For an object placed at a distance $2.4\ \text{m}$ from a lens, a sharp focused image is observed on a screen placed at a distance $12\ \text{cm}$ from the lens. A glass plate of refractive index $1.5$ and thickness $1\ \text{cm}$ is introduced between the lens and the screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen?
Answer: (B) $3.2\ \text{m}$
Focal length: $\dfrac1f = \dfrac1{12} + \dfrac1{240} = \dfrac{21}{240}$.
The plate shifts the image away from the lens by $t\left(1 - \dfrac1\mu\right) = \dfrac13\ \text{cm}$, so the lens itself must now form its image at $v = 12 - \dfrac13 = \dfrac{35}{3}\ \text{cm}$.
$$\frac1u = \frac1v - \frac1f = \frac{3}{35} - \frac{21}{240} = -\frac{1}{560}\ \Rightarrow\ |u| = 560\ \text{cm} = 5.6\ \text{m}$$
The object must be moved $5.6 - 2.4 = 3.2\ \text{m}$ farther from the lens.
Solution by Sreeraj P, M.Sc Physics