Optics Ques 16
- 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 $km h^{-1}$?
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Answer:
Correct Answer: 16.$(3)$
Solution:
Formula:
- Using mirror formula twice,
$ \begin{alignedat} & \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 \text{ m} \text { and } \frac{1}{(+50 / 7)}+\frac{1}{-u_2}=\frac{1}{+10} \\ & \therefore \quad \frac{1}{u _2}=\frac{7}{50}-\frac{5}{50} \text { or } u _2=25 m \end{aligned} $
Speed of object $=\frac{v _1-v _2}{\text { time }}$
$ =\frac{25}{30}$ $ ms^{-1}$ $=3 $ $kmh^{-1} $
$\therefore \quad $ Answer is $ 3 $ .
BETA
