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Image of an object approaching a convex mirror of radius of curvature $20 \mathrm{~m}$ along its optical axis is observed to move from $\frac{25}{3} \mathrm{~m}$ to $\frac{50}{7} \mathrm{~m}$ in $30 \mathrm{~s}$. What is the speed of the object in $\mathrm{km} \mathrm{h}^{-1}$ ?
Solution:
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Verified Answer
The correct answer is:
3
Using mirror formula twice,
$\frac{1}{+25 / 3}+\frac{1}{-u_1}=\frac{1}{+10}$
or $\quad \frac{1}{u_1}=\frac{3}{25}-\frac{1}{10}$ or $u_1=50 \mathrm{~m}$
and, $\quad \frac{1}{(+50 / 7)}+\frac{1}{-u_2}=\frac{1}{+10}$
$\therefore \quad \frac{1}{u_2}=\frac{7}{50}-\frac{1}{10}$ or $u_2=25 \mathrm{~m}$
Speed of object $=\frac{u_1-u_2}{\text { time }}$
$=\frac{25}{30} \mathrm{~ms}^{-1}$
$=3 \mathrm{kmh}^{-1}$
$\therefore$ The answer is 3 .
$\frac{1}{+25 / 3}+\frac{1}{-u_1}=\frac{1}{+10}$
or $\quad \frac{1}{u_1}=\frac{3}{25}-\frac{1}{10}$ or $u_1=50 \mathrm{~m}$
and, $\quad \frac{1}{(+50 / 7)}+\frac{1}{-u_2}=\frac{1}{+10}$
$\therefore \quad \frac{1}{u_2}=\frac{7}{50}-\frac{1}{10}$ or $u_2=25 \mathrm{~m}$
Speed of object $=\frac{u_1-u_2}{\text { time }}$
$=\frac{25}{30} \mathrm{~ms}^{-1}$
$=3 \mathrm{kmh}^{-1}$
$\therefore$ The answer is 3 .
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