SME
Grade 11 Science · Physics
Light: reflection, refraction, mirrors, lenses and interference
Use this for “An object 30 cm from a converging lens with f = 10 cm. Where is the image, and how big is it?”
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
hᵢ
Image height
calculated
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 distances are measured from the lens.
•
dᵢ > 0: a real image on the far side, where light really meets; dᵢ < 0: a virtual one on the object’s side.
•
m < 0 means the image is upside down.
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
,
then
flip.
dᵢ
=
1/(
1
f
−
1/dₒ)
dᵢ
=
1/(
1
10
−
1
30
)
dᵢ
=
1/(
1
15
)
dᵢ = 15 cm
Step 2 · Find magnification (m)
Use m = −dᵢ/dₒ
The magnification: negative means the image is upside down.
m
=
−
15
30
m
=
−0.5
Step 3 · Find image height (hᵢ)
Use hᵢ = m × hₒ
The
image
is
m
times
as
tall
as
the
object.
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 ✓
Topic check
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