{4}{3}} \right )}=\frac{1}{-5}$

$V_{2}=70cm$   and the image is real.

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