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?

An object at $$u = -2.4 \text{ m} = -240 \text{ cm}$$ forms a sharp image at $$v = 12 \text{ cm}$$. A glass plate of refractive index $$\mu = 1.5$$ and thickness $$t = 1 \text{ cm}$$ is introduced between the lens and screen.

Find the focal length of the lens.

Using the lens formula (with sign convention $$u = -240 \text{ cm}$$, $$v = +12 \text{ cm}$$):

$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{12} - \frac{1}{(-240)} = \frac{1}{12} + \frac{1}{240} = \frac{20 + 1}{240} = \frac{21}{240} = \frac{7}{80}$$ $$f = \frac{80}{7} \text{ cm}$$

Find the apparent shift caused by the glass plate.

The glass plate shifts the image towards the lens by:

$$\Delta = t\left(1 - \frac{1}{\mu}\right) = 1\left(1 - \frac{1}{1.5}\right) = 1 \times \frac{1}{3} = \frac{1}{3} \text{ cm}$$

Find the new image distance required.

For the image to still fall on the screen, the lens must form the image at:

$$v' = 12 - \frac{1}{3} = \frac{35}{3} \text{ cm}$$

Find the new object distance.

$$\frac{1}{u'} = \frac{1}{v'} - \frac{1}{f} = \frac{3}{35} - \frac{7}{80} = \frac{240 - 245}{2800} = \frac{-5}{2800} = \frac{-1}{560}$$ $$u' = -560 \text{ cm} = -5.6 \text{ m}$$

Calculate the shift in object position.

The object must be shifted from $$2.4 \text{ m}$$ to $$5.6 \text{ m}$$ away from the lens:

$$\text{Shift} = 5.6 - 2.4 = 3.2 \text{ m}$$

The correct answer is Option B: $$3.2 \text{ m}$$.