### Related Formula Magnification m$m$ of a mirror is given by: m = -fracvu$m = -\frac{v}{u}$ The mirror equation relates focal length to distance positions: frac1f = frac1v + frac1u implies f = fracuvu+v$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \implies f = \frac{uv}{u+v}$ Radius of curvature R = 2f$R = 2f$. ### Core Logic Given, magnification m = -3$m = -3$. This tells us the image is real and inverted[cite: 131, 757]: -3 = -fracvu implies v = 3u$-3 = -\frac{v}{u} \implies v = 3u$ Since both real objects and real images lie on the same side in front of a concave mirror, u$u$ and v$v$ are both negative fields. The physical distance separation between them is: |v| - |u| = 20 implies 3|u| - |u| = 20 implies 2|u| = 20 implies |u| = 10 text cm$|v| - |u| = 20 \implies 3|u| - |u| = 20 \implies 2|u| = 20 \implies |u| = 10 \text{ cm}$ Therefore, object distance u = -10 text cm$u = -10 \text{ cm}$ and image distance v = -30 text cm$v = -30 \text{ cm}$ . Substitute into the focal equation formula: f = frac(-10)(-30)-10 - 30 = frac300-40 = -7.5 text cm$f = \frac{(-10)(-30)}{-10 - 30} = \frac{300}{-40} = -7.5 \text{ cm}$ R = 2 times |f| = 2 times 7.5 = 15 text cm$R = 2 \times |f| = 2 \times 7.5 = 15 \text{ cm}$ ### Step 1: Visual Diagram The visual positioning profile tracking focal path boundaries is given below:

### Pattern Recognition A magnification of -3$-3$ tells you immediately that the object lies between the Focus (F$F$) and Center of Curvature (C$C$), while the image forms beyond C$C$. This geometric layout instantly verifies that the radius value must exceed 10text cm$10\text{ cm}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics