Q 12-09-096JEE MainJEE Main 2024 (8 Apr, Shift 2)Medium
The position of the image formed by the combination of lenses is:
Answer: (D) 30 cm (right of third lens)
First lens ($f_1 = 10\ \text{cm}$, $u = -30\ \text{cm}$):
$$\frac1v = \frac1{10} - \frac1{30} \;\Rightarrow\; v = 15\ \text{cm}$$
This image is $15 - 5 = 10\ \text{cm}$ to the right of the second lens, a virtual object with $u = +10\ \text{cm}$. For $f_2 = -10\ \text{cm}$:
$$\frac1v = -\frac1{10} + \frac1{10} = 0 \;\Rightarrow\; v = \infty$$
Parallel rays reach the third lens, which focuses them at its focal point: $30\ \text{cm}$ to the right of the third lens.
Solution by Sreeraj P, M.Sc Physics