Study the data given below showing the focal length of three concave mirrors A, B and C and the respective distances of objects placed in front of the mirrors:

Case	Mirror	Focal Length (cm)	Object Distance (cm)
1	A	20	45
2	B	15	30
3	C	30	20

(i) In which one of the above cases the mirror will form a diminished image of the object? Justify your answer. (ii) List two properties of the image formed in case 2. (iii) (A) What is the nature and size of the image formed by mirror C? Draw ray diagram to justify your answer. OR (iii) (B) An object is placed at a distance of $18\mathrm{cm}$ from the pole of a concave mirror of focal length $12\mathrm{cm}$ . Find the position of the image formed in this case.

(i) Case 1 (Mirror A) will form a diminished image. Justification: Here, $f = 20\text{ cm}$, so $C = 2f = 40\text{ cm}$. Since the object distance $u = 45\text{ cm}$ is greater than $2f$ ($u > C$), the object is placed beyond the Centre of Curvature. For a concave mirror, an object beyond $C$ forms a diminished image between $F$ and $C$.

(ii) In Case 2, $f = 15\text{ cm}$ and $u = 30\text{ cm}$. Since $u = 2f$, the object is at the Centre of Curvature ($C$). Properties: 1. The image is real and inverted. 2. The image is of the same size as the object.

(iii) (A) For Mirror C, $f = 30\text{ cm}$ and $u = 20\text{ cm}$. Since $u < f$, the object is between the Pole ($P$) and Focus ($F$). Nature: Virtual and erect. Size: Highly enlarged (magnified).

OR

(iii) (B) Given: $u = -18\text{ cm}$, $f = -12\text{ cm}$.

Using mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} = \frac{1}{-12} - \frac{1}{-18} = \frac{-3 + 2}{36} = -\frac{1}{36}$

$v = -36\text{ cm}$.

The image is formed at a distance of $36\text{ cm}$ in front of the mirror.