A concave lens always produces a magnification that is positive and has a magnitude less than 1, regardless o…

Question

A concave lens always produces a magnification that is positive and has a magnitude less than 1, regardless of where the object is placed. Justify both of these statements using the lens formula and the definition of magnification, and describe what they together tell us about the image formed.

Accepted Answer

## Model Answer For a concave lens, *f* is negative and object distance *u* is always negative. From the lens formula: $$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$$ both terms on the right are negative, so *v* is always negative (virtual image on same side as object). Magnification $m = \frac{v}{u}$ = (negative)/(negative) = **positive** (erect image). Since the concave lens diverges rays, |v| < |u| always, so $|m| < 1$ (image is diminished). Together, a concave lens always forms a **virtual, erect, and diminished** image, regardless of object position. *Source: Chapter 9, Sections 9.3.6 (Sign Convention) and 9.3.5 (Image Formation)* --- ## Explanation - Examiners expect you to use the sign convention: for concave lens *f* < 0, *u* < 0, which forces *v* < 0. - Show *m = v/u* explicitly to justify the positive sign. - The key conclusion — virtual, erect, diminished — must be stated; it ties both statements together and is a common direct question.

Science (086) — AI-generated practice question