(a)
Ray Diagram: Object placed at 2F₁ (distance = 2f from lens):
```
|
O | I
↑ ----+---- ↓
2F₁ F₁ O F₂ 2F₂
```
Two rays are drawn: (1) parallel to principal axis, refracts through F₂; (2) through optical centre, passes undeviated. Image forms at 2F₂ — real, inverted, same size as object.
$$\text{Magnification} = -1$$
---
(b)
Given: h = 2 cm, f = 12 cm, u = −16 cm
(i) Distance of image from lens:
Using lens formula: $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$
$$\frac{1}{v} - \frac{1}{-16} = \frac{1}{12}$$
$$\frac{1}{v} = \frac{1}{12} - \frac{1}{16} = \frac{4-3}{48} = \frac{1}{48}$$
$$v = +48 \text{ cm}$$
The image is formed 48 cm from the lens (on the opposite side).
(ii) Length of image:
$$m = \frac{v}{u} = \frac{48}{-16} = -3$$
$$h' = m \times h = -3 \times 2 = -6 \text{ cm}$$
Length of image = 6 cm (real and inverted).
Source: Light – Reflection and Refraction, Section 9.3.4 & 9.3.6
---