A 2.0-cm-tall candle flame is 2.0 m from a wall. You happen to havea lens with focal length of 32 cm.

A. How many places can you put the lens to form a well-focused imageof the candle on the wall?

B. For each position, what is the height and orientation of theimage?

### Answer

*A.*

Take the difference between the object and image distance as `2m`

.

```
\begin{aligned}
v-u &=2 \mathrm{~m} \\
v &=2 \mathrm{~m}+u
\end{aligned}
```

The height of the object is `2.0 cm`

.

The expression for lens equation is as follows:

`\frac{1}{f}=\frac{1}{u}+\frac{1}{v}`

The object is placed on the left side of the lens. Therefore, the above equation is rewritten as follows:

`\begin{aligned} \frac{1}{f} &=\frac{1}{-u}+\frac{1}{v} \\ &=\frac{1}{v}-\frac{1}{u} \end{aligned}`

Substitute `2m+u`

for `v`

.

```
\begin{aligned}
&=\frac{u}{(2 m+u) u}-\frac{(2 m+u)}{(2 m+u) u} \\
&=\frac{u-2 m-u}{(2 m+u) u} \\
&=\frac{-2 m}{(2 m+u) u}
\end{aligned}
```

Again, substitute `32cm`

for `f`

.

```
\begin{aligned}
\frac{1}{32 \mathrm{~cm}\left(\frac{10^{-2} \mathrm{~m}}{1 \mathrm{~cm}}\right)} &=\frac{-2 \mathrm{~m}}{(2 \mathrm{~m}+u) u} \\
\frac{1}{0.32 \mathrm{~m}} &=\frac{-2 \mathrm{~m}}{(2 \mathrm{~m}+u) u} \\
(2 \mathrm{~m}+u) u &=-2 \mathrm{~m}(0.32 \mathrm{~m}) \\
u^{2}+2 u+0.64 &=0
\end{aligned}
```

Solve the quadratic equation.

`u=-0.4,-1.6 \mathrm{~m}`

Hence, the lens can be placed at `40 cm`

and `160 cm`

to form a well-focused image of the candle flame on the wall.

**Thus, The lens can be placed at two locations.**

*B.*

When the distance between the candle and lens was `160 \mathrm{~cm}`

. The expression for lens equation is as follows:

`\frac{1}{f}=\frac{1}{u}+\frac{1}{v}`

Substitute `32 \mathrm{~cm}`

for `f`

and `160 \mathrm{~cm}`

for `u`

.

`\begin{aligned} \frac{1}{32 \mathrm{~cm}} &=\frac{1}{160 \mathrm{~cm}}+\frac{1}{v} \\ \frac{1}{v} &=\frac{1}{32 \mathrm{~cm}}-\frac{1}{160 \mathrm{~cm}} \\ &=\frac{1}{40 \mathrm{~cm}} \\ v &=40 \mathrm{~cm} \end{aligned}`

The expression for magnification of the lens is as follows:

`M=\frac{-v}{u}`

Substitute `40 \mathrm{~cm}`

for `v`

and `160 \mathrm{~cm}`

for `u`

.

`\begin{aligned} M &=\frac{-40 \mathrm{~cm}}{160 \mathrm{~cm}} \\ &=-0.25 \mathrm{~cm} \end{aligned}`

Another expression for magnification is as follows:

`M=\frac{h_{i}}{h_{o}}`

Compare the above two expressions and rearrange the expression for `h_{i}`

as follows:

`\frac{-v}{u}=\frac{h_{\mathrm{i}}}{h_{\mathrm{o}}}`

`h_{\mathrm{i}}=\frac{-(v)}{u} h_{6}`

Substitute `40 \mathrm{~cm}`

for `v, 2 \mathrm{~cm}`

for `h_{0}`

, and `160 \mathrm{~cm}`

for `u`

.

`h_{i}=\frac{-(40 \mathrm{~cm})}{160 \mathrm{~cm}}(2 \mathrm{~cm})`

`=-0.5 \mathrm{~cm}`

When the distance between the candle and lens was `40 \mathrm{~cm}`

.

The expression for lens equation is as follows:

`M=\frac{-v}{u}`

Substitute `160 \mathrm{~cm}`

for `v`

and `40 \mathrm{~cm}`

for `u`

.

`M=\frac{-160 \mathrm{~cm}}{40 \mathrm{~cm}}`

`=-4 \mathrm{~cm}`

Another expression for magnification is as follows:

`M=\frac{h_{i}}{h_{o}}`

Compare the above two expressions and rearrange the expression for `h_{i}`

as follows:

`\frac{-v}{u}=\frac{h_{\mathrm{i}}}{h_{\mathrm{e}}}`

`h=\frac{-(v)}{u} h_{ㅇ}`

Substitute `160 \mathrm{~cm}`

for `$v`

, `2 \mathrm{~cm}`

for `h_{0}`

, and `40 \mathrm{~cm}`

for `u`

.

`h_{i}=\frac{-(160 \mathrm{~cm})}{40 \mathrm{~cm}}(2 \mathrm{~cm})`

`=-8 \mathrm{~cm}`

**Therefore, the height of the image when the distance between the candle and lens are 160 \mathrm{~cm} and 40 \mathrm{~cm} are -0.5 \mathrm{~cm} and -8 \mathrm{~cm}, respectively. In both cases, the magnification is negative, since both are inverted.**