A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance d. Then d

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1.1 cm away from the lens

0.55 cm away from the lens

0.55 cm towards the lens

answer is D.

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Detailed Solution

Case I: u = –10 cm, v = 10 cm, f = ?
Using lens formula, $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{10} - \frac{1}{- 10}$
$\Rightarrow f = 5 cm$
Case-II:- Due to introduction of slab, shift in the source is
$= t (1 - \frac{1}{μ}) = 1 .5 (1 - \frac{2}{3}) = 0 .5 cm$
Now, u = –9.5 cm, v = ?
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{5} - \frac{1}{9 .5} = \frac{4 .5}{5 \times 9 .5} \Rightarrow v = 10 .55 cm ∴ d = 10 .55 - 10 = 0 .55 cm$
away from the lens.