An object $2.4$ m in front of a lens forms a sharp image on a film $12$ cm behind the lens. A glass plate $1$ cm thick, of refractive index $1.50$, is interposed between the lens and the film with its plane faces parallel to the film. At what distance (from the lens) should the object be shifted to be in sharp focus on the film?
Answer: (D) $5.6$ m
Focal length: $\dfrac{1}{f} = \dfrac{1}{12} + \dfrac{1}{240} = \dfrac{21}{240}$, so $f = \dfrac{240}{21}$ cm.
The plate shifts the image away from the lens by $t\left(1 - \dfrac{1}{\mu}\right) = \dfrac{1}{3}$ cm. For the final image to land on the film, the lens alone must form it at $v = 12 - \dfrac{1}{3} = \dfrac{35}{3}$ cm:
$$\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{3}{35} - \frac{21}{240} = -\frac{1}{560} \;\Rightarrow\; u = -560\ \text{cm}$$
The object must be $5.6$ m from the lens.
Solution by Sreeraj P, M.Sc Physics