The power of a lens is +4D. Find the focal length of this lens. An object is placed at a distance of 50 cm from the optical centre of this lens. State the nature and magnification of the image formed by the lens and also draw a ray diagram to justify your answer.
Generated by claude-sonnet-4-6 · 2026-06-15 06:50 · grounding rag
Model Answer
Finding focal length:
$$P = \frac{1}{f} \Rightarrow f = \frac{1}{P} = \frac{1}{+4} = +0.25 \text{ m} = 25 \text{ cm}$$
This is a convex lens.
Applying lens formula (u = −50 cm, f = +25 cm):
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{25} + \frac{1}{(-50)} = \frac{2-1}{50} = \frac{1}{50}$$
$$v = +50 \text{ cm}$$
Magnification:
$$m = \frac{v}{u} = \frac{+50}{-50} = -1$$
Nature of image: Real, inverted, and of the same size as the object (formed at 2F₂, 50 cm on the other side of the lens).
Ray Diagram:
```
|
O | I
2F₁ F₁ O F₂ 2F₂
←50cm→|← 50cm →
```
(Object at 2F₁; image forms at 2F₂, real and inverted.)
Source: Chapter 9, Sections 9.3.7 and 9.3.8
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Explanation
- Since P = +4D (positive), the lens is convex; f = 1/P in metres = 0.25 m = 25 cm.
- Object is placed at u = −50 cm = 2f, so by Table 9.4, image forms at 2F₂ on the other side — real, inverted, same size.
- Lens formula confirms v = +50 cm; magnification m = v/u = −1 (negative sign → real and inverted; |m| = 1 → same size).
- Draw the standard two-ray diagram: one ray parallel to principal axis refracts through F₂; another through optical centre goes straight. Examiners expect the ray diagram to be labelled with 2F₁, F₁, O, F₂, 2F₂.