If we want to obtain a virtual and magnified image of an object by using a concave mirror of focal length 18 cm, where should the object be placed? Use mirror formula to determine the object distance for an image of magnification +2 produced by this mirror to justify your answer.
Generated by claude-sonnet-4-6 · 2026-06-15 06:50 · grounding rag
Model Answer
For a virtual and magnified image, the object must be placed between the pole (P) and the principal focus (F) of the concave mirror, i.e., between 0 and 18 cm from the mirror.
Verification using mirror formula:
Given: $f = -18$ cm, $m = +2$
Since $m = -\dfrac{v}{u}$,
$$+2 = -\frac{v}{u} \implies v = -2u$$
Applying mirror formula:
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{-2u} + \frac{1}{u} = \frac{1}{-18}$$
$$\frac{-1+2}{2u} = \frac{1}{-18}$$
$$\frac{1}{2u} = \frac{-1}{18} \implies u = -9 \text{ cm}$$
The object distance is 9 cm, which lies between P and F (0 to 18 cm). This confirms that placing the object between the pole and focus gives a virtual, magnified image.
Source: Chapter 9, Section 9.2.4 — Mirror Formula and Magnification
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Explanation
- The key concept: a concave mirror gives a virtual, erect, magnified image only when the object is between P and F (refer Table 9.1).
- Magnification is positive for virtual images in mirrors, so $m = +2$ is the right clue.
- Use $m = -v/u$ (mirror formula, not lens formula) to link $v$ and $u$, then substitute into the mirror formula.
- Sign convention: object distance $u$ is negative (object in front of mirror); focal length $f$ of concave mirror is negative.
- The calculated $u = -9$ cm (between 0 and 18 cm) justifies the initial statement — examiners look for this logical closure.