The refractive indices of the two thin lenses are $μ_{2} = 1.2$ and $μ_{3} = 1.5$ respectively. The medium on the two sides of the combined lens are having refractive indices $μ_{1} = 1$ and $μ_{4} = 1.8$ as shown in the figure. A linear object AB of size 0.30 cm is placed at a distance 25 cm from the combined lens as shown in the figure. Then choose the correct option(s).

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The image is formed at a distance 40 cm to the right of the combined lens

The image is formed at a distance 60 cm to the left of the combined lens.

The size of the image formed is 0.40 cm.

The size of the image formed is 0.20 cm.

answer is B, D.

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Detailed Solution

$\frac{μ_{4}}{v} - \frac{μ_{1}}{u} = (\frac{μ_{2} - μ_{1}}{R_{1}}) + (\frac{μ_{3} - μ_{2}}{R_{2}}) + (\frac{μ_{4} - μ_{3}}{R_{3}}) \frac{1.8}{v} - \frac{1}{- 25} = (\frac{1.2 - 1}{10}) + (\frac{1.5 - 1.2}{- 30}) + (\frac{1.8 - 1.5}{\infty}) \frac{1.8}{v} + \frac{1}{25} = \frac{0.2}{10} - \frac{0.3}{30} + 0 \frac{1.8}{v} + \frac{1}{25} = \frac{1}{50} - \frac{1}{100} \frac{1.8}{v} = \frac{1}{100} - \frac{1}{25} \Rightarrow v = - 60 cm$

Lateral magnification, $m = (\frac{μ_{1}}{μ_{4}} \frac{v}{u}) = \frac{1}{1.8} (\frac{- 60}{- 25})$
$m = + \frac{4}{3}$
The size of image formed $= \frac{4}{3} \times 0.3 = 0.40 cm$