One fourth part of an equiconvex lens of focal length 100 cm is removed as shown in the figure. An object of height 1 cm is placed in front of the lens. It is observed that all the images are of equal height. Then

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The product of magnification of both the lenses is negative.

The magnitude of magnification produced by upper and lower part is equal.

The no. of images formed is two.

Object is at a distance of $\frac{400}{3}$ cm from the lens

answer is A, B, C, D.

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Detailed Solution

The two parts of the lens will have different focal lengths. So, there are two images.
$\frac{1}{v_{1}} - \frac{1}{u} = \frac{1}{100}$ ; $m_{1} = \frac{v_{1}}{u}$ , $\frac{1}{v_{2}} - \frac{1}{u} = \frac{1}{200}$ ; $m_{2} = \frac{v_{2}}{u}$
For same height of images $m_{1} = - m_{2}$ $v_{1} = - v_{2}$
$u = \frac{- 400}{3}$ cm, m1 = –3, m2 = 3, m1m2 = –9
∴ (a), (b), (c) and (d)