Question

In figure (i), a lens of focal length $10\text{cm}$ is shown. It is cut into two parts and placed as shown in figure (ii).
An object AB of height $1\text{cm}$ is placed at a distance of $7.5\text{cm}$ from combination in figure (ii). The height of the image will be

• A. $2\text{cm}$
• B. $1\text{cm}$
• C. $1.5\text{cm}$
• D. $3\text{cm}$

Answer 1

The correct option is A $2\text{cm}$

Focal length of lens does not change when cut parallel to principle axis as there is no change in radii of curvature of surfaces.
Hence $f_1=f_2=10\text{cm}$
Effective focal length of combination in figure (ii) is
$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{10}+\frac{1}{10}$
$\therefore F=5\text{cm}$
Using lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{F}$
$\frac{1}{v}-\frac{1}{-7.5}=\frac{1}{+5}$
$\therefore v=+15\text{cm}$
Magnification of lens is
$m=\frac{v}{u}=\frac{+15}{-7.5}=-2$
$h_I=m{h}_O=(-2)1=-2\text{cm}$
$|h_I|=2\text{cm}$