SME
Grade 11 Science · Problem type
A diverging lens
Use this for “An object is 24 cm from a diverging lens with f = −12 cm. Where is the image, and how big?”
Diagram
1/dᵢ
=
1
f
−
1/dₒ
=
1/(−12)
−
1
24
dᵢ = −8 cm
m
=
−dᵢ/dₒ
=
−
−8
24
=
0.333
Virtual, upright and smaller: the rays only seem to come from in front of the lens (dashed).
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/(−12)
=
1
24
+
1/(−8)
m
=
−dᵢ/dₒ
0.333 = −(−8)/24
hᵢ
=
m
×
hₒ
1 = 0.333 × 3
Assumptions
•
A diverging lens spreads rays out: f < 0.
•
Its image is always virtual, upright and smaller, between F and the lens.
•
dᵢ > 0 is a real image (light really meets there); dᵢ < 0 is a virtual one. m < 0 means upside down.
Step-by-step
WE KNOW
f = −12 cm, dₒ = 24 cm, hₒ = 3 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/(−12)
−
1
24
)
dᵢ = 1/(−0.125)
dᵢ = −8 cm
Step 2 · Find magnification (m)
Use m = −dᵢ/dₒ
The magnification: negative means the image is upside down.
m
=
−(−8)/24
m
=
0.333
Step 3 · Find image height (hᵢ)
Use hᵢ = m × hₒ
The
image
is
m
times
as
tall
as
the
object.
hᵢ = 0.333 × 3
hᵢ = 1 cm
Check
Put the numbers back in. Both sides should match.
1/(−12)
=
1
24
+
1/(−8)
✓
0.333 ≈ −(−8)/24 ✓
1 ≈ 0.333 × 3 ✓
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