Using the New Cartesian Sign Convention, an object is placed 20 cm in front of a concave mirror of focal leng…

Question

Using the New Cartesian Sign Convention, an object is placed 20 cm in front of a concave mirror of focal length 15 cm. Calculate the image distance and magnification. State whether the image is real or virtual, and erect or inverted.

Accepted Answer

## Model Answer **Given:** u = −20 cm, f = −15 cm (concave mirror) **Using mirror formula:** $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$ $$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15} - \frac{1}{-20} = -\frac{1}{15} + \frac{1}{20} = \frac{-4+3}{60} = \frac{-1}{60}$$ $$\boxed{v = -60 	ext{ cm}}$$ The image is formed **60 cm in front of the mirror**. **Magnification:** $$m = -\frac{v}{u} = -\frac{(-60)}{(-20)} = -3$$ The image is **real** (negative m) and **inverted** (negative sign), and magnified 3 times. *Source: Chapter 9, Section 9.2.4 — Mirror Formula and Magnification* --- ## Explanation - Always assign signs first: object in front → u is negative; concave mirror → f is negative. - A negative v confirms the image is in front of the mirror → **real**. - A negative magnification always means **real and inverted** for mirrors. - The examiner awards marks for: correct formula use (1), correct v (1), correct m with nature of image (1).

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