---
title: "Lenses, Images, and the Eye"
book: "High School Physics"
subject: physics
language: en
chapter: 10
exercises: 15
source: https://one-course.com/books/physics/2/en/chapter/10-lenses-images-and-the-eye
---

# Chapter 10 — Lenses, Images, and the Eye

Every sharp picture — on a phone sensor, a cinema screen, or the back of your own eye — comes from the same trick: a curved transparent medium bends rays by [refraction](https://one-course.com/books/physics/2/en/chapter/3-refraction-of-light#def-g10-refraction-refraction) ([Chapter 3](https://one-course.com/books/physics/2/en/chapter/3-refraction-of-light#ch-g10-refraction)) so that all the rays leaving one point meet again at another. This chapter tames the trick: one [lens](#def-g11-lenses-and-eye-lens), three rays, one formula — then points it at cameras, spectacles, and the living [lens](#def-g11-lenses-and-eye-lens) reading this sentence.

## 10.1 Converging and diverging lenses

**Definition 10.1 (Thin lens).**

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*.

**Definition 10.2 (Focal points and focal length).**

Rays arriving parallel to the axis of a [converging lens](#def-g11-lenses-and-eye-lens) all cross the axis at one point behind it, the *image focal point* $F'$; symmetrically, rays leaving the *object focal point* $F$ (in front, with $OF = OF'$) exit parallel to the axis. The *focal length* $f' = \overline{OF'}$ is positive for a [converging lens](#def-g11-lenses-and-eye-lens); through a [diverging lens](#def-g11-lenses-and-eye-lens), parallel rays exit as if from a point $F'$ *in front* of the [lens](#def-g11-lenses-and-eye-lens): $f' < 0$.

**Definition 10.3 (Vergence).**

The *vergence* of a [lens](#def-g11-lenses-and-eye-lens) of [focal length](#def-g11-lenses-and-eye-focal) $f'$ (in meters) is $C = 1/f'$, measured in *diopters* (symbol D): $1~\text{D} = 1\,\mathrm{m}^{-1}$; [diverging](#def-g11-lenses-and-eye-lens) lenses have $C < 0$. Opticians prescribe in diopters: vergences of [thin](#def-g11-lenses-and-eye-lens) lenses in contact simply add (admitted here; derived in the Year 1 volume).

**Example 10.4 (Reading a prescription).**

A [lens](#def-g11-lenses-and-eye-lens) with $f' = 0.50\,\mathrm{m}$ has $C = 1/0.50 = +2.0$ D: [converging](#def-g11-lenses-and-eye-lens). A $-4.0$ D prescription means $f' = -0.25\,\mathrm{m}$: [diverging](#def-g11-lenses-and-eye-lens), [focal length](#def-g11-lenses-and-eye-focal) $25\,\mathrm{cm}$.

## 10.2 Constructing the image

**Method 10.5 (The three construction rays).**

Let $AB$ be an object perpendicular to the axis at $A$. Among all rays leaving the tip $B$, three have known paths:

1. the ray through the [optical center](#def-g11-lenses-and-eye-lens) $O$ goes straight on;
2. the ray parallel to the axis exits through $F'$ ;
3. the ray through $F$ exits parallel to the axis.

Any two suffice: their intersection is the image point $B'$, and the image $A'B'$ stands perpendicular to the axis at $A'$. If the outgoing rays diverge, their *backward* extensions (dashed) meet at a [virtual image](#def-g11-lenses-and-eye-images) point.

![The three construction rays: object AB beyond F, image A'B' real and inverted (here OA = -2f': life-size, = -1).](https://one-course.com/images/onecourse/chapters/physics-2/g11-lenses-and-eye/fig-46d807f5b378.svg)

*The three construction rays: object $AB$ beyond $F$, image $A'B'$ [real](#def-g11-lenses-and-eye-images) and inverted (here $\overline{OA} = -2f'$: life-size, $\gamma = -1$).*

**Definition 10.6 (Real and virtual images).**

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](#def-g11-lenses-and-eye-lens) sees it perfectly well. A [converging lens](#def-g11-lenses-and-eye-lens) gives a real, inverted image of any object beyond $F$; slide the object inside the [focal length](#def-g11-lenses-and-eye-focal) 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'.](https://one-course.com/images/onecourse/chapters/physics-2/g11-lenses-and-eye/fig-8ad9403171ff.svg)

*Object inside the [focal length](#def-g11-lenses-and-eye-focal) (the [magnifier](#def-g11-lenses-and-eye-magnifier) setting): the backward extensions build a [virtual](#def-g11-lenses-and-eye-images), upright, enlarged image $A'B'$.*

## 10.3 One formula: the conjugate relation

Constructions show *where* the image is; a formula computes it. Orient the axis along the light and use *algebraic measures*: $\overline{OA}$ is the coordinate of $A$ relative to $O$ (negative for a [real](#def-g11-lenses-and-eye-images) object in front); heights $\overline{AB}$ count positive upward.

**Theorem 10.7 (Thin-lens conjugate relation).**

Through a [thin lens](#def-g11-lenses-and-eye-lens) of [focal length](#def-g11-lenses-and-eye-focal) $f'$, an object point $A$ on the axis images at the point $A'$ with

$$
\frac{1}{\overline{OA'}} - \frac{1}{\overline{OA}} = \frac{1}{f'} = C .
$$

**Proof.** *Admitted at this level.* ∎

**Remark 10.8.**

The relation follows from [Snell’s law](https://one-course.com/books/physics/2/en/chapter/3-refraction-of-light#thm-g10-refraction-snell) applied to rays close to the axis; the derivation — and the limits of the [thin-lens](#def-g11-lenses-and-eye-lens) model — are made honest in the Year 1 volume.

**Definition 10.9 (Magnification).**

The *magnification* of the image is

$$
\gamma = \frac{\overline{A'B'}}{\overline{AB}}
       = \frac{\overline{OA'}}{\overline{OA}} .
$$

$\abs{\gamma}$ compares sizes, its sign orientations ($\gamma < 0$: inverted); the sign of $\overline{OA'}$ reads the nature (positive: [real](#def-g11-lenses-and-eye-images), behind the [lens](#def-g11-lenses-and-eye-lens); negative: [virtual](#def-g11-lenses-and-eye-images), in front).

**Example 10.10 (The formula at work).**

An object stands $30\,\mathrm{cm}$ in front of a [converging lens](#def-g11-lenses-and-eye-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](#def-g11-lenses-and-eye-images) $15\,\mathrm{cm}$ behind the [lens](#def-g11-lenses-and-eye-lens), inverted, half-size — exactly what the three rays draw.

## 10.4 The eye

**Definition 10.11 (The eye as an optical system).**

Optically, the eye is a [converging lens](#def-g11-lenses-and-eye-lens) facing a screen: the *cornea* and *crystalline lens* act as one [converging lens](#def-g11-lenses-and-eye-lens) of adjustable [focal length](#def-g11-lenses-and-eye-focal); the *retina*, about $17\,\mathrm{mm}$ behind, is the fixed screen. Focusing is done by squeezing the [lens](#def-g11-lenses-and-eye-lens), not by moving it as in a camera: *accommodation* is the change of $f'$ produced by the ciliary muscles. The closest point seen sharply at full accommodation is the *near point* — about $25\,\mathrm{cm}$ for the *standard eye* — the farthest, at rest, the *far point* (infinity for the standard eye).

![Accommodation: to focus a near object on the same retina, the crystalline lens bulges — its vergence increases.](https://one-course.com/images/onecourse/chapters/physics-2/g11-lenses-and-eye/fig-8e035e059baa.svg)

![Accommodation: to focus a near object on the same retina, the crystalline lens bulges — its vergence increases.](https://one-course.com/images/onecourse/chapters/physics-2/g11-lenses-and-eye/fig-e75c7c02936d.svg)

*[Accommodation](#def-g11-lenses-and-eye-eye): to focus a near object on the same [retina](#def-g11-lenses-and-eye-eye), the [crystalline lens](#def-g11-lenses-and-eye-eye) bulges — its [vergence](#def-g11-lenses-and-eye-vergence) increases.*

**Example 10.12 (The vergence of your eye).**

The [retina](#def-g11-lenses-and-eye-eye) sits at $\overline{OA'} = +0.017\,\mathrm{m}$. Distant object ($1/\overline{OA} \approx 0$): $C = 1/0.017 \approx 59$ D. [Near point](#def-g11-lenses-and-eye-eye) ($\overline{OA} = -0.25\,\mathrm{m}$): $C = 1/0.017 + 1/0.25 \approx
63$ D. Reading costs about $4$ D: the eye is a $59$ D [lens](#def-g11-lenses-and-eye-lens) with a built-in $+4$ D fine tune.

## 10.5 Defects, corrections, and the magnifier

**Definition 10.13 (Myopia and hyperopia).**

A *myopic* (short-sighted) eye is too [converging](#def-g11-lenses-and-eye-lens) for its length: parallel rays focus *in front of* the [retina](#def-g11-lenses-and-eye-eye), distant objects blur, the [far point](#def-g11-lenses-and-eye-eye) is at finite distance; a [diverging](#def-g11-lenses-and-eye-lens) spectacle [lens](#def-g11-lenses-and-eye-lens) corrects it. A *hyperopic* (far-sighted) eye is not [converging](#def-g11-lenses-and-eye-lens) enough: near objects would focus behind the [retina](#def-g11-lenses-and-eye-eye), the [near point](#def-g11-lenses-and-eye-eye) recedes; a [converging lens](#def-g11-lenses-and-eye-lens) corrects it. Aging stiffens the [crystalline lens](#def-g11-lenses-and-eye-eye) (*presbyopia*), shrinking [accommodation](#def-g11-lenses-and-eye-eye) — same remedy: reading glasses.

**Example 10.14 (Prescribing for a myope).**

A [myopic](#def-g11-lenses-and-eye-defects) eye has its [far point](#def-g11-lenses-and-eye-eye) at $50\,\mathrm{cm}$: the correcting [lens](#def-g11-lenses-and-eye-lens) must show distant objects *at the [far point](#def-g11-lenses-and-eye-eye)* — object at infinity, image at $\overline{OA'} = -0.50\,\mathrm{m}$ ([virtual](#def-g11-lenses-and-eye-images), 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](#def-g11-lenses-and-eye-images) and sees it sharply.

![A myopic eye focuses parallel rays before the retina (red point); a diverging lens lands the focus on the retina.](https://one-course.com/images/onecourse/chapters/physics-2/g11-lenses-and-eye/fig-384c9df3708d.svg)

![A myopic eye focuses parallel rays before the retina (red point); a diverging lens lands the focus on the retina.](https://one-course.com/images/onecourse/chapters/physics-2/g11-lenses-and-eye/fig-a5f055ff034f.svg)

*A [myopic](#def-g11-lenses-and-eye-defects) eye focuses parallel rays before the [retina](#def-g11-lenses-and-eye-eye) (red point); a [diverging lens](#def-g11-lenses-and-eye-lens) lands the focus on the [retina](#def-g11-lenses-and-eye-eye).*

**Definition 10.15 (The magnifier).**

A *magnifier* is a [converging lens](#def-g11-lenses-and-eye-lens) of short [focal length](#def-g11-lenses-and-eye-focal) used with the object *inside* the [focal length](#def-g11-lenses-and-eye-focal): the image is [virtual](#def-g11-lenses-and-eye-images), upright, enlarged ([Definition 10.6](#def-g11-lenses-and-eye-images)), and the eye views it comfortably far away.

**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](#def-g11-lenses-and-eye-images), upright image, five times larger, a comfortable $20\,\mathrm{cm}$ from the [lens](#def-g11-lenses-and-eye-lens).

## 10.6 Exercises

**Exercise 10.1 ★.**

Compute the [vergences](#def-g11-lenses-and-eye-vergence) for $f' = 50\,\mathrm{cm}$, $-20\,\mathrm{cm}$ and $4.0\,\mathrm{mm}$, then the [focal length](#def-g11-lenses-and-eye-focal) of a $+8.0$ D [lens](#def-g11-lenses-and-eye-lens).

**Solution of Exercise 10.1.**

$C = 1/0.50 = +2.0$ D; $1/(-0.20) = -5.0$ D; $1/0.0040 = +250$ D. A $+8.0$ D [lens](#def-g11-lenses-and-eye-lens): $f' = 1/8.0 = 0.125\,\mathrm{m} = 12.5\,\mathrm{cm}$.

**Exercise 10.2 ★.**

A [lens](#def-g11-lenses-and-eye-lens) is thicker at the edges than at the center. [Converging](#def-g11-lenses-and-eye-lens) or [diverging](#def-g11-lenses-and-eye-lens)? Can it serve as a [magnifier](#def-g11-lenses-and-eye-magnifier)? What do you see holding it over a printed page?

**Solution of Exercise 10.2.**

[Diverging](#def-g11-lenses-and-eye-lens) ($f' < 0$). No: for any object distance $d$, $\gamma = f'/(f' - d)$ (shown in [Exercise 10.15](#exo-g11-lenses-and-eye-15)) lies between $0$ and $1$ — the image is always [virtual](#def-g11-lenses-and-eye-images), upright and *reduced*. The print appears upright and shrunk, never magnified.

**Exercise 10.3 ★.**

A [converging lens](#def-g11-lenses-and-eye-lens) has $f' = 10\,\mathrm{cm}$; an object stands $20\,\mathrm{cm}$ in front of it. Construct the image with two of the three rays, then check its position and size with the conjugate relation.

**Solution of Exercise 10.3.**

The object sits at $2f'$; the central and parallel rays cross $20\,\mathrm{cm}$ behind the [lens](#def-g11-lenses-and-eye-lens). Check: $1/\overline{OA'} = 1/10 - 1/20 = 1/20$, so $\overline{OA'} = +20\,\mathrm{cm}$ and $\gamma = 20/(-20) = -1$: [real](#def-g11-lenses-and-eye-images), inverted, life-size — as drawn.

**Exercise 10.4 ★.**

An object stands $24\,\mathrm{cm}$ in front of a [converging lens](#def-g11-lenses-and-eye-lens) with $f' = 8.0\,\mathrm{cm}$. Compute $\overline{OA'}$ and $\gamma$, and describe the image (nature, orientation, size).

**Solution of Exercise 10.4.**

$1/\overline{OA'} = 1/8.0 - 1/24 = 1/12$: $\overline{OA'} = +12\,\mathrm{cm}$, $\gamma = 12/(-24) = -0.50$. [Real image](#def-g11-lenses-and-eye-images) $12\,\mathrm{cm}$ behind the [lens](#def-g11-lenses-and-eye-lens), inverted, half-size.

**Exercise 10.5 ★.**

In the eye: what forms the [converging lens](#def-g11-lenses-and-eye-lens)? What plays the screen? What changes during [accommodation](#def-g11-lenses-and-eye-eye) — and what cannot? Define the [near point](#def-g11-lenses-and-eye-eye) of the [standard eye](#def-g11-lenses-and-eye-eye).

**Solution of Exercise 10.5.**

[Cornea](#def-g11-lenses-and-eye-eye) and [crystalline lens](#def-g11-lenses-and-eye-eye) form the [converging lens](#def-g11-lenses-and-eye-lens); the [retina](#def-g11-lenses-and-eye-eye) is the screen. [Accommodation](#def-g11-lenses-and-eye-eye) changes the [focal length](#def-g11-lenses-and-eye-focal) ([vergence](#def-g11-lenses-and-eye-vergence)) of the [crystalline lens](#def-g11-lenses-and-eye-eye); the lens–retina distance cannot change. [Near point](#def-g11-lenses-and-eye-eye): closest point seen sharply at full [accommodation](#def-g11-lenses-and-eye-eye) — $25\,\mathrm{cm}$ for the [standard eye](#def-g11-lenses-and-eye-eye).

**Exercise 10.6 ★★.**

A camera [lens](#def-g11-lenses-and-eye-lens) has $f' = 50\,\mathrm{mm}$. Where is the sensor when focused on a distant landscape? Compute the lens–sensor distance for a subject $3.0\,\mathrm{m}$ away, and the travel between the two settings.

**Solution of Exercise 10.6.**

Landscape: sensor in the focal plane, $50\,\mathrm{mm}$ from the [lens](#def-g11-lenses-and-eye-lens). At $3.0\,\mathrm{m}$: $1/\overline{OA'} = 1/50 - 1/3000 = 59/3000$, so $\overline{OA'} \approx 50.8\,\mathrm{mm}$. Travel: about $0.8\,\mathrm{mm}$.

**Exercise 10.7 ★★.**

A [magnifier](#def-g11-lenses-and-eye-magnifier) has $f' = 6.0\,\mathrm{cm}$; a stamp lies $5.0\,\mathrm{cm}$ below it. Position, nature, orientation and [magnification](#def-g11-lenses-and-eye-magnification) of the image?

**Solution of Exercise 10.7.**

$1/\overline{OA'} = 1/6.0 - 1/5.0 = -1/30$: $\overline{OA'} = -30\,\mathrm{cm}$ — [virtual](#def-g11-lenses-and-eye-images), upright, $\gamma = (-30)/(-5.0) = +6.0$: six times the stamp, $30\,\mathrm{cm}$ above the [lens](#def-g11-lenses-and-eye-lens).

**Exercise 10.8 ★★.**

An object and a screen face each other; a [converging lens](#def-g11-lenses-and-eye-lens) between them gives a sharp image when both are $40\,\mathrm{cm}$ from the [lens](#def-g11-lenses-and-eye-lens). Compute $f'$ and $\gamma$. Why must this symmetric image be exactly life-size?

**Solution of Exercise 10.8.**

$1/f' = 1/40 + 1/40 = 1/20$: $f' = 20\,\mathrm{cm}$; $\gamma = 40/(-40) = -1$. Symmetry [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) $\abs{\overline{OA'}} = \abs{\overline{OA}}$, hence $\abs{\gamma} = 1$; the image is [real](#def-g11-lenses-and-eye-images) and inverted, so $\gamma = -1$: exactly life-size.

**Exercise 10.9 ★★.**

An object stands $50\,\mathrm{cm}$ in front of a [diverging lens](#def-g11-lenses-and-eye-lens) with $f' = -25\,\mathrm{cm}$. Compute $\overline{OA'}$ and $\gamma$; describe the image. Why do a myope’s eyes look slightly smaller through their glasses?

**Solution of Exercise 10.9.**

$1/\overline{OA'} = -1/25 - 1/50 = -3/50$: $\overline{OA'} \approx -16.7\,\mathrm{cm}$, $\gamma = (-16.7)/(-50) = +0.33$: [virtual](#def-g11-lenses-and-eye-images), upright, one third of the size. A myope’s glasses are [diverging](#def-g11-lenses-and-eye-lens): everything seen through them — including, from outside, the wearer’s eyes — is reduced.

**Exercise 10.10 ★★.**

A [myopic](#def-g11-lenses-and-eye-defects) eye has its [far point](#def-g11-lenses-and-eye-eye) at $40\,\mathrm{cm}$. What [vergence](#def-g11-lenses-and-eye-vergence) must the correcting [lens](#def-g11-lenses-and-eye-lens) have, and where does it put the image of a distant mountain? Why is the mountain then seen effortlessly?

**Solution of Exercise 10.10.**

Image of infinity at the [far point](#def-g11-lenses-and-eye-eye): $f' = -0.40\,\mathrm{m}$, so $C = -2.5$ D. The mountain’s image sits $40\,\mathrm{cm}$ in front of the [lens](#def-g11-lenses-and-eye-lens) — exactly the [far point](#def-g11-lenses-and-eye-eye), which the relaxed eye sees sharply with zero [accommodation](#def-g11-lenses-and-eye-eye).

**Exercise 10.11 ★★.**

A presbyopic reader has their [near point](#def-g11-lenses-and-eye-eye) at $80\,\mathrm{cm}$ but wants to read at $25\,\mathrm{cm}$: the glasses must image the page at $25\,\mathrm{cm}$ onto a [virtual](#def-g11-lenses-and-eye-images) page at $80\,\mathrm{cm}$. Compute the required [vergence](#def-g11-lenses-and-eye-vergence).

**Solution of Exercise 10.11.**

$C = 1/\overline{OA'} - 1/\overline{OA} = 1/(-0.80) - 1/(-0.25)
= -1.25 + 4.00 = +2.75$ D.

**Exercise 10.12 ★★★.**

A projector must throw an image of a slide, magnified $\gamma = -50$, onto a screen $5.1\,\mathrm{m}$ from the slide.

1. Find $\overline{OA}$ , $\overline{OA'}$ , given $\overline{OA'} - \overline{OA} = 5.1\,\mathrm{m}$ .
2. Deduce the [focal length](#def-g11-lenses-and-eye-focal) of the projection [lens](#def-g11-lenses-and-eye-lens) .
3. How tall is the image of a $24\,\mathrm{mm}$ -tall slide?

**Solution of Exercise 10.12.**

*1.* $\gamma = \overline{OA'}/\overline{OA} = -50$ gives $\overline{OA'} = -50\,\overline{OA}$; with $\overline{OA'} - \overline{OA} = 5.1\,\mathrm{m}$, $-51\,\overline{OA} = 5.1\,\mathrm{m}$: $\overline{OA} = -0.10\,\mathrm{m}$, $\overline{OA'} = +5.0\,\mathrm{m}$.

*2.* $1/f' = 1/5.0 + 1/0.10 = 10.2$: $f' \approx 9.8\,\mathrm{cm}$.

*3.* $50 \times 24\,\mathrm{mm} = 1.2\,\mathrm{m}$, inverted — which is why slides are loaded upside down.

**Exercise 10.13 ★★★.**

The [retina](#def-g11-lenses-and-eye-eye) is $17\,\mathrm{mm}$ behind the eye’s [lens](#def-g11-lenses-and-eye-lens).

1. Compute the eye’s [vergence](#def-g11-lenses-and-eye-vergence) focused at infinity, then on the standard [near point](#def-g11-lenses-and-eye-eye) ( $25\,\mathrm{cm}$ ). Deduce the [accommodation](#def-g11-lenses-and-eye-eye) [amplitude](https://one-course.com/books/physics/2/en/chapter/8-signals-and-waves#def-g10-signals-and-waves-amplitude) of the [standard eye](#def-g11-lenses-and-eye-eye) .
2. An older eye has its [near point](#def-g11-lenses-and-eye-eye) at $1.0\,\mathrm{m}$ . Compute its [amplitude](https://one-course.com/books/physics/2/en/chapter/8-signals-and-waves#def-g10-signals-and-waves-amplitude) and the fraction of the standard [amplitude](https://one-course.com/books/physics/2/en/chapter/8-signals-and-waves#def-g10-signals-and-waves-amplitude) remaining.

**Solution of Exercise 10.13.**

*1.* $C_\infty = 1/0.017 \approx 59$ D; $C_{25} = 1/0.017 + 1/0.25 \approx 63$ D. [Amplitude](https://one-course.com/books/physics/2/en/chapter/8-signals-and-waves#def-g10-signals-and-waves-amplitude): $1/0.25 = 4.0$ D.

*2.* [Amplitude](https://one-course.com/books/physics/2/en/chapter/8-signals-and-waves#def-g10-signals-and-waves-amplitude) $= 1/1.0 = 1.0$ D: one quarter ($25\%$) of the [standard eye](#def-g11-lenses-and-eye-eye)’s $4.0$ D remains.

**Exercise 10.14 ★★★.**

A [magnifier](#def-g11-lenses-and-eye-magnifier) sold as “$\times 3$” has its [focal length](#def-g11-lenses-and-eye-focal) given by $3 = 0.25/f'$ ($f'$ in meters).

1. Compute $f'$ .
2. Where must an object sit for its [virtual image](#def-g11-lenses-and-eye-images) to lie $25\,\mathrm{cm}$ in front of the [lens](#def-g11-lenses-and-eye-lens) ? Compute $\gamma$ ; compare with the advertised $\times 3$ .

**Solution of Exercise 10.14.**

*1.* $f' = 0.25/3 \approx 0.083\,\mathrm{m} = 8.3\,\mathrm{cm}$.

*2.* $\overline{OA'} = -25\,\mathrm{cm}$: $1/\overline{OA} = -1/25 - 3/25 = -4/25$, so $\overline{OA} = -6.25\,\mathrm{cm}$ and $\gamma = (-25)/(-6.25) = +4.0$ — better than the advertised $\times 3$, which assumes the image at infinity; pulling it to the [near point](#def-g11-lenses-and-eye-eye) gains one [unit](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit).

**Exercise 10.15 ★★★.**

A [real](#def-g11-lenses-and-eye-images) object stands at distance $d > 0$ in front of a [converging lens](#def-g11-lenses-and-eye-lens) ($\overline{OA} = -d$, $f' > 0$). Show that $\overline{OA'} = \frac{f'd}{d - f'}$ and $\gamma = \frac{f'}{f' - d}$. Deduce: [real](#def-g11-lenses-and-eye-images) and inverted when $d > f'$; [virtual](#def-g11-lenses-and-eye-images), upright, *enlarged* ($\gamma > 1$) when $0 < d < f'$ — every [converging lens](#def-g11-lenses-and-eye-lens) is a [magnifier](#def-g11-lenses-and-eye-magnifier), used close enough.

**Solution of Exercise 10.15.**

$1/\overline{OA'} = 1/f' - 1/d = (d - f')/(f'd)$, so $\overline{OA'} = f'd/(d - f')$ and $\gamma = \overline{OA'}/(-d) = f'/(f' - d)$. If $d > f'$: $\overline{OA'} > 0$ ([real](#def-g11-lenses-and-eye-images)) and $\gamma < 0$ (inverted). If $0 < d < f'$: $\overline{OA'} < 0$ ([virtual](#def-g11-lenses-and-eye-images)) and $\gamma = f'/(f' - d) > 1$: upright and enlarged — the [magnifier](#def-g11-lenses-and-eye-magnifier) setting.

## 10.7 Problem: One formula, three instruments

**Problem 10.1.**

Weekend problem — the same conjugate relation focuses a camera, prescribes reading glasses, designs a loupe, and measures the zoom built into your own eye

A photographer racks a $50\,\mathrm{mm}$ [lens](#def-g11-lenses-and-eye-lens), a grandmother holds the newspaper at arm’s length, a jeweler leans over a diamond: one line of algebra, $1/\overline{OA'} - 1/\overline{OA} = 1/f'$ ([Theorem 10.7](#thm-g11-lenses-and-eye-conjugate)), runs all three trades — and then the eye itself.

**Part I — One formula.** A [converging lens](#def-g11-lenses-and-eye-lens) has $f' = 20\,\mathrm{cm}$.

1. An object at $60\,\mathrm{cm}$ : compute $\overline{OA'}$ and $\gamma$ ; describe the image.
2. Move the object to $30\,\mathrm{cm}$ ; recompute. What got exchanged with question 1? (Light paths are reversible.)
3. Move it to $10\,\mathrm{cm}$ , inside $f'$ : recompute and describe.
4. Summarize: as the object slides in from very far toward $F$ , then past it, how does the image move and change nature?
5. Show that a very distant object ( $1/\overline{OA} \approx 0$ ) always images in the focal plane: $\overline{OA'} = f'$ .

**Part II — The photographer.** A camera [lens](#def-g11-lenses-and-eye-lens) has $f' = 50\,\mathrm{mm}$; the sensor, $24\,\mathrm{mm}$ tall, can be moved relative to the [lens](#def-g11-lenses-and-eye-lens).

6. Where is the sensor when focused on a distant landscape?
7. Now a face $1.0\,\mathrm{m}$ away: compute the new lens–sensor distance and the travel from the landscape setting.
8. The mechanism allows at most $55\,\mathrm{mm}$ of lens–sensor distance. Compute the minimum focusing distance.
9. At that closest focus, compute $\gamma$ . Does the image of a $20\,\mathrm{cm}$ face fit on the sensor?
10. Life-size “macro” means $\gamma = -1$ : where must object and sensor then be, and why does this demand a special [lens](#def-g11-lenses-and-eye-lens) ?

**Part III — The grandmother.** With age the [near point](#def-g11-lenses-and-eye-eye) recedes: grandmother’s is at $1.0\,\mathrm{m}$.

11. Explain the arm’s-length newspaper.
12. To read at $25\,\mathrm{cm}$ , her glasses must turn a page at $25\,\mathrm{cm}$ into a [virtual image](#def-g11-lenses-and-eye-images) at $1.0\,\mathrm{m}$ : compute the required [vergence](#def-g11-lenses-and-eye-vergence) .
13. Compute $\gamma$ . The image is four times larger, yet looks no bigger: explain, comparing sizes *and* distances.
14. Show that print is sharp only between $25\,\mathrm{cm}$ and $f' \approx 33\,\mathrm{cm}$ . What of a *distant* object seen through the glasses?
15. Deduce in one sentence why bifocal lenses exist.

**Part IV — The jeweler, and the eye’s own zoom.** The jeweler’s loupe has $f' = 5.0\,\mathrm{cm}$; his [near point](#def-g11-lenses-and-eye-eye) is at $25\,\mathrm{cm}$; his [retina](#def-g11-lenses-and-eye-eye) is $17\,\mathrm{mm}$ behind his eye’s [lens](#def-g11-lenses-and-eye-lens).

16. A stone $4.0\,\mathrm{cm}$ under the loupe: position, nature, [magnification](#def-g11-lenses-and-eye-magnification) ?
17. Where must the stone sit for its image to fall at his [near point](#def-g11-lenses-and-eye-eye) , and what is $\gamma$ then?
18. Why do experienced jewelers place the stone exactly at $F$ instead (image at infinity)?
19. Now the eye alone: compute its [vergence](#def-g11-lenses-and-eye-vergence) viewing a distant object, then one at $25\,\mathrm{cm}$ : how many [diopters](#def-g11-lenses-and-eye-vergence) of [accommodation](#def-g11-lenses-and-eye-eye) ?
20. Convert both [vergences](#def-g11-lenses-and-eye-vergence) to [focal lengths](#def-g11-lenses-and-eye-focal) : by how much, in millimeters and percent, does the eye change its own [focal length](#def-g11-lenses-and-eye-focal) — the zoom the camera did with travel and grandmother with $+3$ D?

**Solution of Problem 10.1.**

**1.** $1/\overline{OA'} = 1/20 - 1/60 = 1/30$: $\overline{OA'} = +30\,\mathrm{cm}$, $\gamma = -0.50$ — [real](#def-g11-lenses-and-eye-images), inverted, half-size.

**2.** $\overline{OA'} = +60\,\mathrm{cm}$, $\gamma = -2$: the positions $30\,\mathrm{cm}$ and $60\,\mathrm{cm}$ exchanged roles, the [magnifications](#def-g11-lenses-and-eye-magnification) are reciprocal — reversed light follows the same path.

**3.** $1/\overline{OA'} = 1/20 - 1/10 = -1/20$: $\overline{OA'} = -20\,\mathrm{cm}$, $\gamma = +2$ — [virtual](#def-g11-lenses-and-eye-images), upright, doubled.

**4.** From infinity the image starts in the focal plane; as the object approaches $F$ the [real](#def-g11-lenses-and-eye-images), inverted image recedes to infinity and grows; inside $f'$ it turns [virtual](#def-g11-lenses-and-eye-images), upright, enlarged, in front of the [lens](#def-g11-lenses-and-eye-lens).

**5.** $1/\overline{OA} \approx 0$ leaves $1/\overline{OA'} = 1/f'$: $\overline{OA'} = f'$, the focal plane.

**6.** In the focal plane, $50\,\mathrm{mm}$ behind the [lens](#def-g11-lenses-and-eye-lens).

**7.** $1/\overline{OA'} = 1/50 - 1/1000 = 19/1000$: $\overline{OA'} \approx 52.6\,\mathrm{mm}$; travel $\approx 2.6\,\mathrm{mm}$.

**8.** $1/\overline{OA} = 1/55 - 1/50 = -1/550$: minimum focusing distance $550\,\mathrm{mm}$ = $55\,\mathrm{cm}$.

**9.** $\gamma = 55/(-550) = -0.10$: the face images at $20\,\mathrm{mm}$, just inside the $24\,\mathrm{mm}$ sensor.

**10.** $\gamma = -1$ [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) $\overline{OA'} = -\overline{OA}$, so $2/\overline{OA'} = 1/f'$: object $100\,\mathrm{mm}$ ($= 2f'$) in front, sensor $100\,\mathrm{mm}$ behind — double the landscape draw, far beyond the $55\,\mathrm{mm}$ mechanism; macro lenses provide the extra travel.

**11.** Anything nearer than her [near point](#def-g11-lenses-and-eye-eye) ($1.0\,\mathrm{m}$) blurs: stretched arms bring the page as close to $1.0\,\mathrm{m}$ as possible — sharp, if small.

**12.** $C = 1/(-1.0) - 1/(-0.25) = -1.0 + 4.0 = +3.0$ D.

**13.** $\gamma = (-1.0)/(-0.25) = +4.0$: four times larger but four times farther — the same angular size; the gain is not size but *sharpness*, since the image now lies at her [near point](#def-g11-lenses-and-eye-eye).

**14.** Sharp needs the image beyond $1.0\,\mathrm{m}$: from the page at $25\,\mathrm{cm}$ (image at $1.0\,\mathrm{m}$) to the page at $f' = 1/3.0 \approx 0.33\,\mathrm{m}$ (image at infinity). A distant object images $33\,\mathrm{cm}$ *behind* the glasses — a [real image](#def-g11-lenses-and-eye-images) the eye cannot use: the distant world blurs.

**15.** Hence bifocals: reading [power](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-power) in the lower half of the [lens](#def-g11-lenses-and-eye-lens), none (or the distance correction) in the upper half.

**16.** $1/\overline{OA'} = 1/5.0 - 1/4.0 = -1/20$: $\overline{OA'} = -20\,\mathrm{cm}$, [virtual](#def-g11-lenses-and-eye-images), upright, $\gamma = +5.0$.

**17.** $1/\overline{OA} = -1/25 - 1/5.0 = -6/25$: $\overline{OA} \approx -4.2\,\mathrm{cm}$; $\gamma = (-25)/(-25/6) = +6.0$.

**18.** Stone at $F$: image at infinity, viewed with zero [accommodation](#def-g11-lenses-and-eye-eye) — hours of inspection without eye strain.

**19.** $C_\infty = 1/0.017 \approx 59$ D; $C_{25} = 1/0.017 + 1/0.25 \approx 63$ D: about $4$ D of [accommodation](#def-g11-lenses-and-eye-eye).

**20.** $f' = 1/58.8 \approx 17.0\,\mathrm{mm}$ against $1/62.8 \approx 15.9\,\mathrm{mm}$: about $1.1\,\mathrm{mm}$, some $6\%$ — the built-in zoom the camera imitated with $2.6\,\mathrm{mm}$ of travel and grandmother patched with $+3$ D.