A thin convex lens of focal length 15 cm is split into two halves and the two halves are placed at a separation of 120 cm. Between the two halves of convex lens, a plane mirror is placed horizontally and at a distance of 4 mm below the principal axis of the lens halves. An object AB of length 2 mm is placed at a distance of 20 cm from one half lens as shown in figure.
The final image of the point A is formed at a distance of $\frac{n}{3} m m$ from the principle axis. Determine the value of n.

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answer is 8.

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Detailed Solution

The first half–lens forms the image A'B' (= 6 mm).
$∴ \frac{1}{v} - \frac{1}{- 20} = \frac{1}{15} \Rightarrow v = + 60$ cm.
$∴ m_{1} = - 3 .$
The plane mirror forms the image A''B'' with A'' located 8 mm below the principal axis. The second half lens forms the image A''B''. Now u = – 60 cm
$\Rightarrow v = + 20 c m$
$∴ m_{2} = - \frac{1}{3}$
$∴$ $A'''$ is located at $\frac{8}{3}$ mm above the principal axis.
$∴$ n = 8.