An equi-convex lens of $μ = 1.5$ and $R = 20 c m$ is cut into two equal parts along its axis. Two parts are then separated by a distance of 120 cm (as shown in figure). An object of height 3 mm is place at a distance of 30 cm to the left of first half lens. The final image will form at

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150 cm to the right of first half lens, 4 mm in size and erect

150 cm to the right of first half lens, 3 mm in size and erect

120 cm to the right of first half lens, 3 mm in size and inverted

120 cm to the right of first half lens, 4 mm in size and inverted

answer is B.

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Detailed Solution

$\frac{1}{f} = (1.5 - 1) (\frac{1}{20} - \frac{1}{- 20}) \Rightarrow f = + 20 c m$
$\frac{1}{v_{1}} - \frac{1}{- 30} = \frac{1}{+ 20}$
$∴ v_{1} = + 60 c m$
$m_{1} = \frac{v_{1}}{u_{1}} = \frac{+ 60}{- 30} = - 2$
For second lens,
$\frac{1}{v_{2}} - \frac{1}{- 60} = \frac{1}{+ 20}$
$v_{2} = + 30 c m$
$m_{2} = \frac{v_{2}}{u_{2}} = \frac{(+ 30)}{(- 60)} = - \frac{1}{2}$
$m = m_{1} m_{2} = 1$
Final image is 30cm to the right of second lens or 150cm to the right of first lens
$m = 1$
Hence, image height = object height $= 3 m m$