SME
University Physics III: Waves, Optics & Modern
Geometric optics
Use this for “An object 4 cm tall stands 30 cm from a lens of focal length 10 cm. Find the image distance, magnification and height.”
Diagram
1/dᵢ = 1/f − 1/dₒ = 1/10 − 1/30
dᵢ = 15 cm
m = −dᵢ/dₒ = −15/30 = −0.5
Real, inverted and smaller: the rays really meet behind the lens.
This is an example. Type your own values to start a new problem.
1
f (cm)
=
1
dₒ (cm)
+
1
dᵢ (cm)
m
Magnification
calculated
hₒ
Object height
example · cm
hᵢ
Image height
calculated · cm
Formulas
1
f
=
1
dₒ
+
1
dᵢ
1
10
=
1
30
+
1
15
m
=
−
dᵢ
dₒ
−0.5
=
−
15
30
hᵢ
=
m
×
hₒ
−2 = −0.5 × 4
Assumptions
•
A thin converging lens (f > 0), and rays close to the axis, so every ray from one point meets again at one point.
•
dᵢ > 0 is a real image behind the lens; dᵢ < 0 is a virtual image on the object’s side, seen through the lens.
•
m < 0 means the image is upside down; |m| > 1 means it is bigger than the object.
Step-by-step
WE KNOW
f = 10 cm, dₒ = 30 cm, hₒ = 4 cm
FIND
image distance (dᵢ), magnification (m), image height (hᵢ)
Step 1 · Find image distance (dᵢ)
Use 1 ÷ f = 1 ÷ dₒ + 1 ÷ dᵢ
Take
1
dₒ
from
1
f
and
turn
the
result
over:
dᵢ
=
fdₒ
dₒ −
f
.
dᵢ
=
f
×
dₒ
dₒ −
f
dᵢ
=
10
×
30
30 − 10
dᵢ
=
10
×
30
20
dᵢ
=
300
20
dᵢ = 15 cm → real: the rays meet behind the lens.
Step 2 · Find magnification (m)
Use m = −dᵢ ÷ dₒ
The image is as many times the object’s size as it is farther from the lens; the minus sign marks a real image as upside down.
m
=
−
15
30
m
=
−0.5
Step 3 · Find image height (hᵢ)
Use hᵢ = m × hₒ
The magnification scales the height; a negative height points down.
hᵢ = −0.5 × 4
hᵢ = −2 cm
Check
Put the numbers back in. Both sides should match.
1
10
=
1
30
+
1
15
✓
−0.5
=
−
15
30
✓
−2 = −0.5 × 4 ✓
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The energy released in a decay (Q value)
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A lens’s focal length from its surfaces
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