Problems (Review all worksheets with problems done in the class)
10. An object that is $8.5\mathrm{cm}$ high is placed $28.0\mathrm{cm}$ in front of a converging lens that has a focal length of $12\mathrm{cm}$.
a) Calculate the image distance (di). Show your work.
[Ans: $di=21\mathrm{cm}$ ]
b) Calculate the image height (hi). Show your work.
$[\mathrm{Ans}:\mathrm{hi}=-6.4\mathrm{cm}]$
c) Calculate the image magnification (M). Show your work.
d) Is the image a real or virtual image? Explain.
e) Is the image an upright or inverted image? Explain.

Step 1 ：$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$

Step 2 ：$\frac{1}{12}=\frac{1}{28}+\frac{1}{d_i}$

Step 3 ：$\frac{1}{d_i}=\frac{1}{12}-\frac{1}{28}$

Step 4 ：$d_i=\frac{1}{\frac{1}{12}-\frac{1}{28}}$

Step 5 ：$d_i=\boxed{{\displaystyle 21}}\mathrm{cm}$

Step 6 ：$M=\frac{h_i}{h_o}=\frac{d_i}{d_o}$

Step 7 ：$h_i=M\cdot h_o$

Step 8 ：$h_i=\frac{21}{28}\cdot 8.5$

Step 9 ：$h_i=\boxed{{\displaystyle -6.4}}\mathrm{cm}$

Step 10 ：$M=\frac{-6.4}{8.5}$

Step 11 ：$M=\boxed{{\displaystyle -0.75}}$

Step 12 ：The image is real because the image distance is positive.

Step 13 ：The image is inverted because the image height is negative.