An object is placed at a distance of 30 cm from the optical centre of a concave lens of focal length 15 cm. Use lens formula to determine the distance of the image from the optical centre of the lens.
Generated by claude-sonnet-4-6 · 2026-06-15 06:51 · grounding rag
Model Answer
Given:
- Object-distance, $u$ = –30 cm (object on left of lens)
- Focal length, $f$ = –15 cm (concave lens, negative sign)
- Image-distance, $v$ = ?
Using Lens Formula:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$$
$$\frac{1}{v} = \frac{1}{-15} + \frac{1}{-30}$$
$$\frac{1}{v} = -\frac{2}{30} - \frac{1}{30} = \frac{-3}{30} = \frac{-1}{10}$$
$$\boxed{v = -10 \text{ cm}}$$
Result: The image is formed at a distance of 10 cm from the optical centre, on the same side as the object (negative sign).
The image is virtual, erect, and diminished.
Source: Light – Reflection and Refraction, Section 9.3.7
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Explanation
- Always apply sign convention first: for a concave lens, both $f$ and $u$ are negative.
- Rearrange the lens formula as $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$ — this avoids sign errors.
- A negative $v$ confirms the image is virtual and on the same side as the object, which is always the case for a concave lens with a real object.
- Examiners award marks for: correct signs (1 mark), correct substitution (1 mark), correct calculation (1 mark), result with unit (1 mark), and nature of image (1 mark).