---
title: "Mirrors and Thin Lenses"
book: "University Physics — Year 1"
subject: physics
language: en
chapter: 3
exercises: 12
source: https://one-course.com/books/physics/3/en/chapter/3-mirrors-and-thin-lenses
---

# Chapter 3 — Mirrors and Thin Lenses

A make-up mirror shows a face twice its size; the back of a spoon shows it tiny and upright, the bowl shows it tiny and upside down; a magnifying glass held at arm’s length turns the room on its head, held close it enlarges a stamp; a phone [lens](#def-b1-mirrors-thin-lenses-lens) the size of a lentil paints the whole street onto a sensor. One algebraic relation between three distances governs every one of these [images](#def-b1-mirrors-thin-lenses-image), and this chapter establishes it — for curved mirrors first, then for the [thin lens](#def-b1-mirrors-thin-lenses-lens) that almost all optical instruments are built from.

![A converging lens forming the real, inverted image of a candle flame on a screen: object and image distances obey the conjugation relation of this chapter.](https://one-course.com/images/onecourse/chapters/physics-3/b1-mirrors-thin-lenses/img-4e16a71a9752.jpg)

*A [converging lens](#def-b1-mirrors-thin-lenses-lens) forming the real, inverted [image](#def-b1-mirrors-thin-lenses-image) of a candle flame on a screen: object and [image](#def-b1-mirrors-thin-lenses-image) distances obey the conjugation relation of this chapter.*

## 3.1 Images and the conditions of Gauss

**Definition 3.1 (Object, image, stigmatism).**

An optical system receives rays from a point $A$ (the *object point*). If all the emerging rays pass through one point $A'$, or all seem to come from one point $A'$, the system is *stigmatic* for $A$ and $A'$ is the *image* of $A$. The image is *real* if the emerging rays actually cross at $A'$ (it can be caught on a screen), *virtual* if only their backward extensions do (it is seen through the system, not projected). Likewise the object is real if the incoming rays diverge from $A$, virtual if they converge toward $A$ behind the system’s entrance. The pair $(A, A')$ is *conjugate*.

**Definition 3.2 (Centered system; Gauss conditions).**

A *centered system* has an axis of rotational symmetry, the *optical axis*. Rays that stay close to the axis and make small angles with it are *paraxial*; a system used with paraxial rays only is said to work under the *Gauss conditions*. Under those conditions a centered system is approximately stigmatic for every point, and an object plane perpendicular to the axis has an [image](#def-b1-mirrors-thin-lenses-image) plane perpendicular to the axis (*aplanatism*).

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

**Remark 3.3 (Why small angles).**

The [plane mirror](https://one-course.com/books/physics/3/en/chapter/2-geometric-optics-rays-reflection-refraction#prop-b1-rays-reflection-refraction-mirror) is the only simple system that is rigorously stigmatic for every point. A [spherical mirror](#def-b1-mirrors-thin-lenses-mirror) or [lens](#def-b1-mirrors-thin-lenses-lens) is stigmatic only to the extent that $\sin\theta \approx \tan\theta \approx \theta$: that approximation is what turns [Snell’s law](https://one-course.com/books/physics/3/en/chapter/2-geometric-optics-rays-reflection-refraction#thm-b1-rays-reflection-refraction-snell) into linear relations between distances. Outside it the [image](#def-b1-mirrors-thin-lenses-image) of a point is a blur (*aberrations*), and the designer’s job — stops, combinations of lenses, aspheric surfaces — is to push the blur below what the eye or the sensor can resolve.

**Notation 3.4 (Algebraic measures and magnification).**

The [optical axis](#def-b1-mirrors-thin-lenses-gauss) is oriented in the direction of the incoming light; the transverse direction is oriented upward. For points $P$, $Q$ on the axis, $\overline{PQ}$ denotes the [algebraic measure](#rem-b1-mirrors-thin-lenses-why) ($\overline{PQ} > 0$ if $Q$ is downstream of $P$); for an object $AB$ perpendicular to the axis and its [image](#def-b1-mirrors-thin-lenses-image) $A'B'$, the *[transverse magnification](#rem-b1-mirrors-thin-lenses-why)* is

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

negative for an inverted [image](#def-b1-mirrors-thin-lenses-image), of absolute value greater than $1$ for an enlarged one.

## 3.2 Spherical mirrors

**Definition 3.5 (Spherical mirror).**

A *spherical mirror* is a reflecting spherical cap of center $C$ and vertex $S$ (its intersection with the axis). It is *concave* if the light meets the hollow side ($C$ on the side of the incoming light), *convex* otherwise. Its *focus* $F$ is the [image](#def-b1-mirrors-thin-lenses-image) of a point at infinity on the axis; its [focal length](#def-b1-mirrors-thin-lenses-lens) is $f = \overline{SF}$.

**Proposition 3.6 (Focus of a spherical mirror).**

Under the [Gauss conditions](#def-b1-mirrors-thin-lenses-gauss), the [focus](#def-b1-mirrors-thin-lenses-mirror) is the midpoint of $SC$: $\overline{SF} = \overline{SC}/2$. A [concave mirror](#def-b1-mirrors-thin-lenses-mirror) has $f < 0$ (real [focus](#def-b1-mirrors-thin-lenses-mirror), in front); a convex one has $f > 0$ (virtual [focus](#def-b1-mirrors-thin-lenses-mirror), behind).

**Proof.** A ray parallel to the axis at height $h$ meets the mirror at $I$; the normal at $I$ is the radius $CI$, making an angle $\alpha$ with the axis, $\sin\alpha = h/R$. Reflected at the same angle $\alpha$ to the normal, the ray crosses the axis at $F$; triangle $CIF$ has equal angles $\alpha$ at $C$ and at $I$, so $CF = IF$ and, projecting, $CF = R/(2\cos\alpha)$. For [paraxial rays](#def-b1-mirrors-thin-lenses-gauss) $\cos\alpha \to 1$ and $CF = R/2$: all such rays cross the axis at the same point, the midpoint of $SC$. Sign: for a [concave mirror](#def-b1-mirrors-thin-lenses-mirror) $C$ and $F$ lie in front (upstream), $\overline{SC} < 0$. ∎

**Theorem 3.7 (Conjugation for a spherical mirror).**

Under the [Gauss conditions](#def-b1-mirrors-thin-lenses-gauss), the [image](#def-b1-mirrors-thin-lenses-image) $A'$ of a point $A$ on the axis satisfies

$$
\frac{1}{\overline{SA'}} + \frac{1}{\overline{SA}} = \frac{2}{\overline{SC}}
= \frac{1}{\overline{SF}},
\qquad
\gamma = -\frac{\overline{SA'}}{\overline{SA}} .
$$

**Proof.** Take a [concave mirror](#def-b1-mirrors-thin-lenses-mirror) and a real object $A$ beyond $C$; the algebraic form then covers every case. A ray from $A$ hits the mirror at $I$ (height $h$) and returns to $A'$ on the axis. Call $u$, $u'$, $\omega$ the angles of $AI$, $A'I$ and $CI$ with the axis; paraxially $u = h/AS$, $u' = h/A'S$, $\omega = h/CS$ (distances). The law of reflection says the normal $CI$ bisects the angle between $AI$ and $A'I$: $\omega - u = u' - \omega$, i.e. $u + u' = 2\omega$, i.e. $1/AS + 1/A'S = 2/CS$; with $\overline{SA}$, $\overline{SA'}$, $\overline{SC}$ all negative this is the displayed relation. For the magnification, the ray $B \to S \to B'$ reflected at the vertex makes equal angles with the axis: $A'B'/SA' = AB/SA$ in lengths, and the [image](#def-b1-mirrors-thin-lenses-image) is inverted, $\gamma = -\overline{SA'}/\overline{SA}$. ∎

**Method 3.8 (Constructing an image in a mirror).**

For an object point $B$ off the axis, draw two of the four remarkable rays; their intersection (or that of their extensions) is $B'$:

1. parallel to the axis $\to$ reflected through $F$ ;
2. through $F$ $\to$ reflected parallel to the axis;
3. through $C$ $\to$ reflected back on itself;
4. to the vertex $S$ $\to$ reflected symmetrically about the axis.

Then $A'$ is the foot of the perpendicular from $B'$ to the axis.

![A concave mirror (C, focus F at mid-radius). For an object AB beyond C, the ray parallel to the axis returns through F and the ray through F returns parallel: they cross at B' — a real, inverted, reduced image between F and C.](https://one-course.com/images/onecourse/chapters/physics-3/b1-mirrors-thin-lenses/fig-1566047a0650.svg)

*A [concave mirror](#def-b1-mirrors-thin-lenses-mirror) ($C$, [focus](#def-b1-mirrors-thin-lenses-mirror) $F$ at mid-radius). For an object $AB$ beyond $C$, the ray parallel to the axis returns through $F$ and the ray through $F$ returns parallel: they cross at $B'$ — a real, inverted, reduced [image](#def-b1-mirrors-thin-lenses-image) between $F$ and $C$.*

**Example 3.9 (Shaving mirror, rear-view mirror).**

A [concave mirror](#def-b1-mirrors-thin-lenses-mirror) of radius $60\,\mathrm{cm}$ ($f = -30\,\mathrm{cm}$) with a face at $20\,\mathrm{cm}$: $1/\overline{SA'} = -1/30 + 1/20 = 1/60$, $\overline{SA'} = +60\,\mathrm{cm}$ — behind the mirror, virtual — and $\gamma = -60/(-20) = +3$: upright, three times larger, the shaving mirror. A [convex mirror](#def-b1-mirrors-thin-lenses-mirror) of radius $2.0\,\mathrm{m}$ ($f = +1.0\,\mathrm{m}$) with a car $20\,\mathrm{m}$ behind: $1/\overline{SA'} = 1 + 1/20$, $\overline{SA'} = 0.95\,\mathrm{m}$ (virtual, behind), $\gamma = +0.048$: upright, twenty times smaller, and therefore a wide field — the rear-view mirror, with its warning that objects are closer than they appear.

## 3.3 Thin lenses

**Definition 3.10 (Thin lens).**

A *lens* is a [transparent medium](https://one-course.com/books/physics/3/en/chapter/2-geometric-optics-rays-reflection-refraction#def-b1-rays-reflection-refraction-index) 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](#def-b1-mirrors-thin-lenses-image) [focus](#def-b1-mirrors-thin-lenses-mirror)* $F'$ is the [image](#def-b1-mirrors-thin-lenses-image) of a point at infinity on the axis; the *object [focus](#def-b1-mirrors-thin-lenses-mirror)* $F$ is the point whose [image](#def-b1-mirrors-thin-lenses-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* ($1\,\delta = 1\,\mathrm{m}^{-1}$).

**Theorem 3.11 (Conjugation for a thin lens).**

Under the [Gauss conditions](#def-b1-mirrors-thin-lenses-gauss), the [image](#def-b1-mirrors-thin-lenses-image) $A'$ of a point $A$ on the axis satisfies the relation of Descartes

$$
\frac{1}{\overline{OA'}} - \frac{1}{\overline{OA}} = \frac{1}{f'},
\qquad
\gamma = \frac{\overline{OA'}}{\overline{OA}},
$$

and, equivalently, the relation of Newton

$$
\overline{FA}\cdot\overline{F'A'} = -f'^2,
\qquad
\gamma = \frac{f'}{\overline{FA}} = -\frac{\overline{F'A'}}{f'} .
$$

**Partial proof.** The key fact is that a [thin lens](#def-b1-mirrors-thin-lenses-lens) deviates a ray crossing it at height $h$ by the angle $h/f'$ toward the axis, whatever the ray’s direction — each zone of the [lens](#def-b1-mirrors-thin-lenses-lens) acts as a thin prism ([Remark 2.21](https://one-course.com/books/physics/3/en/chapter/2-geometric-optics-rays-reflection-refraction#rem-b1-rays-reflection-refraction-thin)) whose apex angle grows linearly with $h$ for spherical surfaces; this linearity, exact for [paraxial rays](#def-b1-mirrors-thin-lenses-gauss), is admitted here and derived from [Snell’s law](https://one-course.com/books/physics/3/en/chapter/2-geometric-optics-rays-reflection-refraction#thm-b1-rays-reflection-refraction-snell) at two spherical surfaces in the Year 2 volume. Granting it: a ray from $A$ (at distance $p = -\overline{OA} > 0$ for a real object) reaching the [lens](#def-b1-mirrors-thin-lenses-lens) at height $h$ arrives with slope $h/p$, is bent by $h/f'$ and leaves with slope $h/p - h/f'$; it crosses the axis at $A'$ with $h/\overline{OA'} = h/f' - h/p$, i.e. $1/\overline{OA'}
- 1/\overline{OA} = 1/f'$, independently of $h$: all rays from $A$ meet at $A'$. The ray $B \to O \to B'$ through the center is straight, so $\overline{A'B'}/\overline{AB} = \overline{OA'}/\overline{OA}$. Newton’s form follows by substituting $\overline{OA} = \overline{OF} + \overline{FA}
= -f' + \overline{FA}$ and $\overline{OA'} = f' + \overline{F'A'}$. ∎

**Method 3.12 (Constructing an image through a thin lens).**

Three remarkable rays from $B$:

1. through $O$ : undeviated;
2. parallel to the axis: emerges through $F'$ ;
3. through $F$ : emerges parallel to the axis.

Two suffice; for a [diverging lens](#def-b1-mirrors-thin-lenses-lens), use the extensions (the ray aimed at $F$ behind the [lens](#def-b1-mirrors-thin-lenses-lens) emerges parallel; the parallel ray emerges as if from $F'$ in front).

![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.](https://one-course.com/images/onecourse/chapters/physics-3/b1-mirrors-thin-lenses/fig-9d98a81aa150.svg)

*A [converging lens](#def-b1-mirrors-thin-lenses-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](#def-b1-mirrors-thin-lenses-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.](https://one-course.com/images/onecourse/chapters/physics-3/b1-mirrors-thin-lenses/fig-0212246928e4.svg)

*A [diverging lens](#def-b1-mirrors-thin-lenses-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](#def-b1-mirrors-thin-lenses-image), whatever the object position.*

**Example 3.13 (The same lens, three ways).**

A [converging lens](#def-b1-mirrors-thin-lenses-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](#def-b1-mirrors-thin-lenses-image) on a card. Window $4\,\mathrm{m}$ away: $\overline{OA'}
\approx f'$, tiny inverted [image](#def-b1-mirrors-thin-lenses-image) in the focal plane — the camera. Three uses, one relation.

**Proposition 3.14 (Longitudinal magnification).**

If the object moves along the axis by $\dd\overline{OA}$, the [image](#def-b1-mirrors-thin-lenses-image) moves by $\dd\overline{OA'} = \gamma^2\,\dd\overline{OA}$, in the same direction.

**Proof.** Differentiate Descartes’s relation at fixed $f'$: $-\dd\overline{OA'}/\overline{OA'}^2 + \dd\overline{OA}/\overline{OA}^2 = 0$, so $\dd\overline{OA'} = (\overline{OA'}/\overline{OA})^2\,\dd\overline{OA}$. ∎

**Remark 3.15 (Depth of focus).**

A projector with $\gamma = -50$ has $\gamma^2 = 2500$: a slide that buckles by $0.1\,\mathrm{mm}$ under the lamp’s heat throws its [image](#def-b1-mirrors-thin-lenses-image) $25\,\mathrm{cm}$ out of the screen plane. Conversely a camera ($\abs\gamma
\ll 1$) tolerates large object displacements for a tiny [image](#def-b1-mirrors-thin-lenses-image) shift — the depth of field.

## 3.4 Measuring a focal length

**Method 3.16 (Autocollimation).**

Place a [plane mirror](https://one-course.com/books/physics/3/en/chapter/2-geometric-optics-rays-reflection-refraction#prop-b1-rays-reflection-refraction-mirror) right behind the [lens](#def-b1-mirrors-thin-lenses-lens) and move a lit object (a cross on a frosted glass) along the axis until its [image](#def-b1-mirrors-thin-lenses-image), reflected back through the [lens](#def-b1-mirrors-thin-lenses-lens), forms sharp on the object itself, inverted and the same size. The object is then in the focal plane: rays leave the [lens](#def-b1-mirrors-thin-lenses-lens) parallel, return parallel from the mirror, and are refocused in the focal plane. The object–lens distance is $f'$.

**Proposition 3.17 (Bessel’s method).**

An object and a screen are fixed a distance $D$ apart. A [converging lens](#def-b1-mirrors-thin-lenses-lens) forms a sharp [real image](#def-b1-mirrors-thin-lenses-image) on the screen for exactly two positions if $D > 4f'$, for one if $D = 4f'$, for none if $D < 4f'$. The two positions are a distance $d$ apart with

$$
f' = \frac{D^2 - d^2}{4D} .
$$

At $D = 4f'$ (Silbermann) the single position gives $\gamma = -1$.

**Proof.** Let $p = -\overline{OA} > 0$ be the object–lens distance; then $\overline{OA'} = D - p$ and Descartes gives $1/(D - p) + 1/p = 1/f'$, i.e. $p^2 - Dp + Df' = 0$: discriminant $D^2 - 4Df'$, positive iff $D > 4f'$. The two roots $p_{1,2} = (D \mp \sqrt{D^2 - 4Df'})/2$ add up to $D$ (the positions are symmetric: object and [image](#def-b1-mirrors-thin-lenses-image) distances are exchanged) and differ by $d = \sqrt{D^2 - 4Df'}$; solve for $f'$. At $D = 4f'$, $p = 2f' = \overline{OA'}$ and $\gamma = -1$. ∎

![Bessel’s method: object and screen fixed at D > 4f'; the lens gives a sharp image at two symmetric positions a distance d apart (enlarged, reduced), and f' = (D2 - d2)/4D.](https://one-course.com/images/onecourse/chapters/physics-3/b1-mirrors-thin-lenses/fig-8363a3544849.svg)

*[Bessel’s method](#prop-b1-mirrors-thin-lenses-bessel): object and screen fixed at $D > 4f'$; the [lens](#def-b1-mirrors-thin-lenses-lens) gives a sharp [image](#def-b1-mirrors-thin-lenses-image) at two symmetric positions a distance $d$ apart (enlarged, reduced), and $f' = (D^2 - d^2)/4D$.*

## 3.5 Two thin lenses

**Proposition 3.18 (Lenses in contact; afocal doublet).**

Two thin lenses in contact behave as one [thin lens](#def-b1-mirrors-thin-lenses-lens) of [vergence](#def-b1-mirrors-thin-lenses-lens) $V = V_1 + V_2$. Two lenses placed so that $F_1' = F_2$ form an *afocal* system: a parallel beam emerges parallel, and a beam inclined by $\alpha$ emerges inclined by $\alpha' = G\alpha$ with the *angular magnification* $G = -f_1'/f_2'$.

**Proof.** The [image](#def-b1-mirrors-thin-lenses-image) $A_1$ given by $L_1$ is the object for $L_2$. In contact, $O_1 = O_2 = O$: $1/\overline{OA_1} - 1/\overline{OA} = V_1$ and $1/\overline{OA'} - 1/\overline{OA_1} = V_2$; add. Afocal: a parallel beam at angle $\alpha$ focuses in the common focal plane at height $h = f_1'\alpha$ from the axis (ray through $O_1$); seen from $L_2$ this point is in its object focal plane, so the beam emerges parallel, at the angle of the ray from the point through $O_2$: $\alpha' = -h/f_2'$. ∎

**Example 3.19 (A telescope in one line).**

$f_1' = 1000\,\mathrm{mm}$, $f_2' = 25\,\mathrm{mm}$: $G = -40$; the Moon, half a degree wide, becomes a $20^\circ$ disk — the astronomical telescope of the next chapter. With a diverging eyepiece, $f_2' = -25\,\mathrm{mm}$, $G = +40$: upright, Galileo’s spyglass.

## 3.6 Exercises

**Exercise 3.1 ★.**

A [concave mirror](#def-b1-mirrors-thin-lenses-mirror) has $R = 40\,\mathrm{cm}$. Find the [image](#def-b1-mirrors-thin-lenses-image) of an object placed at $60\,\mathrm{cm}$, $30\,\mathrm{cm}$ and $10\,\mathrm{cm}$ from the vertex: position, real or virtual, magnification.

**Solution of Exercise 3.1.**

$\overline{SF} = -20\,\mathrm{cm}$; $1/\overline{SA'} = 1/\overline{SF} -
1/\overline{SA}$. At $60\,\mathrm{cm}$: $\overline{SA'} = -30\,\mathrm{cm}$, real, $\gamma = -0.5$. At $30\,\mathrm{cm}$: $\overline{SA'} = -60\,\mathrm{cm}$, real, $\gamma = -2$. At $10\,\mathrm{cm}$: $1/\overline{SA'} = -1/20 + 1/10
= 1/20$, $\overline{SA'} = +20\,\mathrm{cm}$ behind the mirror, virtual, $\gamma = +2$ (upright, enlarged).

**Exercise 3.2 ★.**

A [converging lens](#def-b1-mirrors-thin-lenses-lens) has $f' = 5.0\,\mathrm{cm}$. Find the [image](#def-b1-mirrors-thin-lenses-image) of an object at $20\,\mathrm{cm}$, then at $3.0\,\mathrm{cm}$; sketch the constructions.

**Solution of Exercise 3.2.**

$\overline{OA} = -20\,\mathrm{cm}$: $1/\overline{OA'} = 1/5 - 1/20 = 3/20$, $\overline{OA'} = +6.7\,\mathrm{cm}$, real inverted, $\gamma = -1/3$. $\overline{OA} = -3.0\,\mathrm{cm}$: $1/\overline{OA'} = 1/5 - 1/3 = -2/15$, $\overline{OA'} = -7.5\,\mathrm{cm}$, virtual upright, $\gamma = +2.5$.

**Exercise 3.3 ★.**

Give the [vergence](#def-b1-mirrors-thin-lenses-lens) of a [lens](#def-b1-mirrors-thin-lenses-lens) of [focal length](#def-b1-mirrors-thin-lenses-lens) $25\,\mathrm{cm}$; the [focal length](#def-b1-mirrors-thin-lenses-lens) of a $-2.0\,\delta$ [lens](#def-b1-mirrors-thin-lenses-lens); the [focal length](#def-b1-mirrors-thin-lenses-lens) of a $+3.0\,\delta$ and a $-1.0\,\delta$ [lens](#def-b1-mirrors-thin-lenses-lens) in contact.

**Solution of Exercise 3.3.**

$V = 1/0.25 = +4.0\,\delta$; $f' = 1/(-2.0) = -50\,\mathrm{cm}$; $V = 3.0 - 1.0 = +2.0\,\delta$, $f' = 50\,\mathrm{cm}$.

**Exercise 3.4 ★.**

With $f' = 10\,\mathrm{cm}$ and an object $15\,\mathrm{cm}$ in front of $F$, use Newton’s relation to locate the [image](#def-b1-mirrors-thin-lenses-image) and find $\gamma$; check with Descartes’s relation.

**Solution of Exercise 3.4.**

$\overline{FA} = -15\,\mathrm{cm}$: $\overline{F'A'} = -f'^2/\overline{FA}
= +100/15 = 6.7\,\mathrm{cm}$ past $F'$, i.e. $\overline{OA'} = 16.7\,\mathrm{cm}$; $\gamma = f'/\overline{FA} = -0.67$. Descartes: $\overline{OA} =
-25\,\mathrm{cm}$, $1/\overline{OA'} = 1/10 - 1/25 = 3/50$, $\overline{OA'} = 16.7\,\mathrm{cm}$, $\gamma = 16.7/(-25) = -0.67$.

**Exercise 3.5 ★★.**

A convex rear-view mirror has $R = 2.0\,\mathrm{m}$. A car $1.5\,\mathrm{m}$ high is $20\,\mathrm{m}$ behind it: find the [image](#def-b1-mirrors-thin-lenses-image) position, its height, and explain the warning “objects in mirror are closer than they appear”.

**Solution of Exercise 3.5.**

$\overline{SF} = +1.0\,\mathrm{m}$, $\overline{SA} = -20\,\mathrm{m}$: $1/\overline{SA'} = 1 + 1/20$, $\overline{SA'} = 0.95\,\mathrm{m}$ behind the mirror (virtual, upright); $\gamma = -0.95/(-20) = +0.048$, [image](#def-b1-mirrors-thin-lenses-image) height $7.1\,\mathrm{cm}$. The brain judges distance from apparent size: a car shown twenty times smaller at $0.95\,\mathrm{m}$ is read as a car far away — farther than $20\,\mathrm{m}$.

**Exercise 3.6 ★★.**

A camera [lens](#def-b1-mirrors-thin-lenses-lens), $f' = 50\,\mathrm{mm}$, focuses on a subject $2.0\,\mathrm{m}$ away. By how much must the [lens](#def-b1-mirrors-thin-lenses-lens) move from its infinity setting? What is the magnification? How far back must the photographer stand for a $1.8\,\mathrm{m}$ person to fit the $24\,\mathrm{mm}$ height of the sensor?

**Solution of Exercise 3.6.**

$1/\overline{OA'} = 1/50 - 1/2000 = 0.0195\,\mathrm{mm}^{-1}$, $\overline{OA'} = 51.3\,\mathrm{mm}$: the [lens](#def-b1-mirrors-thin-lenses-lens) moves $1.3\,\mathrm{mm}$ away from the sensor. $\gamma = 51.3/(-2000) = -0.026$. For $\abs\gamma = 24/1800
= 1/75$: $\overline{OA} = f'(1/\gamma - 1) = 50 \times (-76) =
-3.8\,\mathrm{m}$.

**Exercise 3.7 ★★.**

A projector must show a $36\,\mathrm{mm}$ wide slide as a $1.8\,\mathrm{m}$ wide picture on a screen $5.0\,\mathrm{m}$ from the [lens](#def-b1-mirrors-thin-lenses-lens). Find the required [focal length](#def-b1-mirrors-thin-lenses-lens) and the slide–lens distance. Why is the slide inserted upside down?

**Solution of Exercise 3.7.**

$\gamma = -1.8/0.036 = -50$; $\overline{OA'} = 5.0\,\mathrm{m}$, $\overline{OA} = \overline{OA'}/\gamma = -10\,\mathrm{cm}$; $1/f' = 1/5.0 + 1/0.10 = 10.2\,\mathrm{m}^{-1}$, $f' = 98\,\mathrm{mm}$. The [image](#def-b1-mirrors-thin-lenses-image) is inverted ($\gamma < 0$), top–bottom and left–right: the slide goes in rotated by $180^\circ$.

**Exercise 3.8 ★★.**

Explain, with a ray diagram, why autocollimation locates the focal plane exactly, and why the returned [image](#def-b1-mirrors-thin-lenses-image) is inverted and the same size as the object. What happens to the returned [image](#def-b1-mirrors-thin-lenses-image) if the mirror is tilted slightly?

**Solution of Exercise 3.8.**

Object in the focal plane $\Rightarrow$ rays from each point leave the [lens](#def-b1-mirrors-thin-lenses-lens) parallel; the [plane mirror](https://one-course.com/books/physics/3/en/chapter/2-geometric-optics-rays-reflection-refraction#prop-b1-rays-reflection-refraction-mirror) sends them back parallel (symmetric direction); the [lens](#def-b1-mirrors-thin-lenses-lens) focuses a parallel beam in its focal plane, at the point symmetric about the axis of the source point: [image](#def-b1-mirrors-thin-lenses-image) in the object plane, inverted, $\gamma = -1$. Tilting the mirror by $\alpha$ turns the returning beam by $2\alpha$ and shifts the [image](#def-b1-mirrors-thin-lenses-image) sideways by $2\alpha f'$, still sharp in the same plane.

**Exercise 3.9 ★★.**

Object and screen are $1.00\,\mathrm{m}$ apart; sharp [images](#def-b1-mirrors-thin-lenses-image) are obtained for two [lens](#def-b1-mirrors-thin-lenses-lens) positions $40\,\mathrm{cm}$ apart. Find $f'$. What is the shortest object–screen distance at which this [lens](#def-b1-mirrors-thin-lenses-lens) can still project a sharp [image](#def-b1-mirrors-thin-lenses-image), and with what magnification?

**Solution of Exercise 3.9.**

$f' = (100^2 - 40^2)/(4 \times 100) = 21\,\mathrm{cm}$. Shortest distance $4f' = 84\,\mathrm{cm}$, with $\gamma = -1$.

**Exercise 3.10 ★★★.**

A [converging lens](#def-b1-mirrors-thin-lenses-lens) $L_1$ ($f_1' = 20\,\mathrm{cm}$) is followed, $15\,\mathrm{cm}$ further, by a [diverging lens](#def-b1-mirrors-thin-lenses-lens) $L_2$ ($f_2' = -10\,\mathrm{cm}$). Locate the [image](#def-b1-mirrors-thin-lenses-image) of an object at infinity. Show that a distant object of angular size $\alpha$ gives an [image](#def-b1-mirrors-thin-lenses-image) of height $40\,\mathrm{cm}\times\alpha$, and compare with the length of the arrangement: the principle of the telephoto [lens](#def-b1-mirrors-thin-lenses-lens).

**Solution of Exercise 3.10.**

$L_1$ [images](#def-b1-mirrors-thin-lenses-image) infinity at $F_1'$, $20\,\mathrm{cm}$ past $L_1$, i.e. $5\,\mathrm{cm}$ past $L_2$: a virtual object, $\overline{O_2A_1} = +5\,\mathrm{cm}$. $1/\overline{O_2A'} = -1/10 + 1/5 = 1/10$: final [image](#def-b1-mirrors-thin-lenses-image) $10\,\mathrm{cm}$ past $L_2$. Intermediate [image](#def-b1-mirrors-thin-lenses-image) height $h_1 = f_1'\alpha = 20\alpha$ (cm); $\gamma_2 = 10/5 = 2$, so $h = 40\alpha$: equivalent [focal length](#def-b1-mirrors-thin-lenses-lens) $40\,\mathrm{cm}$, in a barrel $15 + 10 = 25\,\mathrm{cm}$ long — a telephoto.

**Exercise 3.11 ★★★.**

A [concave mirror](#def-b1-mirrors-thin-lenses-mirror) forms, on a screen $80\,\mathrm{cm}$ from a lamp filament, an [image](#def-b1-mirrors-thin-lenses-image) of the filament three times its size. Find the positions of the mirror and its radius of curvature.

**Solution of Exercise 3.11.**

$\gamma = -3 = -\overline{SA'}/\overline{SA}$ gives $\overline{SA'} =
3\overline{SA}$; both real, in front: $\overline{SA} = -x$, $\overline{SA'} = -3x$, and screen–lamp distance $2x = 80\,\mathrm{cm}$: $x = 40\,\mathrm{cm}$. Mirror $40\,\mathrm{cm}$ from the lamp, $120\,\mathrm{cm}$ from the screen; $1/\overline{SF} = -1/40 - 1/120 = -1/30$: $R = 2 \times 30 = 60\,\mathrm{cm}$.

**Exercise 3.12 ★★★.**

The Sun subtends $0.53^\circ$. A [converging lens](#def-b1-mirrors-thin-lenses-lens) of diameter $10\,\mathrm{cm}$ and [focal length](#def-b1-mirrors-thin-lenses-lens) $50\,\mathrm{cm}$ forms its [image](#def-b1-mirrors-thin-lenses-image) on a sheet of paper. Compute the [image](#def-b1-mirrors-thin-lenses-image) diameter, the power collected (solar irradiance $1.0\,\mathrm{kW}/\mathrm{m}^{2}$) and the irradiance in the [image](#def-b1-mirrors-thin-lenses-image); compare with the Sun’s direct irradiance and conclude.

**Solution of Exercise 3.12.**

$\theta = 0.53^\circ = 9.3 \times 10^{-3}\,\mathrm{rad}$; [image](#def-b1-mirrors-thin-lenses-image) diameter $f'\theta =
4.6\,\mathrm{mm}$. Collected power $10^3 \times \pi(0.05)^2 = 7.9\,\mathrm{W}$; [image](#def-b1-mirrors-thin-lenses-image) area $\pi(2.3\times10^{-3})^2 = 1.7 \times 10^{-5}\,\mathrm{m}^{2}$; irradiance $4.7 \times 10^{5}\,\mathrm{W}/\mathrm{m}^{2}$, $470$ times the direct sunlight: paper chars and ignites — the burning glass. In general the gain is $(D/f'\theta)^2$.

## 3.7 Problem: The slide projector

**Problem 3.1.**

Weekend problem — one lens, measured on a bench, then asked to throw a 35-mm slide across a hall: how a projector is designed, focused, and fitted with a long-throw converter

A projection [lens](#def-b1-mirrors-thin-lenses-lens) $L_1$ is a converging [thin lens](#def-b1-mirrors-thin-lenses-lens) of unknown [focal length](#def-b1-mirrors-thin-lenses-lens) $f_1'$. The slide is $24\,\mathrm{mm}$ high and $36\,\mathrm{mm}$ wide.

**Part I — The [lens](#def-b1-mirrors-thin-lenses-lens) on the bench.**

1. State Descartes’s relation and the magnification for a [thin lens](#def-b1-mirrors-thin-lenses-lens) , with the sign conventions.
2. Object and screen are fixed a distance $D$ apart. Show that a [real image](#def-b1-mirrors-thin-lenses-image) on the screen requires $D \geq 4f_1'$ .
3. Show that, when $D > 4f_1'$ , the two [lens](#def-b1-mirrors-thin-lenses-lens) positions giving a sharp [image](#def-b1-mirrors-thin-lenses-image) are symmetric with respect to the midpoint of object and screen, and that $f_1' = (D^2 - d^2)/4D$ with $d$ their separation.
4. With $D = 80.0\,\mathrm{cm}$ and $d = 40.0\,\mathrm{cm}$ , compute $f_1'$ .
5. The [lens](#def-b1-mirrors-thin-lenses-lens) sits in a thick barrel whose [optical center](#def-b1-mirrors-thin-lenses-lens) cannot be located precisely. Why does this not affect the method?
6. The distances carry $u(D) = 0.2\,\mathrm{cm}$ and $u(d) = 0.5\,\mathrm{cm}$ . Compute $u(f_1')$ and write the result.
7. Give the magnification at each of the two positions. Which one corresponds to a projector?
8. What is the smallest $D$ for which a sharp [image](#def-b1-mirrors-thin-lenses-image) exists, and the magnification then (Silbermann)?

**Part II — Throwing the picture.** The picture on the screen must be $1.80\,\mathrm{m}$ wide. Take $f_1' = 15.0\,\mathrm{cm}$.

9. What magnification is required (with its sign)?
10. Compute the lens–screen distance and the slide–lens distance.
11. Why is the slide inserted upside down and mirror-reversed?
12. The screen is moved to $10.0\,\mathrm{m}$ . By how much, and in which direction, must the [lens](#def-b1-mirrors-thin-lenses-lens) be moved relative to the slide to refocus? How wide is the picture now?

**Part III — [Focus](#def-b1-mirrors-thin-lenses-mirror) tolerance and light.**

13. Show that a small displacement $\dd\overline{OA}$ of the slide shifts the [image](#def-b1-mirrors-thin-lenses-image) by $\dd\overline{OA'} = \gamma^2\,\dd\overline{OA}$ .
14. The film buckles by $0.10\,\mathrm{mm}$ under the lamp’s heat. By how much does the [image](#def-b1-mirrors-thin-lenses-image) plane move (screen at $7.65\,\mathrm{m}$ )? Comment.
15. The [lens](#def-b1-mirrors-thin-lenses-lens) aperture is $30\,\mathrm{mm}$ in diameter. If the screen sits $20\,\mathrm{cm}$ off the [image](#def-b1-mirrors-thin-lenses-image) plane, what is the diameter of the blur spot of a point of the slide? Seen from $5\,\mathrm{m}$ , does the eye (angular resolution about $3 \times 10^{-4}\,\mathrm{rad}$ ) notice?
16. The slide receives $2.0\,\mathrm{W}$ of light, all of which reaches the screen. Compute the irradiance (power per unit area) on the slide and on the screen, and show that their ratio is $\gamma^2$ .

**Part IV — The long-throw converter.** In a larger hall the screen is $12.0\,\mathrm{m}$ away. A [diverging lens](#def-b1-mirrors-thin-lenses-lens) $L_2$, $f_2' = -30.0\,\mathrm{cm}$, is clipped on $8.0\,\mathrm{cm}$ behind $L_1$; the screen is $12.0\,\mathrm{m}$ from $L_2$.

17. For the final [image](#def-b1-mirrors-thin-lenses-image) to be real and on the screen, where (from $L_2$ ) must the intermediate [image](#def-b1-mirrors-thin-lenses-image) given by $L_1$ lie? Is it a real or a virtual object for $L_2$ ?
18. Deduce the slide position $\overline{O_1A}$ .
19. Compute $\gamma_1$ , $\gamma_2$ and the total magnification.
20. Give the picture width, and compare with what $L_1$ alone would give on the same screen ( $12.08\,\mathrm{m}$ from $L_1$ ).
21. For an object at infinity, show that the system is equivalent to a single [lens](#def-b1-mirrors-thin-lenses-lens) of [focal length](#def-b1-mirrors-thin-lenses-lens) $f_{\mathrm{eq}}$ with $$\frac{1}{f_{\mathrm{eq}}} = \frac{1}{f_1'} + \frac{1}{f_2'} - \frac{e}{f_1' f_2'},$$ $e = \overline{O_1O_2}$, in the sense that a beam inclined by $\alpha$ focuses at height $f_{\mathrm{eq}}\,\alpha$ from the axis. Compute $f_{\mathrm{eq}}$.
22. Check that the ratio of picture widths (with/without converter) is close to $f_1'/f_{\mathrm{eq}}$ . Why only approximately?
23. Why must the intermediate [image](#def-b1-mirrors-thin-lenses-image) lie less than $\abs{f_2'}$ beyond $L_2$ for the final [image](#def-b1-mirrors-thin-lenses-image) to be real?
24. With the same lamp, by what factor does the screen irradiance change when the converter is fitted (same screen distance)?
25. Summarize: the converter multiplies the [focal length](#def-b1-mirrors-thin-lenses-lens) by which factor, and what does it do to picture size and brightness at a given throw?

**Solution of Problem 3.1.**

**1.** $1/\overline{OA'} - 1/\overline{OA} = 1/f'$, $\gamma = \overline{OA'}/\overline{OA}$; axis oriented along the light, distances algebraic from $O$.

**2.** $p = -\overline{OA}$, $\overline{OA'} = D - p$: $1/(D - p) + 1/p = 1/f_1'$, i.e. $p^2 - Dp + Df_1' = 0$; real roots iff $D^2 - 4Df_1' \geq 0$, i.e. $D \geq 4f_1'$.

**3.** The roots sum to $D$: $p_1 + p_2 = D$, so $p_2 = D - p_1$ — object and [image](#def-b1-mirrors-thin-lenses-image) distances swap, the positions are symmetric about the midpoint. $d = p_2 - p_1 = \sqrt{D^2 - 4Df_1'}$, hence $f_1' = (D^2 -
d^2)/4D$.

**4.** $f_1' = (6400 - 1600)/320 = 15.0\,\mathrm{cm}$.

**5.** Only the displacement $d$ of the barrel between the two sharp settings and the object–screen distance $D$ enter: neither requires knowing where $O$ sits inside the barrel.

**6.** $\partial f_1'/\partial D = 1/4 + d^2/4D^2 = 0.3125$, $\partial f_1'/\partial d = -d/2D = -0.25$; so $u(f_1')^2 = (0.3125 \times 0.2)^2 + (0.25 \times 0.5)^2 =
0.0195\,\mathrm{cm}^{2}$ and $u(f_1') = 0.14\,\mathrm{cm}$: $f_1' = (15.00 \pm 0.14)\,\mathrm{cm}$.

**7.** $p = (80 \mp 40)/2 = 20$ or $60\,\mathrm{cm}$: $\gamma =
-60/20 = -3$ (slide near the [lens](#def-b1-mirrors-thin-lenses-lens), enlarged: the projector) and $\gamma = -1/3$.

**8.** $D_{\min} = 4f_1' = 60.0\,\mathrm{cm}$, $\gamma = -1$.

**9.** $\gamma = -1.80/0.036 = -50$.

**10.** $\overline{OA'} = f_1'(1 - \gamma) = 15 \times 51 =
765\,\mathrm{cm} = 7.65\,\mathrm{m}$; $\overline{OA} = \overline{OA'}/\gamma =
-15.3\,\mathrm{cm}$.

**11.** $\gamma < 0$: the [real image](#def-b1-mirrors-thin-lenses-image) is rotated by $180^\circ$ about the axis, so the slide is inserted rotated by $180^\circ$ (upside down and mirror-reversed).

**12.** $\overline{OA'} = 1000\,\mathrm{cm}$: $1/\overline{OA} = 1/1000
- 1/15$, $\overline{OA} = -15.23\,\mathrm{cm}$: the [lens](#def-b1-mirrors-thin-lenses-lens) moves $0.7\,\mathrm{mm}$ toward the slide. $\gamma = 1000/(-15.23) = -65.7$: picture $2.36\,\mathrm{m}$ wide.

**13.** Differentiating Descartes: $-\dd\overline{OA'}/\overline{OA'}^2
+ \dd\overline{OA}/\overline{OA}^2 = 0$, so $\dd\overline{OA'} =
(\overline{OA'}/\overline{OA})^2\,\dd\overline{OA} = \gamma^2\,\dd\overline{OA}$.

**14.** $2500 \times 0.10\,\mathrm{mm} = 25\,\mathrm{cm}$: the [image](#def-b1-mirrors-thin-lenses-image) drifts out of [focus](#def-b1-mirrors-thin-lenses-mirror) as the film warms — hence glass-mounted slides or an autofocus that tracks the film.

**15.** Similar triangles from the aperture to the [image](#def-b1-mirrors-thin-lenses-image) point: blur $= 30 \times 0.20/7.65 = 0.78\,\mathrm{mm}$; from $5\,\mathrm{m}$ this subtends $1.6 \times 10^{-4}\,\mathrm{rad} < 3 \times 10^{-4}\,\mathrm{rad}$: not noticed.

**16.** Slide area $8.64\times10^{-4}\ \mathrm{m}^{2}$: $E_{\text{slide}}
= 2.0/8.64\times10^{-4} = 2.3 \times 10^{3}\,\mathrm{W}/\mathrm{m}^{2}$; screen area $\gamma^2$ times larger, $2.16\,\mathrm{m}^{2}$: $E_{\text{screen}} = 0.93\,\mathrm{W}/\mathrm{m}^{2}$; same power, areas in ratio $\gamma^2 = 2500$.

**17.** $\overline{O_2A'} = 1200\,\mathrm{cm}$: $1/\overline{O_2A} =
1/1200 + 1/30 = 0.03417$, $\overline{O_2A} = +29.3\,\mathrm{cm}$: the intermediate [image](#def-b1-mirrors-thin-lenses-image) lies $29.3\,\mathrm{cm}$ *beyond* $L_2$ — a virtual object for $L_2$ (the rays from $L_1$ converge toward it).

**18.** $\overline{O_1A_1'} = 8.0 + 29.3 = 37.3\,\mathrm{cm}$; $1/\overline{O_1A} = 1/37.3 - 1/15 = -0.0398$, $\overline{O_1A} =
-25.1\,\mathrm{cm}$.

**19.** $\gamma_1 = 37.3/(-25.1) = -1.49$; $\gamma_2 = 1200/29.3 =
41.0$; $\gamma = -60.9$.

**20.** Width $60.9 \times 36 = 2.19\,\mathrm{m}$. $L_1$ alone: $\gamma = 1 - \overline{OA'}/f_1' = 1 - 1208/15 = -79.5$, width $2.86\,\mathrm{m}$.

**21.** Beam at $\alpha$: $L_1$ focuses it at height $h_1 = f_1'\alpha$ in its focal plane, which is $\overline{O_2A_1} = f_1' - e$ from $L_2$; $L_2$ [images](#def-b1-mirrors-thin-lenses-image) this point at $\overline{O_2A'}$ with $1/\overline{O_2A'}
= 1/f_2' + 1/(f_1' - e)$ and magnifies by $\gamma_2 = \overline{O_2A'}/(f_1'
- e)$, so $h = \gamma_2 f_1'\alpha$ and $f_{\mathrm{eq}} = \gamma_2 f_1'
= \dfrac{f_1' f_2'}{f_1' + f_2' - e}$, whose inverse is the stated formula. $1/f_{\mathrm{eq}} = 1/15 - 1/30 + 8/450 = 0.0511\,\mathrm{cm}^{-1}$, $f_{\mathrm{eq}} = 19.6\,\mathrm{cm}$.

**22.** $2.19/2.86 = 0.77$ and $15/19.6 = 0.77$. Only approximately because the slide is at finite distance and the two throws are measured from different lenses; for an object at infinity the ratio would be exact.

**23.** Virtual object at $p > 0$ beyond a [diverging lens](#def-b1-mirrors-thin-lenses-lens): $1/\overline{O_2A'} = 1/p + 1/f_2' = 1/p - 1/\abs{f_2'}$, positive ([real image](#def-b1-mirrors-thin-lenses-image)) iff $p < \abs{f_2'}$.

**24.** Same power on a picture of area $(2.19/2.86)^2 = 0.59$ times smaller: irradiance $\times 1.7$.

**25.** The converter multiplies the [focal length](#def-b1-mirrors-thin-lenses-lens) by $f_{\mathrm{eq}}/f_1' = 1.3$; at a given throw the picture is $1.3$ times smaller in each direction and $1.7$ times brighter — a long-throw [lens](#def-b1-mirrors-thin-lenses-lens).