1. Given data

• Object distance from lens: $u = 2.4\,\text{m} = 240\,\text{cm}$
• Image distance from lens (screen position): $v = 12\,\text{cm}$
• Glass plate refractive index: $\mu = 1.5$
• Thickness of glass plate: $t = 1\,\text{cm}$

2. Effect of introducing a glass plate

When a glass plate of thickness $t$ and refractive index $\mu$ is inserted between the lens and screen, the image shifts away from the lens by: $\Delta = t\left(1-\frac{1}{\mu}\right)$

So, $\Delta = 1\left(1-\frac{1}{1.5}\right) = 1\left(1-\frac{2}{3}\right)=\frac{1}{3}\,\text{cm}$

Thus, to keep the image sharp on the same screen, the lens must now form the image at a distance: $v' = 12 - \frac{1}{3} = \frac{35}{3}\,\text{cm}$

3. Find focal length of the lens from initial condition

Using lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ Taking sign convention is unnecessary here if we use magnitudes for a real object and real image with convex lens in the standard form: $\frac{1}{f} = \frac{1}{12} + \frac{1}{240}$ $= \frac{20+1}{240} = \frac{21}{240} = \frac{7}{80}$ Hence, $f = \frac{80}{7}\,\text{cm}$

4. New object distance for image at $v' = 35/3$ cm

Again using lens formula: $\frac{1}{f} = \frac{1}{v'} + \frac{1}{u'}$ $\frac{7}{80} = \frac{3}{35} + \frac{1}{u'}$ Now, $\frac{1}{u'} = \frac{7}{80} - \frac{3}{35}$ Taking LCM $560$: $\frac{1}{u'} = \frac{49 - 48}{560} = \frac{1}{560}$ So, $u' = 560\,\text{cm} = 5.6\,\text{m}$

5. Required shift in object position

Initial object distance was $2.4\,\text{m}$, new object distance is $5.6\,\text{m}$.

Hence the object must be shifted by: $5.6 - 2.4 = 3.2\,\text{m}$

6. Final answer

The object should be shifted by $\boxed{3.2\,\text{m}}$ which corresponds to Option B.