SME
Grade 11 Science · Problem type
A concave mirror
Use this for “A candle is 45 cm from a concave mirror with f = 15 cm. Where does its image form, and how big is it?”
Diagram
1/dᵢ = 1/f − 1/dₒ = 1/15 − 1/45
dᵢ = 22.5 cm
m = −dᵢ/dₒ = −22.5/45 = −0.5
Real, inverted and smaller: the rays really meet in front of the mirror.
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
15
=
1
45
+
1/22.5
m
=
−dᵢ/dₒ
−0.5 = −22.5/45
hᵢ
=
m
×
hₒ
−2 = −0.5 × 4
Assumptions
•
A concave mirror brings light together: f > 0. Beyond F the image is real, in front; inside F it is virtual, behind the mirror.
•
Rays: parallel then through F, through F then parallel, and through C straight back.
•
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 = 15 cm, dₒ = 45 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
15
−
1
45
)
dᵢ
=
1/(
2
45
)
dᵢ = 22.5 cm
Step 2 · Find magnification (m)
Use m = −dᵢ/dₒ
The magnification: negative means the image is upside down.
m
=
−22.5/45
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
15
=
1
45
+
1/22.5
✓
−0.5 = −22.5/45 ✓
−2 = −0.5 × 4 ✓
Topic check
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