Concave 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 concave mirror of focal length 20 cm.

**Part 2:** A 1 cm tall arrowhead is placed 30 cm from a concave mirror of focal length 15 cm.

**Part 3:** A 5 cm tall arrowhead is placed 20 cm from a concave mirror of focal length 15 cm.

**Part 4:** A 1.5 cm tall arrowhead is placed 10 cm from a concave mirror of focal length 10 cm.

**Part 5:** A 1.5 cm tall arrowhead is placed 10 cm from a concave 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.

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

30

c
m

d
i

=
30

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}{30 \; cm} \\[5 pt]
d\_i &= 30 \; 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

=
−

30

c
m

60

c
m

=
−
0.5

h
i

=
M

h
o

=
−
0.5
×
5

c
m

=
−
2.5

c
m

{\displaystyle \begin{align}
M &= -\frac{30 \; cm}{60 \; 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 real image (inverted image) located in front of the concave mirror behind the focal length. The height of the image is diminished. This is illustrated by ray diagram (b) in Figure 1.

**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

20

c
m

1

d
i

=

1

15

c
m

1

d
i

=

1

60

c
m

d
i

=
60

c
m

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

=
−

20

c
m

60

c
m

=
−

1
3

h
i

=
M

h
o

=
−

1
3

×
5

c
m

=
−
1.67

c
m

{\displaystyle \begin{align}
M &= -\frac{20 \; cm}{60 \; cm} = -\frac{1}{3} \\[5 pt]
h\_i &= M h\_o \\[5 pt]
&= -\frac{1}{3} \times 5 \; cm \\[5 pt]
&= -1.67 \;cm
\end{align} }

The image is a real image (inverted image) located in front of the concave mirror beyond the mirror's radius of curvature. The height of the image is reduced. This is illustrated by ray diagram (c) in Figure 1.

**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

30

c
m

1

d
i

=

1

15

c
m

1

d
i

=

1

30

c
m

d
i

=
30

c
m

{\displaystyle \begin{align}
\frac{1}{30 \; cm} + \frac{1}{d\_i} &= \frac{1}{15 \; cm} \\[5 pt]
\frac{1}{d\_i} &= \frac{1}{30 \; cm} \\[5 pt]
d\_i &= 30 \; 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

=
−

30

c
m

30

c
m

=
−
1

h
i

=
M

h
o

=
−
1
×
1

c
m

=
−
1

c
m

{\displaystyle \begin{align}
M &= -\frac{30 \; cm}{30 \; cm} = -1 \\[5 pt]
h\_i &= M h\_o \\[5 pt]
&= -1 \times 1 \; cm \\[5 pt]
&= -1 \;cm
\end{align} }

The image is a real image (inverted image) located in front of the concave mirror on twice the focal length (the mirror's radius of curvature). The height of the image is unchanged. This is illustrated by ray diagram (d) in Figure 1.

**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

10

c
m

1

d
i

=
0

d
i

=
∞

{\displaystyle \begin{align}
\frac{1}{10 \; cm} + \frac{1}{d\_i} &= \frac{1}{10 \; cm} \\[5 pt]
\frac{1}{d\_i} &= 0 \\[5 pt]
d\_i &= \infty
\end{align} }

No image is formed along the principle axis and the image is located very far away; the image will form at infinity. This is illustrated by ray diagram (e) in Figure 1.

**Part 5**

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

=
−

1

20

c
m

d
i

=
−
20

c
m

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

=
−

−
20

c
m

10

c
m

=
2

h
i

=
M

h
o

=
2
×
1.5

c
m

=
3

c
m

{\displaystyle \begin{align}
M &= -\frac{-20 \; cm}{10 \; cm} = 2 \\[5 pt]
h\_i &= M h\_o \\[5 pt]
&= 2 \times 1.5 \; cm \\[5 pt]
&= 3 \;cm
\end{align} }

The image is a virtual image located behind the concave mirror. The height of the image is doubled (magnified 2x). This is illustrated by ray diagram (f) in Figure 1.