Question

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, the new size of the image is

• A. 1.25 cm
• B. 2.5 cm
• C. 1.05 cm
• D. 2 cm

Answer 1

Answer: (B)

Convex lens forms the image at $I_{1} \cdot I_{1}$ is at the second focus of convex lens. Size of $I_{1}=2 \,cm$.
$I_{1}$ acts as virtual object for concave lens. Concave lens forms the image of $I_{1}$ at $I_{2}$.

Lens formula : $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
For concave lens,
$\frac{1}{v}-\frac{1}{4}=-\frac{1}{2}$
or $\frac{1}{v}=-\frac{1}{20}+\frac{1}{4}=\frac{4}{20}=\frac{1}{5}$
or $v=5 \,cm $= Distance of $I_{2}$ from concave lens.
$\therefore $ Magnification of concave lens $=\frac{v}{u}=\frac{5}{4}=1.25$
As size of image at $I_{1}$ is $2 cm$, therefore size of image at
$I_{2}=2 \times 1.25=2.5 \,cm$