An object is placed at $30\,\text{cm}$ in front of a concave mirror of focal length $20\,\text{cm}$. The image distance from the mirror is

1. Mirror formula: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$. 2. With Cartesian sign convention, $u = -30\,\text{cm}$ (object in front), $f = -20\,\text{cm}$ (concave). 3. $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{-20} - \dfrac{1}{-30} = -\dfrac{1}{20} + \dfrac{1}{30}$. 4. Common denominator: $-\dfrac{3}{60} + \dfrac{2}{60} = -\dfrac{1}{60}$. So $v = -60\,\text{cm}$. 5. NEGATIVE means in FRONT of the mirror (same side as the object) — a REAL image. So $60\,\text{cm}$ in front. 6. Sanity check: object lies between $f$ and $2f$ (between 20 cm and 40 cm) — concave mirrors produce real, inverted, magnified images here. Image at 60 cm > $2f$ ✓. 7. Other options come from sign-confusion in the substitution. _Source: NCERT Class 12 Physics Part 2, Ch 9, §9.2.5 (Mirror formula application, Example 9.2 method), p. 5._