The position of the image formed by the lens combination given in the fig

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30cm to the right of 3^rd lens

15cm to the right of 3^rd lens

10cm to the right of 3^rd lens

5cm to the right of 3^rd lens

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{1}{f} = \frac{1}{ϑ} - \frac{1}{4} \\ 1 \to \frac{1}{V_{1}} - (\frac{1}{30}) = \frac{1}{10} \Rightarrow V_{1} = 15 \\ 2 \to \frac{1}{V_{2}} - \frac{1}{10} = \frac{- 1}{10} \Rightarrow V_{2} = 00 \\ 3 \to \frac{1}{V_{3}} - \frac{1}{43} = \frac{1}{f_{3}} \Rightarrow \frac{1}{V_{3}} = \frac{1}{f_{3}} \end{array}$

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