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.
