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 lens and 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?

Given,
\(u=2.4~\text m, v=12~\text{cm}, \mu=1.5, t=1~\text{cm}\)
The shift produced by the glass plate is
\(d=t\left(1-\dfrac{1}{\mu}\right)=1\times\left(1-\dfrac{1}{1.5}\right)=\dfrac{1}{3}~\text{cm}\)
So, final image must be produced at
\(\left(12-\dfrac{1}{3}\right)~\text{cm}=\dfrac{35}{3}~\text{cm},\) from lens so that glass plate must shift it to produce image at screen.
So, we have
\(\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\)
or, \(\dfrac{1}{12}-\dfrac{1}{-240}=\dfrac{1}{f}=\dfrac{1}{\dfrac{35}{3}}-\dfrac{1}{u}\)
or, \(\dfrac{1}{u}=\dfrac{3}{35}-\dfrac{1}{12}-\dfrac{1}{240}\\=\dfrac{144-140-7}{1680}=\dfrac{-3}{1680}=\dfrac{-1}{560}~\text{cm}\)
or \(u = – 560 ~ \text{cm}=-5.6~\text m\)
Therefore, object should be shifted to \(= 5.6 – 2.4 = 3.2~\text m\), for sharp image.
Hence, option \((B)\) is correct.