(i) Define principal axis of a lens. Draw a ray diagram to show what happens when a ray of light parallel to the principal axis of a concave lens passes through it.
(ii) The focal length of a concave lens is 20 cm. At what distance from the lens should a 5 cm tall object be placed so that its image is formed at a distance of 15 cm from the lens ? Also calculate the size of the image formed.
Generated by claude-sonnet-4-6 · 2026-06-15 06:49 · grounding rag
Model Answer
(i) Principal Axis: The straight line passing through the two centres of curvature of the two surfaces of a lens (i.e., through its optical centre) is called the principal axis.
Ray Diagram – Concave Lens:
```
|
\ | /
\ | /
----\---O---------/-----→ Principal Axis
F₁ | /
| /
```
A ray parallel to the principal axis, after refraction through a concave lens, appears to diverge from the principal focus (F₁) located on the same side as the incident ray.
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(ii) Given:
- Focal length of concave lens, f = −20 cm
- Object height, h = 5 cm
- Image distance, v = −15 cm (image is virtual, on same side as object)
- Object distance, u = ?
Using lens formula:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{-15} - \frac{1}{u} = \frac{1}{-20}$$
$$\frac{1}{u} = \frac{1}{-15} + \frac{1}{20} = \frac{-4+3}{60} = \frac{-1}{60}$$
$$\therefore u = -60 \text{ cm}$$
The object should be placed 60 cm from the lens.
Size of image:
$$m = \frac{v}{u} = \frac{-15}{-60} = +\frac{1}{4}$$
$$h' = m \times h = \frac{1}{4} \times 5 = 1.25 \text{ cm}$$
The image is 1.25 cm tall, virtual, erect, and diminished.
Source: Chapter 9, Sections 9.3.5 and 9.3.6
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Explanation
- Part (i): Examiners expect a one-line definition and a ray diagram clearly showing the incident ray parallel to the principal axis diverging after refraction, with the diverged ray appearing to come from F₁ on the same side. Label the lens, principal axis, and F₁.
- Part (ii): For a concave lens, both f and v are negative by sign convention (real-is-positive/new Cartesian). Apply the lens formula correctly. The magnification formula $m = v/u$ gives the image height directly. A common error is using wrong signs — always state the signs used.
- The image formed by a concave lens is always virtual, erect, and diminished — confirm this matches your answer.