AI-generated practice question — model-generated for extra practice, not a previous-year CBSE board question.
(a) Ray Diagrams:
Case (i) — Object beyond C:
```
Object C F P
↓ | | |
----[ray 1→ parallel to axis → reflects through F]
----[ray 2→ through C → reflects back along same path]
Image (between F and C, inverted, diminished)
```
Draw two rays: one parallel to the principal axis (reflects through F) and one through C (reflects back along the same path). They meet between F and C to form a real, inverted, diminished image.
Case (ii) — Object between P and F:
Draw one ray parallel to the axis (reflects through F) and one through C. The reflected rays diverge; when extended behind the mirror they meet to form a virtual, erect, enlarged image behind the mirror.
(b) Nature, position and size of image:
| Position of Object | Position of Image | Size | Nature |
|---|---|---|---|
| Beyond C | Between F and C | Diminished | Real, Inverted |
| Between P and F | Behind the mirror | Enlarged | Virtual, Erect |
(c) Calculation:
Given: u = −2 cm, f = −8 cm
Using mirror formula: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-8} - \frac{1}{-2} = -\frac{1}{8} + \frac{1}{2} = \frac{3}{8}$$
$$v = +\frac{8}{3} \approx +2.67 \text{ cm}$$
Magnification: $m = -\dfrac{v}{u} = -\dfrac{+8/3}{-2} = +\dfrac{4}{3} \approx +1.33$
The image is virtual, erect, and magnified (m > 1). Since the tooth is between P and F, the concave mirror forms an enlarged, erect, virtual image behind the mirror, allowing the dentist to see a clear, magnified view of the tooth. This makes it very suitable for dental examination.
Source: Chapter 9, Section 9.2.1 and 9.2.2
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