Question

A point object $O$ is placed at a distance of $0.3m$ from a convex lens (focal length $0.2m)$ cut into two halves each of which is displaced by $0.0005m$ as shown in the figure.
What will be the location of the image?

• A. $30cm$, right of lens
• B. $60cm$, right of lens
• C. $70cm$, left of lens
• D. $40cm$, left of lens

Answer 1

The correct option is B $60cm$, right of lens

As explained in the each half lens will form an image in the same place. The optic axes of the lenses are displaced,
$\frac{1}{v}-\frac{1}{(-30)}=\frac{1}{20},v=60cm$
From similar triangles $O{I}_1{I}_2$ and $O{P}_1{P}_2$, we have
$\frac{I_1{I}_2}{P_1{P}_2}=\frac{u+v}{u}{I}_1{I}_2=\frac{90}{30}(2\times 0.05)=0.3cm$
Thus, the two images are $0.3cm$ apart.
Alternatively, imagine two arrows (see figure) that act as object for the lens.
Magnification, $m=\frac{v}{u}=\frac{(+60)}{(-30)}=-2$
Image of height of arrow is
$y=2\times \left(0.05\right)=0.10cm$
Thus, two inverted images are formed whose tips are at $I_1$ and $I_2$, respectively.
Thus, $I_1{I}_2=2y+△=(2\times 0.1)+0.1=0.3cm$.