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: $d i=21 \mathrm{~cm}$ ] 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 ：\( d_i = \boxed{21} \mathrm{cm} \)

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 ：\( h_i = \boxed{-6.4} \mathrm{cm} \)

Step 10 ：\( M = \frac{-6.4}{8.5} \)

Step 11 ：\( M = \boxed{-0.75} \)

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

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