A lens forms a virtual, erect image of an object placed 8 cm in front of it. The image is located 4.8 cm from the lens on the same side as the object. (a) Using the lens formula, calculate the focal length of this lens and state what type of lens it is. (b) With the help of sign convention, explain why the focal length carries the sign you obtained.
Generated by claude-sonnet-4-6 · 2026-06-26 01:15 · grounding rag
Model Answer
Given: u = −8 cm, v = −4.8 cm (image on same side as object, so negative)
Using lens formula:
$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-4.8} - \frac{1}{-8} = -\frac{1}{4.8} + \frac{1}{8}$$
$$\frac{1}{f} = \frac{-5 + 3}{24} = \frac{-2}{24} = \frac{-1}{12}$$
∴ f = −12 cm
(a) The focal length is −12 cm. Since f is negative, this is a concave lens. (This is consistent with the concave lens always forming a virtual, erect, diminished image on the same side as the object.)
(b) By sign convention, all distances are measured from the optical centre. The image is virtual and forms on the same side as the object, so v is negative. Applying the lens formula gives a negative f. According to sign convention, the focal length of a concave (diverging) lens is always negative.
Source: Chapter 9, Sections 9.3.6 and 9.3.7
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Explanation
- Signs are critical: Object is always on the left → u is negative. Virtual image on the same side as the object → v is also negative. Plugging both negatives into the lens formula yields a negative f.
- Lens type identification: Negative focal length = concave lens. Examiners expect you to state this explicitly.
- Part (b) is conceptual: just link the sign of v (virtual image, same side) to the resulting negative f and the rule that concave lenses have negative focal lengths.
- Don't confuse v being negative here (virtual image) with the mirror convention — lens formula is $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, not the mirror formula.