# A convex lens of focal length 20 cm and a concave lens of focal length 15cm are kept 30cm apart with their principal axes coincident. When an object is placed 30cm in front of the convex lens, calculate the position of the final image formed by the combination. Would this result change if the object were placed 30cm in front of the concave lens? Give reason.

We have given,

$f_{1}=20cm$ = focal length for convex lens

$f_{2}=15cm$ = focal length for concave lens

Now,

The image formed by the convex lens can be determined as -

$\frac{1}{f_{1}}= \frac{1}{v}-\frac{1}{u}$             $\left (\because u = +30 \right )$

So,

$\frac{1}{20}= \frac{1}{v}+\frac{1}{30}$

$\frac{1}{v}=\frac{1}{20}-\frac{1}{30}$

$\frac{1}{v}=\frac{30-20}{20\times 30}$

$\frac{1}{v}=\frac{10}{600}$

$v = \frac{600}{10}$

$v = 60cm$

Now, image distance for the concave lens, we have u = +30cm for concave lens

$\frac{1}{f_{2}}= \frac{1}{v}-\frac{1}{u}$

$\frac{1}{-15}= \frac{1}{v}-\frac{1}{30}$

$\frac{1}{v}=\frac{1}{-15}+\frac{1}{30}$

$v=-30cm$

therefore, $v=-30cm$

Hence, according to the principle of reversibility, the result will not change, if the object is placed 30cm in front of the concave lens.

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