The size of the image of an object, which is at infinity, as formed by a convex lens of focal length $30cm$ is $2cm$. If a concave lens of focal length $20cm$ is placed between the convex lens and the image at a distance of $26cm$ from the convex lens, calculate the new size of the images.

The correct option is A $2.5cm$
For the image formed by convex lens in first case,

Magnification=$\frac{v}{u}=\frac{30}{u}=\frac{2}{y_0}$

When concave lens is placed between the convex lens and image,

$u=30-26=4$

Thus $\frac{1}{v^{\prime }}-\frac{1}{4}=\frac{1}{-20}$

$v^{\prime }=5$

Total magnification in second case= Magnification by convex lens$\times$ Magnification by concave lens

$=\frac{30}{u}\times \frac{5}{4}=\frac{y}{y_0}$

Thus $\frac{y}{2}=\frac{5}{4}$

$y=2.5cm$