Prentice's Rule Prism Calculator

Prism dioptres = decentration in centimetres × lens power. A 4 mm centration error on a 4.00 D lens induces 1.60Δ — more than twice ANSI Z80.1’s horizontal imbalance tolerance.

Last updated 8 Oct 2026

Decentration ⇄ induced prism

Prentice's rule

Lens power in the meridian (D)The power in the meridian the decentration is measured along — the horizontal power for a horizontal displacement, the vertical for a vertical one. The sign is dropped by the rule, so −4.00 and +4.00 induce the same amount of prism; what the sign decides is the direction of the base. On a sphero-cylinder the power in an oblique meridian is not the sphere or the sphere-plus-cylinder, and this page cannot compute it: that needs a trigonometric term the engine does not have. Use the power in the 180 or 90 meridian, from the prescription in either cylinder form.

Decentration (mm)How far the patient’s visual axis sits from the lens’s optical centre. For a horizontal error that is the difference between the monocular centration distance the lens was glazed to and the patient’s actual monocular pupillary distance; for a vertical one it is the difference between the fitted and the specified optical centre height.

Target prism, for the inverse (Δ)Used only by the inverse figure in the list below, which gives the decentration that would produce this much prism in a lens of the power entered — the deliberate use of Prentice’s rule, where a small prescribed prism is obtained by decentring a stock lens instead of grinding it. The figure disappears at a power of zero, because no amount of decentring a plano lens produces prism.

1.60Δ inducedExample

A −4.00 D lens, optical centre 4 mm from the visual axis, target prism 1.00Δ

Prentice’s rule

P = d × D, with P in prism dioptres, d the decentration in CENTIMETRES and D the absolute lens power in dioptres · in millimetres, P = d(mm) × D ÷ 10

the units are the trap: the rule as published takes the decentration in CENTIMETRES, and optical centres are measured and specified in millimetres. The source read for this page states it as “Prism = Distance × Power, where the distance is the decentration in centimeters” and gives the millimetre form as the product divided by ten. Using millimetres without dividing produces an answer ten times too large
the sign is dropped: quoted: “We do not use power signs in prism calculations”. A −4.00 D and a +4.00 D lens decentred the same way induce the same 1.60Δ at 4 mm. What the sign decides is the BASE DIRECTION: looking through a point temporal to the optical centre of a plus lens gives base-in prism, and of a minus lens base-out
the prism dioptre: defined by the displacement it produces: “a prism of power 1Δ would produce 1 unit of displacement for an object held 100 units from the prism” — one centimetre at one metre. So the induced prism in dioptres is also the image displacement in centimetres per metre of viewing distance, which is why the displacement at six metres is printed above
which meridian: the power used must be the power in the meridian along which the lens is decentred. For a horizontal decentration that is the horizontal power, for a vertical one the vertical power. On a sphero-cylinder those are the two principal meridian powers only when the axis is at 180 or 90; at an oblique axis the meridional power needs a trigonometric term and this page cannot compute it
ANSI Z80.1, as reported: the study guide read for this page gives the maximum allowable vertical prismatic imbalance as 0.33Δ and the horizontal as 0.67Δ, with alternative placement limits of 1.0 mm vertically and ±2.5 mm horizontally (±1.0 mm above ±2.75 D). THE STANDARD ITSELF WAS NOT READ — it is a purchased document — and the figures are stated as that guide reports them

Worked example

A −4.00 D lens, optical centre 4 mm from the visual axis, target prism 1.00Δ
The sign of the power is dropped: |−4.00| = 4.00 D
4 mm is 0.4 cm, which is the unit the rule is written in
P = 0.4 × 4.00 = 1.60Δ
Against the tolerances. That is 4.8 times the 0.33Δ vertical imbalance limit and 2.4 times the 0.67Δ horizontal one, as the study guide reports them from ANSI Z80.1 — if the fellow lens is correctly centred, so that the prism in this lens is the whole imbalance
Per millimetre. At 4.00 D each millimetre of decentration is 0.4Δ, so the vertical tolerance is reached at 0.83 mm and the horizontal at 1.68 mm. Sub-millimetre centration is not a refinement at this power; it is the tolerance
The inverse. For 1.00Δ from a 4.00 D lens: 1.00 × 10 ÷ 4.00 = 2.5 mm of decentration. That is Prentice's rule used on purpose
The published worked cases, reproduced. A +3.00 D blank decentred 0.3 cm gives 0.9Δ; at 0.5 cm, 1.5Δ; at 1.0 cm, 3Δ; at 1.5 cm, 4.5Δ. And the source's patient case — a −2.00 DS right lens glazed to 30 mm for a 34 mm monocular pupillary distance, so 0.4 cm of decentration — gives 0.8Δ base out. All five agree exactly
And the displacement. 1.60Δ displaces the image 1.60 cm at one metre and 9.6 cm at the six metres of an acuity chart — which is roughly a letter and a half of lateral shift on the 6/12 line

Induced prism by power and decentration, against the ANSI limits

Every figure is Prentice’s rule. The two right-hand columns are the inverse, and they are the row that explains the standard: ANSI Z80.1’s horizontal centration tolerance tightens from ±2.5 mm to ±1.0 mm above ±2.75 D, and at 2.75 D a 2.44 mm error is exactly the 0.67Δ horizontal imbalance limit. The tolerance and the prism limit are the same requirement written two ways. At 10.00 D the numbers are memorable: one millimetre is one prism dioptre.
Lens power	1 mm	2 mm	3 mm	4 mm	Decentration reaching 0.33Δ	Decentration reaching 0.67Δ
0.50 D	0.05Δ	0.10Δ	0.15Δ	0.20Δ	6.60 mm	13.40 mm
1.00 D	0.10Δ	0.20Δ	0.30Δ	0.40Δ	3.30 mm	6.70 mm
2.00 D	0.20Δ	0.40Δ	0.60Δ	0.80Δ	1.65 mm	3.35 mm
2.75 D	0.28Δ	0.55Δ	0.82Δ	1.10Δ	1.20 mm	2.44 mm
4.00 D	0.40Δ	0.80Δ	1.20Δ	1.60Δ	0.83 mm	1.68 mm
6.00 D	0.60Δ	1.20Δ	1.80Δ	2.40Δ	0.55 mm	1.12 mm
10.00 D	1.00Δ	2.00Δ	3.00Δ	4.00Δ	0.33 mm	0.67 mm

Why a lens becomes a prism away from its centre

A spherical lens is a stack of prisms. Everywhere except the optical centre its two surfaces are not parallel, so light passing through is deviated as well as focused, and the further from the centre the greater the deviation. Prentice’s rule quantifies it with a single multiplication: the prism in dioptres is the decentration in centimetres times the absolute power in dioptres. It holds exactly for a thin lens, in whichever meridian the decentration is measured, and the sign of the power is dropped — a +4.00 D and a −4.00 D lens decentred identically induce the same amount of prism, and differ only in which way the base points.

The unit is where it goes wrong. The rule as published takes centimetres, and every optical centre in dispensing is measured and specified in millimetres, so the practical form is the product divided by ten. Forget the division and the answer is ten times too large; use the full binocular pupillary distance halved instead of each eye’s own monocular centration distance and the decentration is whatever the facial asymmetry happens to be. At 4.00 D every millimetre is 0.4Δ, which is why sub-millimetre centration on a high-powered lens is a tolerance rather than a refinement.

The tolerances make the arithmetic concrete. The study guide read for this page reports ANSI Z80.1 as allowing a maximum vertical prismatic imbalance of 0.33Δ and a horizontal of 0.67Δ, with placement alternatives of 1.0 mm vertically and ±2.5 mm horizontally, tightening to ±1.0 mm above ±2.75 D. The two halves of that are the same requirement: run Prentice’s rule backwards at 2.75 D and a 2.44 mm error is exactly 0.67Δ, which is why the placement limit tightens at that power. The vertical allowance is half the horizontal because vertical fusional reserve is far smaller — the eyes will absorb a horizontal imbalance that a vertical one of the same size would not survive. ANSI Z80.1 itself is a purchased standard and was not read for this page, so the figures are given as that guide reports them.

Two honest limits, both of them the calculator engine’s rather than the rule’s. The power used must be the power in the meridian the lens is decentred along, and on a sphero-cylinder at an oblique axis that is neither the sphere nor the sphere-plus-cylinder but a trigonometric combination of the two, which this engine cannot form. And the degree equivalent of a prism dioptre is not printed anywhere on this page: it is arctan(Δ ÷ 100), there is no inverse-trigonometric function available, and the familiar 0.57° per prism dioptre is a small-angle approximation that is already more than a degree out by 30Δ. A tolerance that depends on a number is better served by its absence than by an approximation of it. The induced prism also interacts with vertex distance, because a lens sitting further from the eye is being looked through at a different point.

Frequently asked questions

What is Prentice’s rule?

The prism induced by looking through a lens away from its optical centre, in prism dioptres, equals the decentration in centimetres times the absolute value of the lens power in dioptres. In millimetres it is the product divided by ten: a 4 mm decentration on a 4.00 D lens gives 1.60Δ.

Does the sign of the lens power matter?

Not for the amount. The published rule drops the sign, so +4.00 D and −4.00 D give the same prism for the same decentration. The sign decides the base direction: looking through a point temporal to the optical centre of a plus lens gives base-in prism, and of a minus lens base-out.

How much decentration does ANSI Z80.1 allow?

As the study guide read for this page reports it: a maximum vertical prismatic imbalance of 0.33Δ and a horizontal of 0.67Δ, or alternatively the prism reference point within 1.0 mm of its specified vertical position and the total horizontal deviation from the distance pupillary distance within ±2.5 mm, tightening to ±1.0 mm for powers over ±2.75 D. The standard itself is a purchased document and was not read.

How do I decentre a lens to get a prescribed prism?

Run the rule backwards: the decentration in millimetres is the wanted prism times ten, divided by the absolute power. 1.00Δ from a 4.00 D lens needs 2.5 mm. It only works where there is power to work with — no amount of decentring a plano lens produces prism, which is why the inverse figure on this page disappears at zero power.

How many degrees is a prism dioptre?

The exact relation is arctan(Δ ÷ 100), and this page deliberately does not compute it: the calculator engine behind these pages has no trigonometric or inverse-trigonometric function, and the small-angle figure of about 0.57° per prism dioptre is more than a degree wrong by 30Δ. What a prism dioptre is defined by is displacement, not angle: 1Δ shifts the image one centimetre at one metre, so 1.60Δ shifts it 9.6 cm at six metres.

Related calculators

References

The Many Facets of Prism in Ophthalmic Lenses — Part 2. 20/20 Magazine continuing education, 2020mag.com. Accessed 9 October 2026.
Prism correction. Wikipedia, citing Carlton J. Frames and Lenses. 2000:53ff.
Open Exam Prep. Section 15.1: ANSI Z80.1 Prescription Tolerances, National Opticianry Competency Examination study guide. open-exam-prep.com. Accessed 9 October 2026. (ANSI Z80.1 itself is a purchased standard and was not read.)

Not medical advice. For healthcare professionals and education. Reference intervals vary by laboratory and assay — always use your own laboratory's. Never base a dose or a treatment decision on this page alone. Full disclaimer at calcengines.com/disclaimer/