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 $kmh⁻¹?$

Question

Dr. Abubacker · Accepted Answer

(a) Using mirror formula,
$\frac{1}{+ 25/3} + \ \frac{1}{- u₁}\  = \ \frac{1}{+ 10}$
Or $\frac{1}{u_{1}} = \ \frac{3}{25}\  - \ \frac{1}{10}$
Or $u_{1}\  = \ 50\ m$
And $\frac{1}{( + 50/7)\ } + \ \frac{1}{- u_{1}}\  = \ \frac{1}{+ 10}$
$	herefore\frac{1}{u_{2}}\  = \ \frac{7}{50}\  - \ \frac{1}{10}$
Or $u_{2}\  = \ 25\ m$
Speed of object $= \frac{u_{1} - \ u_{2}}{time}$
$= \frac{25}{30}{ms}^{- 1}$
$= \ 3\ kmh^{- 1}$

Q. 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 $kmh⁻¹?$

Explanation

Reviewed by Dr. Abubacker

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