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An object \(5 \mathrm{~cm}\) tall is placed \(1 \mathrm{~m}\) from a concave spherical mirror which has a radius of curvature of \(20 \mathrm{~cm}\). The size of the image is
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The correct answer is:
\(0.55 \mathrm{~cm}\)
$: h_0=5 \mathrm{~cm}, h_i=?$
$u=-100 \mathrm{~cm}, R=-20 \mathrm{~cm} . \quad \therefore f=-10 \mathrm{~cm}$
Using mirror formula,
$\begin{aligned} & \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}-\frac{1}{100}=-\frac{1}{10} \\ & \frac{1}{v}=\frac{1}{100}-\frac{1}{10}=\frac{1-10}{100}=-\frac{9}{100} \\ & \therefore v=-\frac{100}{9} \mathrm{~cm} . \\ & \frac{h_i}{h_0}=\frac{v}{u} \Rightarrow \frac{h_i}{5 \mathrm{~cm}}=\frac{(100 / 9)}{100}=\frac{1}{9} \\ & \therefore \quad h_i=5 / 9=0.55 \mathrm{~cm} .\end{aligned}$
$u=-100 \mathrm{~cm}, R=-20 \mathrm{~cm} . \quad \therefore f=-10 \mathrm{~cm}$
Using mirror formula,
$\begin{aligned} & \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}-\frac{1}{100}=-\frac{1}{10} \\ & \frac{1}{v}=\frac{1}{100}-\frac{1}{10}=\frac{1-10}{100}=-\frac{9}{100} \\ & \therefore v=-\frac{100}{9} \mathrm{~cm} . \\ & \frac{h_i}{h_0}=\frac{v}{u} \Rightarrow \frac{h_i}{5 \mathrm{~cm}}=\frac{(100 / 9)}{100}=\frac{1}{9} \\ & \therefore \quad h_i=5 / 9=0.55 \mathrm{~cm} .\end{aligned}$
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