Optics 1 Question 14
14. Image of an object approaching a convex mirror of radius of curvature $20 m$ along its optical axis is observed to move from $\frac{25}{3} m$ to $\frac{50}{7} m$ in $30 s$. What is the speed of the object in $\underset{(2010)}{km h^{-1} \text { ? }}$
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Answer:
Correct Answer: 14. 3
Solution:
- Using mirror formula twice,
$$ \begin{aligned} & \frac{1}{+25 / 3}+\frac{1}{-u _1}=\frac{1}{+10} \\ & \text { or } \quad \frac{1}{u _1}=\frac{3}{25}-\frac{1}{10} \\ & \text { or } \quad u _1=50 m \text { and } \frac{1}{(+50 / 7)}+\frac{1}{-u _2}=\frac{1}{+10} \\ & \therefore \quad \frac{1}{u _2}=\frac{7}{50}-\frac{1}{10} \text { or } u _2=25 m \end{aligned} $$
Speed of object $=\frac{u _1-u _2}{\text { time }}$
$$ =\frac{25}{30} ms^{-1}=3 kmh^{-1} $$
$\therefore$ Answer is 3 .
