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Sagot :
Answer:
The magnification = 0.52
Explanation:Power, P = -4.10D
Object distance, u = 22.5 cm
Focal length, f = 100/power
f = 100/-4.10
f = -24.39 cm
Using the formula below, calculate the image distance (v)
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]Substitute u = 22.5, f = -24.39, and solve for v
[tex]\begin{gathered} \frac{1}{-24.39}=\frac{1}{22.5}+\frac{1}{v} \\ \\ \frac{1}{v}=-\frac{1}{24.39}-\frac{1}{22.5} \\ \\ \frac{1}{v}=-0.041-0.044 \\ \\ \frac{1}{v}=-0.085 \\ \\ v=-\frac{1}{0.085} \\ \\ v=-11.8 \end{gathered}[/tex]The magnification, M = |v/u|
M = |-11.8/22.5|
M = |-0.52|
M = 0.52
The magnification = 0.52
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