SME
University Physics III: Waves, Optics & Modern · Problem type
A lens’s focal length from its surfaces
Use this for “A glass lens (n = 1.5) has radii +20 cm and −20 cm. Find f.”
Diagram
1/dᵢ = 1/f − 1/dₒ = 1/20 − 1/40
dᵢ = 40 cm
m = −dᵢ/dₒ = −40/40 = −1
Real, inverted and the same size: the rays really meet behind the lens.
This is an example. Type your own values to start a new problem.
1
f (cm)
=
(
n
−
1)
1
R₁ (cm)
−
1
R₂ (cm)
P
Lens power
worked out (not typed) · m⁻¹
Formulas
1
f
=
(
n
−
1)
×
(
1
R₁
−
1
R₂
)
1
20
=
(1.5
−
1)
×
(
1
20
−
1
−20
)
P
=
100
f
5
=
100
20
Assumptions
•
A thin lens in air with two round faces; light meets the face of radius R₁ first, then R₂.
•
R > 0 when the face’s centre of curvature is beyond the lens (on the far side); a flat face has R = ∞, so type a radius like 10⁶ cm.
•
Converging lenses only (f > 0, thicker in the middle). The power P = 1 ÷ f with f in metres, in diopters: 1 D = 1 m⁻¹.
Step-by-step
WE KNOW
n = 1.5, R₁ = 20 cm, R₂ = −20 cm
FIND
focal length (f), lens power (P)
Step 1 · Find focal length (f)
Use 1 ÷ f = (n − 1) × (1 ÷ R₁ − 1 ÷ R₂)
Over
one
denominator,
1
R₁
−
1
R₂
is
R₂
−
R₁
R₁R₂
;
times
n
−
1,
then
turn
it
over.
f
=
R₁
×
R₂
(
n
− 1) × (
R₂
−
R₁
)
f
=
20
×
−20
(1.5 − 1) × (−20 − 20)
f
=
20
×
−20
0.5 × (−40)
f
=
20
×
−20
−20
f
=
−
400
−20
f
=
20
cm
Step 2 · Find lens power (P)
Use P = 100 ÷ f
The
power
is
1
f
with
f
in
metres:
f
in
cm
is
f
100
m,
so
P
=
100
f
,
in
diopters
(
m
-1
).
P
=
100
20
P
=
5
m⁻¹
Check
Put the numbers back in. Both sides should match.
1
20
=
(1.5
−
1)
×
(
1
20
−
1
−20
)
✓
5
=
100
20
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.