The refractive indices of the two thin lenses are $μ_{2} = 1.2 a n d μ_{3} = 1.5$ respectively. The medium on the two sides of the combined lens are having refractive indices $μ_{1} = 1 a n d μ_{4} = 1.8$ 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 size of the image formed is 0.40 cm

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

The size of the image formed is 0.20 cm

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

answer is B, D.

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Detailed Solution

$\begin{array}{l} \frac{μ_{4}}{v} - \frac{μ_{1}}{u} = (\frac{μ_{2} - μ_{1}}{R_{1}}) + (\frac{μ_{3} - μ_{2}}{R_{2}}) + (\frac{μ_{4} - μ_{3}}{R_{3}}) \\ \frac{1.8}{v} - \frac{1}{- 25} = (\frac{1.2 - 1}{10}) + (\frac{1.5 - 1.2}{- 30}) + (\frac{1.8 - 1.5}{\infty}) \end{array} \begin{array}{l} \frac{1.8}{v} + \frac{1}{25} = \frac{0.2}{10} - \frac{0.3}{30} + 0 \\ \frac{1.8}{v} + \frac{1}{25} = \frac{1}{50} - \frac{1}{100} \\ \frac{1.8}{v} = \frac{1}{100} - \frac{1}{25} \\ \Rightarrow v = - 60 c m \end{array} \begin{array}{l} L a t e r a l M a g n i f i c a t i o n m = (\frac{μ_{1}}{μ_{4}} \frac{v}{u}) = \frac{1}{1.8} (\frac{- 60}{- 25}) \end{array} m = + \frac{4}{3} T h e s i z e o f i m a g e f o r m e d = \frac{4}{3} x 0.3 = 0.40 c m$