What is Real and virtual images? — Physics

Definition 10.6 High School Physics · Chapter 10 — Lenses, Images, and the Eye

An image is real when the outgoing rays actually pass through it: a screen placed there catches it; virtual when only their backward extensions meet: no screen catches it, but an eye looking into the lens sees it perfectly well. A converging lens gives a real, inverted image of any object beyond $F$ ; slide the object inside the focal length and the image turns virtual, upright, enlarged.

Object inside the focal length (the magnifier setting): the backward extensions build a virtual, upright, enlarged image A'B'. — Object inside the focal length (the magnifier setting): the backward extensions build a virtual, upright, enlarged image $A'B'$ .

Examples

Example 10.10 (The formula at work)

An object stands $30\,\mathrm{cm}$ in front of a converging lens with $f' = 10\,\mathrm{cm}$ : $\overline{OA} = -30\,\mathrm{cm}$ , so

\frac{1}{\overline{OA'}} = \frac{1}{10} - \frac{1}{30} = \frac{1}{15}, \qquad \overline{OA'} = +15\,\mathrm{cm}, \qquad \gamma = \frac{15}{-30} = -0.50 :

a real image $15\,\mathrm{cm}$ behind the lens, inverted, half-size — exactly what the three rays draw.

Example 10.14 (Prescribing for a myope)

A myopic eye has its far point at $50\,\mathrm{cm}$ : the correcting lens must show distant objects at the far point — object at infinity, image at $\overline{OA'} = -0.50\,\mathrm{m}$ (virtual, in front). Then $1/f' = 1/\overline{OA'}$ : $f' = -0.50\,\mathrm{m}$ , $C = -2.0$ D; the relaxed eye looks at that virtual image and sees it sharply.

Example 10.16 (A jeweler’s loupe)

Loupe with $f' = 5.0\,\mathrm{cm}$ , stone at $4.0\,\mathrm{cm}$ : $1/\overline{OA'} = 1/5.0 - 1/4.0 = -1/20$ , so $\overline{OA'} = -20\,\mathrm{cm}$ and $\gamma = (-20)/(-4.0) = +5.0$ : a virtual, upright image, five times larger, a comfortable $20\,\mathrm{cm}$ from the lens.

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Examples

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