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A point object is placed at a distance of $10 \mathrm{~cm}$ and its real image is formed at a distance of $20 \mathrm{~cm}$ from a concave mirror. If the object is moved by $0.1 \mathrm{~cm}$ towards the mirror, the image will shift by about
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The correct answer is:
$0.4 \mathrm{~cm}$ away from the mirror
Mirror formula
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u} \Rightarrow \frac{1}{f}=\frac{1}{-20}+\frac{1}{(-10)} \Rightarrow f=\frac{20}{3} \mathrm{~cm}$. If object moves towards the mirror by $0.1 \mathrm{~cm}$ then
$u=(10-0.1)=9.9 \mathrm{~cm}$. Hence again from mirror formula $\frac{1}{-20 / 3}=\frac{1}{v}+\frac{1}{-9.9} \Rightarrow v^{\prime}=20.4 \mathrm{~cm}$ i.e. image
shifts away from the mirror by $0.4 \mathrm{~cm}$.
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u} \Rightarrow \frac{1}{f}=\frac{1}{-20}+\frac{1}{(-10)} \Rightarrow f=\frac{20}{3} \mathrm{~cm}$. If object moves towards the mirror by $0.1 \mathrm{~cm}$ then
$u=(10-0.1)=9.9 \mathrm{~cm}$. Hence again from mirror formula $\frac{1}{-20 / 3}=\frac{1}{v}+\frac{1}{-9.9} \Rightarrow v^{\prime}=20.4 \mathrm{~cm}$ i.e. image
shifts away from the mirror by $0.4 \mathrm{~cm}$.
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