Convex Mirrors

## **Problem**

Determine the image distance from the mirror, magnification, and height of the image in each scenario.

**Part 1:** A 5 cm tall arrowhead is placed 60 cm from a convex mirror of focal length 20 cm.

**Part 2:** A 5 cm tall arrowhead is placed 30 cm from a convex mirror of focal length 20 cm.

**Part 3:** A 5 cm tall arrowhead is placed 20 cm from a convex mirror of focal length 20 cm.

**Part 4:** A 5 cm tall arrowhead is placed 10 cm from a convex mirror of focal length 20 cm.

## **Hint**

(1) Mirror equation:

1

d
o

1

d
i

=

1
f

{\displaystyle \frac{1}{d\_o} + \frac{1}{d\_i} = \frac{1}{f} }

.

(2) Magnification formula:

# M
−

d
i

d
o

=

h
i

h
o

{\displaystyle M = -\frac{d\_i}{d\_o} = \frac{h\_i}{h\_o} }

.

## **Solution**

**Part 1**

Use the mirror equation to calculate the image distance. Remember the focus of a convex mirror is negative.

1

d
o

1

d
i

=

1
f

{\displaystyle \frac{1}{d\_o} + \frac{1}{d\_i} = \frac{1}{f} }

1

60

c
m

1

d
i

=

1

−
20

c
m

1

d
i

=
−

1

15

c
m

d
i

=
−
15

c
m

{\displaystyle \begin{align}
\frac{1}{60 \; cm} + \frac{1}{d\_i} &= \frac{1}{-20 \; cm} \\[5 pt]
\frac{1}{d\_i} &= -\frac{1}{15 \; cm} \\[5 pt]
d\_i &= -15 \; cm
\end{align} }

Use the magnification formula to determine the magnification and the height of the image.

# M
−

d
i

d
o

=

h
i

h
o

{\displaystyle M = -\frac{d\_i}{d\_o} = \frac{h\_i}{h\_o} }

M

=
−

−
15

c
m

60

c
m

=
0.25

h
i

=
M

h
o

=
0.25
×
5

c
m

=
1.25

c
m

{\displaystyle \begin{align}
M &= -\frac{-15 \; cm}{60 \; cm} = 0.25 \\[5 pt]
h\_i &= M h\_o \\[5 pt]
&= 0.25 \times 5 \; cm \\[5 pt]
&= 1.25 \;cm
\end{align} }

The image is a virtual image (inverted image) located behind the concave mirror. The height of the image is diminished.

**Part 2**

Use the mirror equation to calculate the image distance.

1

d
o

1

d
i

=

1
f

{\displaystyle \frac{1}{d\_o} + \frac{1}{d\_i} = \frac{1}{f} }

1

30

c
m

1

d
i

=

1

−
20

c
m

1

d
i

=
−

1

12

c
m

d
i

=
−
12

c
m

{\displaystyle \begin{align}
\frac{1}{30 \; cm} + \frac{1}{d\_i} &= \frac{1}{-20 \; cm} \\[5 pt]
\frac{1}{d\_i} &= -\frac{1}{12 \; cm} \\[5 pt]
d\_i &= -12 \; cm
\end{align} }

Use the magnification formula to determine the magnification and the height of the image.

# M
−

d
i

d
o

=

h
i

h
o

{\displaystyle M = -\frac{d\_i}{d\_o} = \frac{h\_i}{h\_o} }

M

=
−

−
12

c
m

30

c
m

=
0.4

h
i

=
M

h
o

=
0.4
×
5

c
m

=
2

c
m

{\displaystyle \begin{align}
M &= -\frac{-12 \; cm}{30 \; cm} = 0.4 \\[5 pt]
h\_i &= M h\_o \\[5 pt]
&= 0.4 \times 5 \; cm \\[5 pt]
&= 2 \;cm
\end{align} }

The image is a virtual image (inverted image) located behind the concave mirror. The height of the image is diminished.

**Part 3**

Use the mirror equation to calculate the image distance.

1

d
o

1

d
i

=

1
f

{\displaystyle \frac{1}{d\_o} + \frac{1}{d\_i} = \frac{1}{f} }

1

20

c
m

1

d
i

=

1

−
20

c
m

1

d
i

=
−

1

10

c
m

d
i

=
−
10

c
m

{\displaystyle \begin{align}
\frac{1}{20 \; cm} + \frac{1}{d\_i} &= \frac{1}{-20 \; cm} \\[5 pt]
\frac{1}{d\_i} &= -\frac{1}{10 \; cm} \\[5 pt]
d\_i &= -10 \; cm
\end{align} }

Use the magnification formula to determine the magnification and the height of the image.

# M
−

d
i

d
o

=

h
i

h
o

{\displaystyle M = -\frac{d\_i}{d\_o} = \frac{h\_i}{h\_o} }

M

=
−

−
10

c
m

20

c
m

=
0.5

h
i

=
M

h
o

=
0.5
×
5

c
m

=
2.5

c
m

{\displaystyle \begin{align}
M &= -\frac{-10 \; cm}{20 \; cm} = 0.5 \\[5 pt]
h\_i &= M h\_o \\[5 pt]
&= 0.5 \times 5 \; cm \\[5 pt]
&= 2.5 \;cm
\end{align} }

The image is a virtual image (inverted image) located behind the concave mirror. The height of the image is diminished.

**Part 4**

Use the mirror equation to calculate the image distance.

1

d
o

1

d
i

=

1
f

{\displaystyle \frac{1}{d\_o} + \frac{1}{d\_i} = \frac{1}{f} }

1

10

c
m

1

d
i

=

1

−
20

c
m

1

d
i

=
−

3

20

c
m

d
i

=
−
6.67

c
m

{\displaystyle \begin{align}
\frac{1}{10 \; cm} + \frac{1}{d\_i} &= \frac{1}{-20 \; cm} \\[5 pt]
\frac{1}{d\_i} &= -\frac{3}{20 \; cm} \\[5 pt]
d\_i &= -6.67 \; cm
\end{align} }

Use the magnification formula to determine the magnification and the height of the image.

# M
−

d
i

d
o

=

h
i

h
o

{\displaystyle M = -\frac{d\_i}{d\_o} = \frac{h\_i}{h\_o} }

M

=
−

−
6.67

c
m

10

c
m

=
0.6667

h
i

=
M

h
o

=
0.6667
×
5

c
m

=
3.3

c
m

{\displaystyle \begin{align}
M &= -\frac{-6.67 \; cm}{10 \; cm} = 0.6667 \\[5 pt]
h\_i &= M h\_o \\[5 pt]
&= 0.6667 \times 5 \; cm \\[5 pt]
&= 3.3 \;cm
\end{align} }

The image is a virtual image (inverted image) located behind the concave mirror. The height of the image is diminished.