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For an object placed at a distance 2.4 m from a lens, a sharp focused image is observed on a screen placed at a distance 12 cm from the lens. A glass plate of refractive index 1.5 and thickness 1 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?

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1.2 m

5.6 m

0.8 m

3.2 m

answer is B.

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Detailed Solution

Applying lens formula

$\frac{1}{0 .12} + \frac{1}{2 .4} = \frac{1}{f} \Rightarrow \frac{1}{f} = \frac{210}{24}$

Upon putting the glass slab, shift of image is

$Δx = t (1 - \frac{1}{μ}) = \frac{1}{3} cm Now v = 12 - \frac{1}{3} = \frac{35}{3} cm$

Again apply lens formula

$\frac{3}{0.35} - \frac{1}{u} = \frac{1}{f} = \frac{210}{24}$

Solving $u = - 5.6 m$

Thus shift of object is

$5 .6 - 2 .4 = 3 .2 m$

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