University Physics — Year 1 · Bachelor Year 1

4Optical Instruments and the Eye

From the back row of a concert hall a pair of binoculars pulls the singer’s face to within arm’s reach; a microscope turns a drop of pond water into a zoo; a phone the thickness of a finger photographs the Milky Way. Every one of these instruments is a few thin lenses arranged for one purpose: to present to the eye, or to a sensor, an image larger, brighter or sharper than the eye could form alone. So the chapter begins with the eye itself — what it can and cannot do — and then builds the magnifier, the microscope, the telescope and the camera out of the conjugation relation of the previous chapter.

The 40-inch refractor of Yerkes Observatory (1897), still the largest lens telescope ever built: an objective of one metre and a tube of nineteen. Beyond this size a lens sags under its own weight, and astronomy turned to mirrors.

A small refractor at dusk: an objective of long focal length and an eyepiece of short one — the afocal telescope of this chapter, aimed at the Moon of its weekend problem.

4.1 The eye

Definition 4.1 (Reduced eye)

The reduced eye models the cornea and the crystalline lens as a single thin converging lens of variable focal length, and the retina as a screen at a fixed distance $d \approx 17\,\mathrm{mm}$ behind it. Muscles change the lens’s curvature to keep the image of the object looked at on the retina: this is accommodation. The iris in front of the lens is a variable aperture of diameter $2\,$ to $8\,\mathrm{mm}$ , the pupil.

Definition 4.2 (Far point, near point)

The far point (punctum remotum) is the object point seen sharply without accommodating; the near point (punctum proximum) the closest point seen sharply with maximal accommodation. For a normal (emmetropic) eye the far point is at infinity and the near point at $d_m \approx 25\,\mathrm{cm}$ , the conventional distance of distinct vision. The difference of vergences needed for the two is the amplitude of accommodation: about $4\,\delta$ for a young adult, falling with age.

Example 4.3 (The eye’s vergence)

Distant object: image at $d = 17\,\mathrm{mm}$ means $f' = 17\,\mathrm{mm}$ , $V = 1/0.017 = 59\,\delta$ . Object at $25\,\mathrm{cm}$ : $V = 1/0.017 + 1/0.25 = 59 + 4 = 63\,\delta$ . Four diopters of accommodation out of sixty: the crystalline lens adds only a few percent to the cornea’s job, but those few percent are everything for reading.

Proposition 4.4 (Defects and corrections)

A myopic eye is too convergent: its far point is at a finite distance $D_r$ ; a diverging lens of focal length $f' = -D_r$ (worn close to the eye) images infinity at the far point and restores distant vision. A hyperopic eye is not convergent enough: its far point is virtual (behind the eye) and its near point too far; a converging lens corrects it. Presbyopia is the loss of accommodation with age: the near point recedes and reading needs converging lenses, whatever the far point.

Proof. A lens close to the eye and the eye form a system whose object must be what the eye can see: the lens must turn an object at infinity into a virtual image at the far point, $\overline{OA'} = -D_r$ for $\overline{OA} = -\infty$ , so $f' = \overline{OA'} = -D_r$ . The other cases are the same argument with the relevant points. ∎

The reduced eye: a thin lens of variable focal length and a retina 17\, mm behind it. Relaxed, it focuses distant objects; accommodating, it shortens f' to bring a near object’s image onto the retina. The image is inverted — the brain turns it back. — The reduced eye: a thin lens of variable focal length and a retina $17\,\mathrm{mm}$ behind it. Relaxed, it focuses distant objects; accommodating, it shortens $f'$ to bring a near object’s image onto the retina. The image is inverted — the brain turns it back.

Definition 4.5 (Angular size and resolution)

The angular size of an object of height $h$ at distance $D$ is $\theta \approx h/D$ (small angles). The eye distinguishes two points only if their angular separation exceeds its angular resolution, about $\epsilon \approx 3 \times 10^{-4}\,\mathrm{rad}$ (one arcminute): the spacing of the cone cells and diffraction at the pupil both set this limit.

Remark 4.6 (What an instrument is for)

Brought to the near point, a $1\,\mathrm{mm}$ detail subtends $1/250 = 4 \times 10^{-3}\,\mathrm{rad}$ : thirteen times the resolution limit; a $0.05\,\mathrm{mm}$ detail subtends $2 \times 10^{-4}\,\mathrm{rad}$ and is invisible. Instruments that look at near objects (magnifier, microscope) increase the angular size beyond what bringing the object closer can do; instruments for distant objects (telescope) increase the angular size of things one cannot approach; the camera replaces the retina by a sensor that can integrate light for seconds and be enlarged afterwards.

4.2 The magnifier

Definition 4.7 (Angular magnification)

For an instrument used by the eye, the angular magnification is $G = \theta'/\theta$ , where $\theta'$ is the angular size of the image seen through the instrument and $\theta$ the angular size of the object seen with the naked eye — at the near point $d_m = 25\,\mathrm{cm}$ for a near object, in place for a distant one.

Proposition 4.8 (The magnifier)

A converging lens of focal length $f' < d_m$ with the object in its object focal plane gives an image at infinity, seen without accommodation, with

G = \frac{d_m}{f'} = d_m V .

Proof. Object $AB$ of height $h$ in the focal plane: rays from $B$ leave parallel, at the angle of the ray through $O$ , $\theta' = h/f'$ . Naked eye at best: $\theta = h/d_m$ . Ratio $d_m/f'$ . ∎

Example 4.9 (A watchmaker’s loupe)

$f' = 25\,\mathrm{mm}$ : $G = 250/25 = 10$ , sold as “ $10\times$ ”. A $0.05\,\mathrm{mm}$ detail then subtends $2 \times 10^{-3}\,\mathrm{rad}$ , comfortably resolved. Pushing $G$ higher means $f'$ of a few millimeters, a tiny lens held against the eye — the limit of the single lens, and the reason the microscope exists.

4.3 The microscope

Definition 4.10 (Compound microscope)

A microscope is an objective $L_1$ of short focal length $f_1'$ forming a real, enlarged intermediate image of a nearby object in the object focal plane of an eyepiece $L_2$ (focal length $f_2'$ ), which acts as a magnifier. The distance $\Delta = \overline{F_1'F_2}$ between the objective’s image focus and the eyepiece’s object focus is the optical interval, a standardized $160\,\mathrm{mm}$ on classical instruments.

Proposition 4.11 (Magnification of a microscope)

With the final image at infinity,

\gamma_1 = -\frac{\Delta}{f_1'}, \qquad G_2 = \frac{d_m}{f_2'}, \qquad G = \abs{\gamma_1}\,G_2 = \frac{\Delta\,d_m}{f_1' f_2'} .

Proof. The intermediate image $A_1B_1$ lies in the eyepiece’s object focal plane, i.e. at $\overline{F_1'A_1} = \Delta$ ; Newton’s relation gives $\gamma_1 = -\overline{F_1'A_1}/f_1' = -\Delta/f_1'$ . The eyepiece then shows $A_1B_1$ (height $\abs{\gamma_1} h$ ) at infinity under the angle $\theta' = \abs{\gamma_1} h/f_2'$ , against $\theta = h/d_m$ for the naked eye. ∎

The microscope: the objective forms a real, enlarged, inverted intermediate image A_1B_1 in the object focal plane of the eyepiece; the eyepiece sends it to infinity, where the relaxed eye views it under a large angle. — The microscope: the objective forms a real, enlarged, inverted intermediate image $A_1B_1$ in the object focal plane of the eyepiece; the eyepiece sends it to infinity, where the relaxed eye views it under a large angle.

Example 4.12 (A student microscope)

Objective $f_1' = 4.0\,\mathrm{mm}$ , eyepiece $f_2' = 25\,\mathrm{mm}$ , $\Delta = 160\,\mathrm{mm}$ : $\gamma_1 = -40$ , $G_2 = 10$ , $G = 400$ . The object sits at $\overline{F_1A} = -f_1'^2/\Delta = -16/160 = -0.10\,\mathrm{mm}$ from $F_1$ : $4.1\,\mathrm{mm}$ from the objective. Focusing means moving the whole tube by fractions of a millimeter.

Remark 4.13 (The resolution limit)

Magnification without limit would be useless: light is a wave, and an objective accepting rays within a half-angle $u$ (in a medium of index $n$ ) cannot separate two points closer than about $\lambda/(2n\sin u)$ — the wave treatment of the Year 2 volume derives it. With $n\sin u \approx 1.4$ (oil immersion) and $\lambda = 0.5\,\text{µ}\mathrm{m}$ , the limit is near $0.2\,\text{µ}\mathrm{m}$ . To make that detail visible to the eye ( $\epsilon = 3 \times 10^{-4}\,\mathrm{rad}$ at $25\,\mathrm{cm}$ , i.e. $75\,\text{µ}\mathrm{m}$ ), a magnification of about $400$ suffices; beyond $1000$ the image is only larger, not richer: empty magnification.

4.4 The astronomical telescope

Proposition 4.14 (Refracting telescope)

An objective $L_1$ and an eyepiece $L_2$ with $F_1' = F_2$ form an afocal system: a star (at infinity) is seen at infinity, the relaxed eye views it, and

G = -\frac{f_1'}{f_2'} .

The image of the objective’s rim through the eyepiece is the exit pupil: a bright disk of diameter $D/\abs G$ located $\overline{O_2P'} = f_2'(f_1' + f_2')/f_1'$ behind the eyepiece, where the eye’s pupil must be placed to receive all the light.

Proof. $G$ was derived in Proposition 3.18. The objective, seen from $L_2$ , is an object at $\overline{O_2O_1} = -(f_1' + f_2')$ ; Descartes gives $1/\overline{O_2P'} = 1/f_2' - 1/(f_1' + f_2') = f_1'/[f_2'(f_1' + f_2')]$ , and the magnification $\overline{O_2P'}/ \overline{O_2O_1} = -f_2'/f_1' = 1/G$ scales the diameter $D$ to $D/\abs G$ . Every ray entering the objective leaves through the exit pupil, since each point of the objective is imaged there. ∎

The astronomical telescope: a parallel beam from a star is focused by the objective in the common focal plane and sent back to infinity by the eyepiece, at an angle

\abs G

times larger. All the light collected by the objective of diameter

D

passes through the exit pupil, of diameter

D/\abs G

Proposition 4.15 (What a telescope gains)

Compared with the naked eye (pupil $d_p$ ):

angular size: multiplied by $\abs G$ ;
light gathered from a point source: multiplied by $(D/d_p)^2$ , provided the exit pupil is not larger than the eye’s pupil, i.e. $\abs G \geq D/d_p$ ;
angular resolution: diffraction at the objective limits it to $\theta_{\min} \approx 1.22\,\lambda/D$ (Rayleigh criterion), far below the eye’s $\epsilon$ ; the magnification that makes this limit visible is $G_r = \epsilon/\theta_{\min}$ , and larger magnifications only enlarge the blur.

Proof. Admitted at this level. ∎

Example 4.16 (An amateur’s 150-millimeter refractor)

$D = 150\,\mathrm{mm}$ , $f_1' = 1200\,\mathrm{mm}$ , eyepiece $25\,\mathrm{mm}$ : $G = -48$ ; exit pupil $3.1\,\mathrm{mm}$ ; light gain $(150/7)^2 \approx 460$ against a dark-adapted pupil of $7\,\mathrm{mm}$ ; resolution $1.22 \times 550\,\mathrm{nm}/0.15\,\mathrm{m} = 4.5 \times 10^{-6}\,\mathrm{rad} = 0.9''$ , so craters $1.7\,\mathrm{km}$ wide on the Moon; $G_r = 3\times10^{-4}/4.5\times 10^{-6} \approx 67$ — the $6\,\mathrm{mm}$ eyepiece ( $G = 200$ ) shows no more detail, only a dimmer, bigger disk. The weekend problem designs this instrument.

Remark 4.17 (Reflectors, binoculars, spyglasses)

A concave mirror of focal length $f_1'$ replaces the objective lens in the reflecting telescope (no chromatic dispersion, and a mirror can be made meters wide and supported from behind): everything above holds with $f_1'$ the mirror’s focal length. The image is inverted — harmless for stars; for terrestrial use, binoculars insert a pair of total-reflection prisms to erect it, and Galileo’s spyglass uses a diverging eyepiece ( $f_2' < 0$ , $G > 0$ ), shorter but with a narrow field.

4.5 The camera

Definition 4.18 (Camera; f-number)

A camera is a converging lens of focal length $f'$ forming a real image on a sensor, at $\overline{OA'} \approx f'$ for distant subjects; focusing moves the lens. A diaphragm of diameter $D$ limits the beam; the f-number is $N = f'/D$ (written $f/N$ ). The angle of view is $2\arctan(\ell/2f')$ for a sensor of width $\ell$ .

Proposition 4.19 (Exposure and depth of field)

The irradiance on the sensor scales as $1/N^2$ , so the exposure time for a given image brightness scales as $N^2$ . A point at a distance other than the focused one forms a blur disk; requiring it to stay below a tolerated diameter $c$ , a lens focused at the hyperfocal distance

H = \frac{f'^2}{N c}

renders everything from $H/2$ to infinity acceptably sharp.

Proof. Irradiance $=$ power/area $\propto D^2/f'^2 = 1/N^2$ (the image of a given subject has a size fixed by $f'$ , and the collected power grows as $D^2$ ). Focus at distance $p$ with the image at $q \approx f'$ ; a point at infinity focuses at $f'$ , i.e. $q - f'$ before the sensor; its cone of rays, of base $D$ at the lens, has diameter $D(q - f')/q$ on the sensor (similar triangles). With $1/q = 1/f' - 1/p$ paraxially, $q - f' \approx f'^2/p$ for $p \gg f'$ , so the blur is $Df'/p = f'^2/(Np)$ ; it equals $c$ for $p = H$ . The same geometry on the near side gives the limit $H/2$ . ∎

Depth of field: the lens is focused on a point at distance p (image on the sensor); a point at infinity focuses at F', before the sensor, and leaves a blur disk of diameter c = Df'/p — smaller for a small aperture (large N), hence the trade-off between sharpness and exposure time. — Depth of field: the lens is focused on a point at distance $p$ (image on the sensor); a point at infinity focuses at $F'$ , before the sensor, and leaves a blur disk of diameter $c = Df'/p$ — smaller for a small aperture (large $N$ ), hence the trade-off between sharpness and exposure time.

Example 4.20 (Full frame and phone)

A $50\,\mathrm{mm}$ lens on a $36\,\mathrm{mm}$ wide sensor: angle of view $2\arctan(18/50) = 40^\circ$ ; at $f/8$ with $c = 0.03\,\mathrm{mm}$ , $H = 2500/(8 \times 0.03) = 10\,\mathrm{m}$ : focused at $10\,\mathrm{m}$ , everything beyond $5\,\mathrm{m}$ is sharp. A phone lens, $f' = 4.3\,\mathrm{mm}$ at $f/1.8$ on a $6\,\mathrm{mm}$ sensor with $c = 5\,\text{µ}\mathrm{m}$ : same angle of view ( $70{}^{\circ}$ wide), $H = 18.5/(1.8 \times 0.005) = 2.1\,\mathrm{m}$ — almost everything is in focus at once, the reason phones fake background blur in software.

An optometrist’s trial lenses and trial frame: a converging or diverging lens of the right vergence in front of the eye moves its far point back to infinity.

4.6 Exercises

Exercise 4.1 ★

With the retina $17\,\mathrm{mm}$ behind the lens, compute the vergence of a normal eye looking at infinity, then at $25\,\mathrm{cm}$ ; deduce the amplitude of accommodation. What focal lengths are these?

Solution

Solution of Exercise 4.1.

Infinity: $f' = 17\,\mathrm{mm}$ , $V = 1/0.017 = 58.8\,\delta$ . $25\,\mathrm{cm}$ : $V = 58.8 + 1/0.25 = 62.8\,\delta$ , $f' = 15.9\,\mathrm{mm}$ . Amplitude $4.0\,\delta$ .

Exercise 4.2 ★

A myopic eye has its far point at $40\,\mathrm{cm}$ . What lens corrects it (focal length, vergence)? A hyperopic eye can accommodate on nothing closer than $1.0\,\mathrm{m}$ : what lens lets it read at $25\,\mathrm{cm}$ ?

Solution

Solution of Exercise 4.2.

Myope: infinity must be imaged at the far point, $f' = -40\,\mathrm{cm}$ , $V = -2.5\,\delta$ . Hyperope: an object at $25\,\mathrm{cm}$ must be imaged (virtually) at $1.0\,\mathrm{m}$ : $V = 1/\overline{OA'} - 1/\overline{OA} = -1 + 4 = +3.0\,\delta$ .

Exercise 4.3 ★

A magnifier has $f' = 4.0\,\mathrm{cm}$ . Give its angular magnification, where the stamp must be held, and the angular size of a $2\,\mathrm{mm}$ detail through it, compared with the naked eye at $25\,\mathrm{cm}$ .

Solution

Solution of Exercise 4.3.

$G = 25/4.0 = 6.3$ ; stamp in the focal plane, $4.0\,\mathrm{cm}$ from the lens. Detail of $2\,\mathrm{mm}$ : $\theta' = 2/40 = 0.05\,\mathrm{rad}$ , against $2/250 = 0.008\,\mathrm{rad}$ at $25\,\mathrm{cm}$ .

Exercise 4.4 ★

A refractor has $f_1' = 900\,\mathrm{mm}$ . Compute $G$ with a $25\,\mathrm{mm}$ and with a $9\,\mathrm{mm}$ eyepiece, the tube length in each case, and the apparent size of the Moon ( $0.52^\circ$ ).

Solution

Solution of Exercise 4.4.

$G = 900/25 = 36$ and $900/9 = 100$ ; tube lengths $f_1' + f_2' = 925\,\mathrm{mm}$ and $909\,\mathrm{mm}$ ; Moon: $36 \times 0.52^\circ = 19^\circ$ and $52^\circ$ .

Exercise 4.5 ★★

Microscope: $f_1' = 4.0\,\mathrm{mm}$ , $f_2' = 20\,\mathrm{mm}$ , $\Delta = 160\,\mathrm{mm}$ . Compute $\gamma_1$ , $G_2$ , $G$ , and the object–objective distance. By how much must the tube move to bring the object $10\,\text{µ}\mathrm{m}$ deeper into focus?

Solution

Solution of Exercise 4.5.

$\gamma_1 = -160/4 = -40$ ; $G_2 = 250/20 = 12.5$ ; $G = 500$ . $\overline{F_1A} = -f_1'^2/\Delta = -16/160 = -0.10\,\mathrm{mm}$ : object $4.1\,\mathrm{mm}$ from the objective. The object must stay at that distance from the objective, so the whole tube moves by $10\,\text{µ}\mathrm{m}$ — the fine-focus knob.

Exercise 4.6 ★★

Two headlights are $1.5\,\mathrm{m}$ apart. From what distance can the naked eye ( $\epsilon = 3 \times 10^{-4}\,\mathrm{rad}$ ) tell them apart? A telescope with $D = 100\,\mathrm{mm}$ : compute its diffraction limit ( $\lambda = 550\,\mathrm{nm}$ ), the distance at which it separates the headlights, and the magnification $G_r$ beyond which it shows nothing new.

Solution

Solution of Exercise 4.6.

Naked eye: $1.5/3\times10^{-4} = 5\,\mathrm{km}$ . Telescope: $\theta_{\min} = 1.22 \times 5.5\times10^{-7}/0.100 = 6.7 \times 10^{-6}\,\mathrm{rad}$ ; headlights separated up to $1.5/6.7\times10^{-6} \approx 220\,\mathrm{km}$ (in principle — air and curvature intervene first). $G_r = 3\times10^{-4}/ 6.7\times10^{-6} = 45$ .

Exercise 4.7 ★★

A $50\,\mathrm{mm}$ lens at $f/2$ : what is the aperture diameter? A scene is correctly exposed at $f/2$ in $1/500$ $\mathrm{s}$ ; what exposure time at $f/8$ ? What does the smaller aperture buy?

Solution

Solution of Exercise 4.7.

$D = 50/2 = 25\,\mathrm{mm}$ . Exposure $\propto N^2$ : $(8/2)^2 = 16$ times longer, $16/500 \approx 1/30$ $\mathrm{s}$ . It buys depth of field (and fewer aberrations) at the price of motion blur or a tripod.

Exercise 4.8 ★★

Compute the hyperfocal distance of a $35\,\mathrm{mm}$ lens at $f/11$ with $c = 0.03\,\mathrm{mm}$ , and the range of sharp distances when it is focused at $H$ . Street photographers preset this: why?

Solution

Solution of Exercise 4.8.

$H = 35^2/(11 \times 0.03) = 3.7\,\mathrm{m}$ ; sharp from $H/2 = 1.9\,\mathrm{m}$ to infinity. Preset at $H$ , the camera needs no focusing: the picture can be taken the instant the scene happens.

Exercise 4.9 ★★

Binoculars “ $8\times30$ ” have $G = 8$ and $D = 30\,\mathrm{mm}$ . Compute the exit pupil; is all the light used by a $7\,\mathrm{mm}$ night pupil? For “ $7\times50$ ”? Which pair is better at dusk, and why does “ $20\times30$ ” make a poor choice?

Solution

Solution of Exercise 4.9.

$8\times30$ : exit pupil $30/8 = 3.8\,\mathrm{mm} < 7\,\mathrm{mm}$ , all light enters the eye. $7\times50$ : $7.1\,\mathrm{mm}$ , matching the night pupil, and $(50/30)^2 = 2.8$ times more light collected: the dusk pair. $20\times30$ : exit pupil $1.5\,\mathrm{mm}$ , dim image, narrow field and hand shake magnified twenty times.

Exercise 4.10 ★★★

A myope’s far point is $25\,\mathrm{cm}$ from the eye. Compute the vergence of the correcting contact lens, then of spectacle lenses worn $15\,\mathrm{mm}$ in front of the eye. Which correction is stronger, and why do the two differ?

Solution

Solution of Exercise 4.10.

Contact lens: $f' = -25\,\mathrm{cm}$ , $V = -4.0\,\delta$ . Spectacles: the far point is $25 - 1.5 = 23.5\,\mathrm{cm}$ from the lens, so $f' = -23.5\,\mathrm{cm}$ , $V = -4.3\,\delta$ — stronger. The lens must form the image at the eye’s far point, whose distance from the lens depends on where the lens sits.

Exercise 4.11 ★★★

Galileo’s spyglass: $f_1' = 30\,\mathrm{cm}$ , $f_2' = -10\,\mathrm{cm}$ . Find the separation of the lenses for an afocal system, $G$ , and the orientation of the image. Locate the exit pupil and explain why the field of view is narrow.

Solution

Solution of Exercise 4.11.

Afocal: $F_1' = F_2$ , separation $f_1' + f_2' = 20\,\mathrm{cm}$ (shorter than the $40\,\mathrm{cm}$ of a Keplerian of the same $G$ ). $G = -30/(-10) = +3$ : upright. Exit pupil $\overline{O_2P'} = f_2'(f_1' + f_2')/f_1' = -10 \times 20/30 = -6.7\,\mathrm{cm}$ : virtual, inside the tube, where the eye cannot be; the eye behind the eyepiece catches only the beams that happen to reach it — a narrow field, the classic complaint about opera glasses.

Exercise 4.12 ★★★

A $200\,\mathrm{mm}$ telescope and a $7\,\mathrm{mm}$ pupil. By what factor is the light from a star increased? Astronomers grade brightness by magnitudes, $m_2 - m_1 = 2.5\log_{10}(F_1/F_2)$ ; the eye reaches magnitude $6$ : what limiting magnitude does the telescope give? How many times farther can a given lamp be seen?

Solution

Solution of Exercise 4.12.

$(200/7)^2 = 820$ ; $\Delta m = 2.5\log_{10}820 = 7.3$ : limiting magnitude $13.3$ . Flux $\propto 1/r^2$ , so the lamp is seen $\sqrt{820} = 29$ times farther.

The near side of the Moon (Lunar Reconnaissance Orbiter, NASA): the target of the weekend problem’s telescope, 3476\, km across at 384\,000\, km. — The near side of the Moon (Lunar Reconnaissance Orbiter, NASA): the target of the weekend problem’s telescope, $3476\,\mathrm{km}$ across at $384\,000\,\mathrm{km}$ .

4.7 Problem: Designing a telescope for the Moon

Problem 4.1

Weekend problem — a 150-millimeter refractor on a garden lawn: choose the eyepieces, place the eye, count the light, and find out how small a crater it can show

The objective is a converging lens $L_1$ of focal length $f_1' = 1200\,\mathrm{mm}$ and diameter $D = 150\,\mathrm{mm}$ ; eyepieces $L_2$ of focal lengths $25\,\mathrm{mm}$ and $6\,\mathrm{mm}$ are available. The Moon is $3.84 \times 10^{5}\,\mathrm{km}$ away and subtends $0.52^\circ$ ; the eye’s night pupil is $7\,\mathrm{mm}$ , its resolution $\epsilon = 3.0 \times 10^{-4}\,\mathrm{rad}$ ; $\lambda = 550\,\mathrm{nm}$ .

Part I — The afocal arrangement.

Where must the eyepiece be placed for a star to be seen without accommodation? Give the objective–eyepiece distance for each eyepiece.
Check with Descartes’s relation, applied to each lens in turn, that a point at infinity on the axis is imaged at infinity.
Show that a beam from a star at angle $\alpha$ from the axis emerges as a parallel beam at angle $\alpha' = -(f_1'/f_2')\alpha$ .
Compute $G$ for each eyepiece and the apparent diameter of the Moon through each.
The image is inverted. Why is that acceptable here, and what would a terrestrial spotting scope add?
Where is the real intermediate image of the Moon, and what is its diameter? (It can be photographed there.)

Part II — Where the eye goes.

Show that the eyepiece forms a real image of the objective’s rim at $\overline{O_2P'} = f_2'(f_1' + f_2')/f_1'$ behind it, of diameter $D/\abs G$ (the exit pupil).
Compute the position and diameter of the exit pupil for each eyepiece.
Why must the eye’s pupil be placed at the exit pupil? What happens to the field if it is placed farther back?
For the light of a star to be fully used, the exit pupil must not exceed the eye’s pupil: deduce the minimum useful magnification of this telescope at night.

Part III — Resolution.

Compute the diffraction limit $\theta_{\min} = 1.22\lambda/D$ in radians and arcseconds.
What is the smallest crater this angle corresponds to on the Moon? And for the naked eye?
Define and compute the resolving magnification $G_r = \epsilon/\theta_{\min}$ . Which eyepiece is closest to it?
With the $6\,\mathrm{mm}$ eyepiece the image looks blurrier and dimmer than with the $25\,\mathrm{mm}$ . Explain both effects.
Atmospheric turbulence smears star images to about $1''$ . For which objective diameters does the Rayleigh limit stop mattering from the ground, and what does a larger $D$ still buy?

Part IV — Light.

By what factor does the objective collect more light than the night pupil?
Convert this to magnitudes ( $\Delta m = 2.5\log_{10}$ of the ratio); if the eye reaches magnitude $6$ , what magnitude does the telescope reach?
The full Moon delivers an irradiance of about $2 \times 10^{-3}\,\mathrm{W}/\mathrm{m}^{2}$ at the ground. What power enters the objective, and how is it spread over the retina compared with naked-eye viewing (same power per unit solid angle argument: compare surface brightness through the two eyepieces)?
Why does a telescope not make the Moon’s surface brighter per unit area than the naked eye, whereas it does make stars brighter?

Part V — Field of view. The eyepiece carries, in its focal plane, a circular field stop of diameter $20\,\mathrm{mm}$ ( $25\,\mathrm{mm}$ eyepiece) or $5\,\mathrm{mm}$ ( $6\,\mathrm{mm}$ eyepiece).

Show that the true field of view is $2\arctan(\phi/2f_1') \approx \phi/f_1'$ for a stop of diameter $\phi$ .
Compute it for both eyepieces and say whether the whole Moon fits in the view.
Compute the apparent field (the angle under which the eye sees the stop through the eyepiece) for both.
The Earth turns at $15''$ per second of time: how long does the Moon take to cross the field of each eyepiece without tracking?
Summarize the design: which eyepiece for the whole Moon, which for craters, and the smallest crater size the instrument can reveal — its one number.
The same objective is replaced by a $150\,\mathrm{mm}$ concave mirror of the same focal length. Which of the above results change?

Solution

Solution of Problem 4.1.

1. $F_2 = F_1'$ : separation $f_1' + f_2' = 1225\,\mathrm{mm}$ ( $25\,\mathrm{mm}$ eyepiece) or $1206\,\mathrm{mm}$ ( $6\,\mathrm{mm}$ ).

2. $L_1$ : $\overline{O_1A_1} = f_1'$ (image at $F_1'$ ). $L_2$ : $\overline{O_2A_1} = -f_2'$ , so $1/\overline{O_2A'} = 1/f_2' + 1/(-f_2') = 0$ : image at infinity.

3. The beam focuses in the common focal plane at height $h = f_1'\alpha$ (ray through $O_1$ ); from there the ray through $O_2$ sets the emerging direction, $\alpha' = -h/f_2' = -(f_1'/f_2')\alpha$ .

4. $G = -48$ and $-200$ ; Moon: $25^\circ$ and $104^\circ$ .

5. Up and down have no meaning in the sky; a spotting scope adds an erecting prism pair (or lens relay).

6. In the focal plane of $L_1$ , diameter $f_1'\theta = 1200 \times 9.1\times10^{-3} = 11\,\mathrm{mm}$ .

7. Object $O_1$ at $\overline{O_2O_1} = -(f_1' + f_2')$ : $1/\overline{O_2P'} = 1/f_2' - 1/(f_1' + f_2') = f_1'/[f_2'(f_1' + f_2')]$ ; magnification $\overline{O_2P'}/\overline{O_2O_1} = -f_2'/f_1' = 1/G$ , diameter $D/\abs G$ .

8. $25\,\mathrm{mm}$ : $\overline{O_2P'} = 25 \times 1225/1200 = 25.5\,\mathrm{mm}$ , diameter $150/48 = 3.1\,\mathrm{mm}$ . $6\,\mathrm{mm}$ : $6.0\,\mathrm{mm}$ behind, diameter $0.75\,\mathrm{mm}$ .

9. Every ray that entered the objective passes through the exit pupil: an eye placed there receives the whole field. Farther back it intercepts only the central beams: the field shrinks to a keyhole.

10. $D/\abs G \leq 7\,\mathrm{mm}$ : $\abs G \geq 150/7 \approx 21$ .

11. $\theta_{\min} = 1.22 \times 5.5\times10^{-7}/0.150 = 4.5 \times 10^{-6}\,\mathrm{rad} = 0.92''$ .

12. $4.5\times10^{-6} \times 3.84\times10^5 = 1.7\,\mathrm{km}$ ; naked eye $3\times10^{-4} \times 3.84\times10^5 = 115\,\mathrm{km}$ .

13. $G_r = 3\times10^{-4}/4.5\times10^{-6} = 67$ ; the $25\,\mathrm{mm}$ eyepiece ( $48$ ) is closest, slightly under; the $6\,\mathrm{mm}$ ( $200$ ) is three times over.

14. At $G = 200$ the diffraction blur subtends $200 \times 4.5 \times10^{-6} = 9 \times 10^{-4}\,\mathrm{rad} > \epsilon$ : visibly fuzzy. The Moon’s light is spread over an area $(200/48)^2 = 17$ times larger on the retina: dimmer.

15. $1'' = 4.85 \times 10^{-6}\,\mathrm{rad}$ : $D = 1.22\lambda/\theta = 140\,\mathrm{mm}$ . Beyond that, seeing limits resolution from the ground (barring adaptive optics); a larger $D$ still collects more light, reaching fainter objects.

16. $(150/7)^2 = 460$ .

17. $\Delta m = 2.5\log_{10}460 = 6.7$ : magnitude $12.7$ .

18. $P = 2\times10^{-3} \times \pi(0.075)^2 = 3.5 \times 10^{-5}\,\mathrm{W}$ , $460$ times the naked eye’s. The retinal image is $\abs G^2$ times larger in area: surface brightness ratio $460/G^2 = 0.20$ ( $G = 48$ ), $0.012$ ( $G = 200$ ) — equal to $(D/\abs G)^2/d_p^2$ , the squared ratio of exit pupil to eye pupil.

19. For an extended object the collected light grows as $D^2$ but the image area as $G^2 \geq (D/d_p)^2$ : the surface brightness can at best equal the naked-eye value (exit pupil $=$ eye pupil). A star is unresolved: all its light lands in one diffraction spot, so its brightness grows as $(D/d_p)^2$ .

20. A point at angle $\beta$ images at height $f_1'\beta$ in the focal plane; the stop passes $\abs{f_1'\beta} \leq \phi/2$ : full field $\phi/f_1'$ (small angles).

21. $20/1200 = 0.0167\,\mathrm{rad} = 0.95^\circ$ : the whole Moon fits. $5/1200 = 0.24^\circ$ : it does not.

22. Apparent field $\phi/f_2'$ : $20/25 = 0.8\,\mathrm{rad} = 46^\circ$ ; $5/6 = 48^\circ$ .

23. $0.95^\circ = 3420''$ : $3420/15 = 230\,\mathrm{s}$ , nearly four minutes; $0.24^\circ = 864''$ : $58\,\mathrm{s}$ .

24. $25\,\mathrm{mm}$ eyepiece ( $G = 48$ , near $G_r$ , bright, whole Moon) for the disk; $6\,\mathrm{mm}$ ( $G = 200$ ) for a closer look at craters, dimmer and fuzzier. The instrument’s number: craters down to $1.7\,\mathrm{km}$ .

25. Same $f_1'$ and $D$ : same $G$ , fields, exit pupils, resolution and light (minus any central obstruction). Gains: no chromatic aberration. Practical change: the focus lies in front of the mirror, so a small secondary mirror must fold the beam out to the eyepiece.

Terms defined in this chapter

See all 393 terms in the glossary