Q 12-09-172JEE MainJEE Main 2021 (27 Aug, Shift 1)Medium
Find the distance of the image from object $O$, formed by the combination of lenses in the figure:
Answer: (A) $75$ cm
**First lens** ($f = +10$ cm, $u = -30$ cm): $\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{30} \Rightarrow v = 15$ cm.
**Second lens** ($f = -10$ cm), $5$ cm further: the first image is $10$ cm beyond it, a virtual object with $u = +10$ cm.
$$\frac{1}{v} = -\frac{1}{10} + \frac{1}{10} = 0 \Rightarrow v = \infty$$
The light leaves as a parallel beam.
**Third lens** ($f = +30$ cm): a parallel beam is focused $30$ cm beyond it.
Distance from $O$: $30 + 5 + 10 + 30 = 75$ cm.
Solution by Sreeraj P, M.Sc Physics