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, calculate the new size of the image.

• 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 $I_1. I_1$ 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}{20} $
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 $ =\frac{v}{u}$
$ = \frac{\text{size of image}}{\text{size of object}} = \frac{5}{4}$
$\therefore \frac{\text{size of image}}{2} = 1.25$
$\therefore $ size of image due to concave lens $= 2.5 \,cm$