The document discusses the mirror equation, which expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f) of a spherical mirror. It provides examples of using the mirror equation to calculate di given do and f for concave mirrors. In one example, an image is formed 30 cm beyond the concave mirror when the object is 15 cm from the mirror and f is 10 cm. In another example, a virtual image is formed 7.78 cm behind the concave mirror when the object is 5 cm from the mirror and the radius is 28 cm, making f 14 cm.

1. Subtitle

2. Mirror Equation
There is an important relationship between the
distance of an object, distance of an image, and the
focal length of the spherical mirror. This relationship is
known as the mirror equation. The mirror equation
expresses the quantitative relationship between the object
distance (do), the image distance (di), and the focal length
(f). In symbols,
𝟏
𝒇
=
𝟏
𝒅 𝒐
+
𝟏
𝒅𝒊

3. This mirror equation is both applicable to concave
and convex mirrors. In a concave mirror, the focal length
is positive (+) while in a convex mirror it is negative (-).
The distance of the object and the distance of the image
is taken as positive (+) because they are both located in
front of the mirror and taken as negative (-) for virtual
images because they are formed behind the mirror.

4. Example 1:
A flower vase is placed 15 cm in front of a concave
mirror whose focal length is 10 cm. Where is the image
of the flower vase located? Describe the image formed.
Given:
do = 15 cm
f = 10 cm
Solution:
Using the basic equation, derive the formula for
the distance of the image.

5. 𝟏
𝒇
=
𝟏
𝒅 𝒐
+
𝟏
𝒅𝒊
f do di
𝟏
𝒇
=
𝟏
𝒅 𝒐
+
𝟏
𝒅𝒊
f do di
do di = fdi + fdo
dodi – fdi = fdo
𝒅𝒊
(𝒅𝒐 − −𝒇)
𝒅 𝒐
−𝒇
=
𝒇 𝒅𝒐
(𝒅𝒐 − 𝒇)
di =
𝒇 𝒅𝒐
𝒅𝒐 − 𝒇
di =
𝒇 𝒅𝒐
(𝒅𝒐 − 𝒇)
di =
𝟏𝟎 𝒄𝒎 (𝟏𝟓 𝒄𝒎)
(𝟏𝟓 𝒄𝒎 −𝟏𝟎 𝒄𝒎)
di =
𝟏𝟓𝟎 𝒄𝒎 𝟐
𝟓𝒄𝒎
di = 30 cm
The image is form beyond the C,
inverted, larger in size and real.

6. do di = fdi + fdo
do di - fdo = fdi
do (di – f) = fdi
do (di − f)
di − f
=
fdi
di − f
do =
fdi
di − f
do di = fdi + fdo
do di = di + do(f)
do di
di + do
=
di + do(f)
di + do
do di
di + do
= f

7. Example 2.
A bottle is held 5cm from a concave mirror
whose radius is 28 cm. At what distance will the
image be formed?
Given:
do = 5 cm
r = 28 cm
Solution:
Determine the focal point, which is one-half of
the radius, of the curvature.

8. Using the basic equation, solve for the distance of the
image di.
di =
𝒇 𝒅𝒐
𝒅𝒐 − 𝒇
di =
𝟏𝟒 𝒄𝒎 (𝟓 𝒄𝒎)
𝟓𝒄𝒎 −𝟏𝟒 𝒄𝒎
di =
70𝑐𝑚2
−9 𝑐𝑚
di = -7.78 cm
The negative sign indicates that the object is located
behind the mirror and it is virtual image, and larger
than the object.

9. Example 2. A cup is placed in front of a concave mirror
whose focal point is 8 cm, and formed an image 24cm
beyond the C. At what point does the cup is located?
do =
fdi
di − f
do =
(8 cm)(24 cm)
24 cm − 8 𝑐𝑚
do =
192 𝑐𝑚
16 𝑐𝑚
do = 12 cm
The cup located 12 cm in front of the mirror.