Physics - Geometrical Optics Question with Solution | TestHub

When concave lens is put in contact with convex lens, real/image is formed i.e., equivalent lens is convex. The focal length of equivalent lens F is greater than focal length of convex lens.

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

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

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

As f increases,$\frac{u}{f}$decreases

$\left(\frac{u}{f}-1\right)$also decreases

Thus, v increases.