An object $AB$ is placed $15$ cm on the left of a convex lens $P$ of focal length $10$ cm. Another convex lens $Q$ is now placed $15$ cm right of lens $P$. If the focal length of lens $Q$ is $15$ cm, the final image is ______.
Answer: (B) real, formed at $7.5$ cm right of lens $Q$, with a size same as that of $AB$
Lens $P$: $u = -15$, $f = 10$: $\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15} = \dfrac{1}{30}$, so $v = 30$ cm and $m_1 = \dfrac{30}{-15} = -2$.
This image is $15$ cm beyond $Q$, a virtual object for $Q$: $u = +15$, $f = 15$:
$$\frac{1}{v} = \frac{1}{15} + \frac{1}{15} \;\Rightarrow\; v = 7.5\ \text{cm}, \qquad m_2 = \frac{7.5}{15} = 0.5$$
Final image: real, $7.5$ cm right of $Q$, total magnification $-2 \times 0.5 = -1$ (same size, inverted).
Solution by Sreeraj P, M.Sc Physics