Focal length
Exact opticsf (m) = 1 / F (D) F = 1 / f
Example: A +4.00 D lens → f = 0.25 m (25 cm)
Convert centimetres to metres before dividing: 25 cm = 0.25 m.
Reference
Every example below is worked by the same engine that grades the questions. “Exam method” marks the workbook approximation the exams expect.
ABO optics formulas
f (m) = 1 / F (D) F = 1 / f
Example: A +4.00 D lens → f = 0.25 m (25 cm)
Convert centimetres to metres before dividing: 25 cm = 0.25 m.
nominal power = front (base) curve + back curve
Example: Front curve +6.00 D, back curve −8.00 D → −2.00 D
F = (n₂ − n₁) / r
Where n₁ = index light leaves, n₂ = index it enters, r = radius in metres.
Example: Air to n = 1.50, radius 0.10 m → +5.00 D
Awaiting SME confirmation against a primary source.
true power = clock reading × (n − 1) / (1.53 − 1)
Where n = index of the lens being measured; the clock is calibrated for 1.53.
Example: Clock reads +5.00 D on a 1.586 lens → +5.53 D
Awaiting SME confirmation against a primary source.
new sphere = S + C new cyl = −C new axis = axis ± 90 (keep 1–180)
Example: +1.00 −2.50 × 090 → −1.50 +2.50 × 180
Add, change the sign, change the axis. An axis of 0 is written 180.
SE = S + C / 2
Example: −2.00 −1.50 × 180 → −2.75 D
F = S + C × {0, 25, 50, 75, 100 %} for 0, 30, 45, 60, 90° away from the axisExample: −1.00 −2.00 × 030, power in the 180 meridian (30° off axis) → −1.50 D
On the axis the cylinder adds nothing; 90° away it adds all of its power.
F(θ) = S + C × sin²(θ − axis)
Example: −1.00 −2.00 × 020, power in the 180 meridian → −1.23 D
Gives the table values at 0/30/45/60/90° and works at any angle.
Δ = F × d(mm) / 10 (or Δ = F × c, with c in cm)
Where F = power in the meridian of decentration, d = distance of the OC from the pupil.
Example: +4.00 D lens, OC 3 mm out from the pupil → 1.20Δ BO
Plus lens: base goes the same way the OC moved. Minus lens: the opposite way.
d (mm) = 10 × Δ / F
Example: 2Δ BI wanted from a −5.00 D lens → 4 mm, OC moved out
P = −F · r F = [[S + C sin²α, −C sinα cosα], [−C sinα cosα, S + C cos²α]]
Where α = cylinder axis, r = position of the pupil from the OC in cm.
Example: OD +2.00 −1.00 × 045, pupil 4 mm nasal and 2 mm below the OC → 0.50Δ BO and 0.10Δ BU
An oblique cylinder adds vertical prism to a horizontal decentration; the exam table does not.
Awaiting SME confirmation against a primary source.
P = √(H² + V²) θ = atan(V / H) H = P cos θ V = P sin θ
Where 360° notation: 0° at the examiner’s right, 90° = base up.
Example: OD 4Δ BI and 3Δ BU → 5.00Δ at 36.9°
Awaiting SME confirmation against a primary source.
imbalance = |Δ OD − Δ OS|, each Δ = vertical power × reading depth (mm) / 10
Example: OD −1.00 D, OS −3.50 D, reading 10 mm below the OCs → 2.50Δ
Slab-off is commonly considered from about 1.5Δ (awaiting SME confirmation).
Awaiting SME confirmation against a primary source.
DC = mm × F² / 1000
Example: +10.00 D lens moved 3 mm → +0.30 D
moved out (away from the eye): EP = F + DC moved in: EP = F − DC
Example: +10.00 D refracted at 12 mm, worn at 15 mm → +10.30 D
Effective power is what the patient gets from the unchanged lens in its new place.
moved out: CP = F − DC moved in: CP = F + DC
Example: +10.00 D refracted at 12 mm, worn at 15 mm → +9.70 D
Compensated power is what to order so the patient still gets the refracted power.
F new = F / (1 − d × F)
Where d in metres, positive when the lens moves toward the eye.
Example: −10.00 D at 12 mm, moved to the eye → −8.93 D
compensate each principal meridian, round each, rebuild in minus cylinder
Example: −8.00 −2.00 × 180 at 12 mm, moved to the eye → −7.25 −1.50 × 180
frame PD = A + DBL
Example: A 52 mm, DBL 18 mm → 70 mm
decentration = (frame PD − patient PD) / 2
Example: Frame PD 70 mm, patient PD 64 mm → 3 mm in, each lens
Divide by two: each lens moves half of the difference.
OC offset per lens = (fitted PD − patient PD) / 2
Example: Lenses made at 68 mm for a 62 mm PD → 3 mm out, each lens
Monocular OC heights are not halved.
MBS = ED + total decentration (= ED + 2 × per-lens)
Example: ED 56 mm, frame PD 70 mm, patient PD 64 mm → 62 mm
Many labs add about 2 mm for edging (awaiting SME confirmation).
position = height − B / 2 (+ above, − below the datum line)
Example: Seg height 20 mm, B 44 mm → 2 mm below the datum line
n = c / v
Where c = speed of light in a vacuum, v = speed in the material.
Example: Light travels at 199,861,639 m/s in the material → n = 1.5
TCA = Δ / Abbe value
Example: 4Δ at the viewing point, Abbe value 30 → 0.13Δ
Awaiting SME confirmation against a primary source.
R = ((n − 1) / (n + 1))²
Example: Uncoated n = 1.50 surface → 4 %
Awaiting SME confirmation against a primary source.