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

Convex lens will form image ${\mathrm{I}}_1$, at its focus which acts like a virtual object for concave lens.

Hence, for concave lens u = + 4 cm, f = -20 cm. So by the lens formula, 

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

$\begin{array}{rcl} & \Rightarrow & \frac{1}{-20}=\frac{1}{\mathrm{v}}-\frac{1}{4}\\ & \Rightarrow & \frac{1}{\mathrm{v}}=-\frac{1}{20}+\frac{1}{4}\\ & \Rightarrow & \mathrm{v} = 5 \mathrm{cm}\end{array}$

The final image is formed at a distance of 5 cm from the concave lens.

$\text{B}y\mathrm{using}, \frac{\mathrm{v}}{\mathrm{u}}=\frac{\mathrm{I}}{\mathrm{O}}\Rightarrow \frac{5}{4}=\frac{\mathrm{I}}{2}$

Where I = size of the image, O = size of the object = 2 cm

On solving, I = 2.5 cm

The size of the new image is 2.5 cm