A converging lens of focal length $20\,\text{cm}$ produces a real image at $30\,\text{cm}$ from the lens. The object distance is closest to

1. Thin lens formula: $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$. 2. With Cartesian sign convention: $f = +20\,\text{cm}$ (converging), $v = +30\,\text{cm}$ (real image on far side, positive). 3. Solve for $u$: $\dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f} = \dfrac{1}{30} - \dfrac{1}{20} = \dfrac{2 - 3}{60} = -\dfrac{1}{60}$. 4. So $u = -60\,\text{cm}$. The negative sign means object is on the SIDE the light comes from (the standard side for real objects). Magnitude: $60\,\text{cm}$. 5. Sanity check: object at $60\,\text{cm} = 3f$ is outside $2f$, image should form between $f$ and $2f$ on the far side. Image at $30\,\text{cm} = 1.5f$ ✓. 6. Other options come from sign-handling errors or using the wrong formula (mirror formula vs lens formula). _Source: NCERT Class 12 Physics Part 2, Ch 9, §9.6 (Thin Lens Formula, Eq. 9.18), p. 17 + Example 9.10._