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The lens combination as shown in the figure, consists of two lenses, $L_1$ and $L_2$, of the focal lengths +10 cm and -10 cm, respectively. The position of the image formed is:
Detailed Solution
First lens: u = -30 cm, f = +10 cm, so $v=\frac{uf}{u+f}=\frac{-30\times10}{-30+10}=15$ cm. This image lies 12 cm beyond the concave lens (placed 3 cm away) and acts as a virtual object for it. Second lens: u = +12 cm, f = -10 cm, so $v=\frac{12\times(-10)}{12-10}=-60$ cm (virtual image). The negative sign means the image is 60 cm to the left of the concave lens.