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A U-shaped wire is placed before a concave mirror having radius of curvature 20cm. Find the total length of the image.
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$\text{R}=20\text{cm}, \ \text{f}=\frac{\text{R}}{2}=-10\text{cm}$ For part AB, PB = 30 + 10 = 40cm So, $\text{u}=-40\text{cm}$$\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}=-\frac{1}{10}-\Big(\frac{1}{-40}\Big)=-\frac{3}{40}$
$\Rightarrow\text{v}=-\frac{40}{3}=-13.3\text{cm}$ So, PB' = 13.3cm $\text{m}=\frac{\text{A}'\text{B}'}{\text{AB}}=-\Big(\frac{\text{v}}{\text{u}}\Big)=-\Big(\frac{-13.3}{-40}\Big)=-\frac{1}{3}$ $\Rightarrow\text{A}'\text{B}'=-\frac{10}{3}=-3.33\text{cm}$ For part CD, PC = 30, So, u = -30cm $\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}=-\frac{1}{10}+\frac{1}{-30}=-\frac{1}{15}$ $\Rightarrow\text{v}=-15\text{cm}=\text{PC}'$ So, $\text{m}=\frac{\text{C}'\text{D}'}{\text{CD}}=-\frac{\text{v}}{\text{u}}=-\Big(\frac{-15}{-30}\Big)=-\frac{1}{2}$ $\Rightarrow\text{C}'\text{D}'=5\text{cm}$ B'C' = PC' - PB' = 15 - 13.3 = 17cm So, total length A'B' + B'C' + C'D' = 3.3 + 1.7 + 5 = 10cm.
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