What is Thin lens? — Physics

Definition 10.1 High School Physics · Chapter 10 — Lenses, Images, and the Eye

A lens is a transparent medium bounded by two surfaces, at least one curved. Thicker at the center than at the edge, it is converging: parallel rays are bent toward each other; thicker at the edge, diverging: they spread apart. A thin lens has negligible thickness: we draw a segment with outward (converging) or inward (diverging) arrowheads. Its center $O$ is the optical center; the perpendicular line through $O$ , the principal axis.

Examples

Example 10.4 (Reading a prescription)

A lens with $f' = 0.50\,\mathrm{m}$ has $C = 1/0.50 = +2.0$ D: converging. A $-4.0$ D prescription means $f' = -0.25\,\mathrm{m}$ : diverging, focal length $25\,\mathrm{cm}$ .

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 10.12 (The vergence of your eye)

The retina sits at $\overline{OA'} = +0.017\,\mathrm{m}$ . Distant object ( $1/\overline{OA} \approx 0$ ): $C = 1/0.017 \approx 59$ D. Near point ( $\overline{OA} = -0.25\,\mathrm{m}$ ): $C = 1/0.017 + 1/0.25 \approx 63$ D. Reading costs about $4$ D: the eye is a $59$ D lens with a built-in $+4$ D fine tune.

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Definition 3.10 University Physics — Year 1 · Chapter 3 — Mirrors and Thin Lenses

A lens is a transparent medium bounded by two spherical (or plane) surfaces; it is thin when its thickness is negligible against the radii and against the distances in play, so that the two vertices merge into one point, the optical center $O$ . A ray through $O$ is not deviated. The image focus $F'$ is the image of a point at infinity on the axis; the object focus $F$ is the point whose image is at infinity; both are at the same distance from $O$ : $\overline{OF'} = -\overline{OF} = f'$ , the focal length. The lens is converging if $f' > 0$ ( $F'$ real, downstream; edges thinner than the center), diverging if $f' < 0$ . Its vergence $V = 1/f'$ is measured in diopters ( $1\,\delta = 1\,\mathrm{m}^{-1}$ ).

A converging lens: object AB beyond F. The ray through O goes straight, the parallel ray bends through F', the ray through F leaves parallel; all three meet at B': real, inverted image. Here OA = -1.5f', so OA' = 3f' and = -2. — A converging lens: object $AB$ beyond $F$ . The ray through $O$ goes straight, the parallel ray bends through $F'$ , the ray through $F$ leaves parallel; all three meet at $B'$ : real, inverted image. Here $\overline{OA} = -1.5f'$ , so $\overline{OA'} = 3f'$ and $\gamma = -2$ .

A diverging lens (F' upstream, F downstream): the parallel ray emerges as if from F', the central ray goes straight; their backward extensions meet at B' — a virtual, upright, reduced image, whatever the object position. — A diverging lens ( $F'$ upstream, $F$ downstream): the parallel ray emerges as if from $F'$ , the central ray goes straight; their backward extensions meet at $B'$ — a virtual, upright, reduced image, whatever the object position.

Examples

Example 3.13 (The same lens, three ways)

A converging lens, $f' = 5.0\,\mathrm{cm}$ . Stamp at $3.0\,\mathrm{cm}$ : $1/\overline{OA'} = 1/5 - 1/3 = -2/15$ , $\overline{OA'} = -7.5\,\mathrm{cm}$ , virtual, $\gamma = +2.5$ : the magnifier. Lamp at $20\,\mathrm{cm}$ : $\overline{OA'} = 6.7\,\mathrm{cm}$ , real, $\gamma = -1/3$ : a reduced inverted image on a card. Window $4\,\mathrm{m}$ away: $\overline{OA'} \approx f'$ , tiny inverted image in the focal plane — the camera. Three uses, one relation.

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