Image with a diverging lens

Problem

A diverging lens has focal length $f = -2\;\mathrm{m}$. A real object is placed at $p = 0{,}5\;\mathrm{m}$ in front of the lens. Find the position $q$ of the image and classify the image (real or virtual, upright or inverted).

Solution

Setup. Thin-lens equation:

$\frac{1}{p} + \frac{1}{q} = \frac{1}{f}$

Substitution. Isolate $1/q$:

$\frac{1}{q} = \frac{1}{f} - \frac{1}{p} = \frac{1}{-2} - \frac{1}{0{,}5} = -0{,}5 - 2 = -2{,}5\;\mathrm{m^{-1}}$

$q = \frac{1}{-2{,}5} = -0{,}4\;\mathrm{m}$

Classification. Magnification:

$I = -\frac{q}{p} = -\frac{-0{,}4}{0{,}5} = +0{,}8$

$q<0$ → virtual image, on the same side as the object (in front of the lens). $I>0$ → upright; $|I|<1$ → reduced.

$\ev{q = -0{,}4\;\mathrm{m};\quad \text{virtual, upright, reduced}}$

Diverging lenses always give virtual, upright, reduced images of real objects: this is the typical behaviour of glasses for short-sightedness.

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Topics: Ottica Concepts: Lenti e specchi Skills: Analisi di casi limite e fantafisica