High School Physics · Grades 10–12

10Lenses, 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 (Chapter 3) so that all the rays leaving one point meet again at another. This chapter tames the trick: one lens, three rays, one formula — then points it at cameras, spectacles, and the living 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 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; through a diverging lens, parallel rays exit as if from a point $F'$ in front of the lens: $f' < 0$ .

Definition 10.3 (Vergence)

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

Example 10.4 (Reading a prescription)

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

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:

the ray through the optical center $O$ goes straight on;
the ray parallel to the axis exits through $F'$ ;
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 point.

The three construction rays: object AB beyond F, image A'B' real and inverted (here OA = -2f': life-size, = -1). — The three construction rays: object $AB$ beyond $F$ , image $A'B'$ real 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 sees it perfectly well. A converging lens gives a real, inverted image of any object beyond $F$ ; slide the object inside the focal length and the image turns virtual, upright, enlarged.

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

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 object in front); heights $\overline{AB}$ count positive upward.

Theorem 10.7 (Thin-lens conjugate relation)

Through a thin lens of focal length $f'$ , 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 applied to rays close to the axis; the derivation — and the limits of the thin-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, behind the lens; negative: virtual, in front).

Example 10.10 (The formula at work)

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

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

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

10.4 The eye

Definition 10.11 (The eye as an optical system)

Optically, the eye is a converging lens facing a screen: the cornea and crystalline lens act as one converging lens of adjustable focal length; the retina, about $17\,\mathrm{mm}$ behind, is the fixed screen. Focusing is done by squeezing the 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.

Example 10.12 (The vergence of your eye)

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

10.5 Defects, corrections, and the magnifier

Definition 10.13 (Myopia and hyperopia)

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

Example 10.14 (Prescribing for a myope)

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

A myopic eye focuses parallel rays before the retina (red point); a diverging lens lands the focus on the retina.

Definition 10.15 (The magnifier)

A magnifier is a converging lens of short focal length used with the object inside the focal length: the image is virtual, upright, enlarged (Definition 10.6), 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, upright image, five times larger, a comfortable $20\,\mathrm{cm}$ from the lens.

10.6 Exercises

Exercise 10.1 ★

Compute the vergences for $f' = 50\,\mathrm{cm}$ , $-20\,\mathrm{cm}$ and $4.0\,\mathrm{mm}$ , then the focal length of a $+8.0$ D lens.

Solution

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: $f' = 1/8.0 = 0.125\,\mathrm{m} = 12.5\,\mathrm{cm}$ .

Exercise 10.2 ★

A lens is thicker at the edges than at the center. Converging or diverging? Can it serve as a magnifier? What do you see holding it over a printed page?

Solution

Solution of Exercise 10.2.

Diverging ( $f' < 0$ ). No: for any object distance $d$ , $\gamma = f'/(f' - d)$ (shown in Exercise 10.15) lies between $0$ and $1$ — the image is always virtual, upright and reduced. The print appears upright and shrunk, never magnified.

Exercise 10.3 ★

A converging 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

Solution of Exercise 10.3.

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

Exercise 10.4 ★

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

Solution

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 $12\,\mathrm{cm}$ behind the lens, inverted, half-size.

Exercise 10.5 ★

In the eye: what forms the converging lens? What plays the screen? What changes during accommodation — and what cannot? Define the near point of the standard eye.

Solution

Solution of Exercise 10.5.

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

Exercise 10.6 ★★

A camera 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

Solution of Exercise 10.6.

Landscape: sensor in the focal plane, $50\,\mathrm{mm}$ from the 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 has $f' = 6.0\,\mathrm{cm}$ ; a stamp lies $5.0\,\mathrm{cm}$ below it. Position, nature, orientation and magnification of the image?

Solution

Solution of Exercise 10.7.

$1/\overline{OA'} = 1/6.0 - 1/5.0 = -1/30$ : $\overline{OA'} = -30\,\mathrm{cm}$ — virtual, upright, $\gamma = (-30)/(-5.0) = +6.0$ : six times the stamp, $30\,\mathrm{cm}$ above the lens.

Exercise 10.8 ★★

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

Solution

Solution of Exercise 10.8.

$1/f' = 1/40 + 1/40 = 1/20$ : $f' = 20\,\mathrm{cm}$ ; $\gamma = 40/(-40) = -1$ . Symmetry forces $\abs{\overline{OA'}} = \abs{\overline{OA}}$ , hence $\abs{\gamma} = 1$ ; the image is real and inverted, so $\gamma = -1$ : exactly life-size.

Exercise 10.9 ★★

An object stands $50\,\mathrm{cm}$ in front of a diverging 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

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, upright, one third of the size. A myope’s glasses are diverging: everything seen through them — including, from outside, the wearer’s eyes — is reduced.

Exercise 10.10 ★★

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

Solution

Solution of Exercise 10.10.

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

Exercise 10.11 ★★

A presbyopic reader has their near point 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 page at $80\,\mathrm{cm}$ . Compute the required vergence.

Solution

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.

Find $\overline{OA}$ , $\overline{OA'}$ , given $\overline{OA'} - \overline{OA} = 5.1\,\mathrm{m}$ .
Deduce the focal length of the projection lens.
How tall is the image of a $24\,\mathrm{mm}$ -tall slide?

Solution

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 is $17\,\mathrm{mm}$ behind the eye’s lens.

Compute the eye’s vergence focused at infinity, then on the standard near point ( $25\,\mathrm{cm}$ ). Deduce the accommodation amplitude of the standard eye.
An older eye has its near point at $1.0\,\mathrm{m}$ . Compute its amplitude and the fraction of the standard amplitude remaining.

Solution

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: $1/0.25 = 4.0$ D.

2. Amplitude $= 1/1.0 = 1.0$ D: one quarter ( $25\%$ ) of the standard eye’s $4.0$ D remains.

Exercise 10.14 ★★★

A magnifier sold as “ $\times 3$ ” has its focal length given by $3 = 0.25/f'$ ( $f'$ in meters).

Compute $f'$ .
Where must an object sit for its virtual image to lie $25\,\mathrm{cm}$ in front of the lens? Compute $\gamma$ ; compare with the advertised $\times 3$ .

Solution

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 gains one unit.

Exercise 10.15 ★★★

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

Solution

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) and $\gamma < 0$ (inverted). If $0 < d < f'$ : $\overline{OA'} < 0$ (virtual) and $\gamma = f'/(f' - d) > 1$ : upright and enlarged — the 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, 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), runs all three trades — and then the eye itself.

Part I — One formula. A converging lens has $f' = 20\,\mathrm{cm}$ .

An object at $60\,\mathrm{cm}$ : compute $\overline{OA'}$ and $\gamma$ ; describe the image.
Move the object to $30\,\mathrm{cm}$ ; recompute. What got exchanged with question 1? (Light paths are reversible.)
Move it to $10\,\mathrm{cm}$ , inside $f'$ : recompute and describe.
Summarize: as the object slides in from very far toward $F$ , then past it, how does the image move and change nature?
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 has $f' = 50\,\mathrm{mm}$ ; the sensor, $24\,\mathrm{mm}$ tall, can be moved relative to the lens.

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

Part III — The grandmother. With age the near point recedes: grandmother’s is at $1.0\,\mathrm{m}$ .

Explain the arm’s-length newspaper.
To read at $25\,\mathrm{cm}$ , her glasses must turn a page at $25\,\mathrm{cm}$ into a virtual image at $1.0\,\mathrm{m}$ : compute the required vergence.
Compute $\gamma$ . The image is four times larger, yet looks no bigger: explain, comparing sizes and distances.
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?
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 is at $25\,\mathrm{cm}$ ; his retina is $17\,\mathrm{mm}$ behind his eye’s lens.

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

Solution

Solution of Problem 10.1.

1. $1/\overline{OA'} = 1/20 - 1/60 = 1/30$ : $\overline{OA'} = +30\,\mathrm{cm}$ , $\gamma = -0.50$ — real, inverted, half-size.

2. $\overline{OA'} = +60\,\mathrm{cm}$ , $\gamma = -2$ : the positions $30\,\mathrm{cm}$ and $60\,\mathrm{cm}$ exchanged roles, the magnifications 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, upright, doubled.

4. From infinity the image starts in the focal plane; as the object approaches $F$ the real, inverted image recedes to infinity and grows; inside $f'$ it turns virtual, upright, enlarged, in front of the 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.

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 $\overline{OA'} = -\overline{OA}$ , 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 ( $1.0\,\mathrm{m}$ ) 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.

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 the eye cannot use: the distant world blurs.

15. Hence bifocals: reading power in the lower half of the 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, 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 — 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.

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.