Question

5.
Image of an object approaching a convex mirror of radius of
curvature 20 m along its optical axis is observed to move from
25
mto = min 30 s. What is the speed of the object in km hı?
​

Answer 1

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$\huge\bf{Answer:-}$

$Speed = \frac{25 \: m \: }{30 \: s} = \frac{3 \: km}{hr}$

$\huge\bf{Solution:}$

Let us first find the initial position of the object.

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \frac{3}{25} + \frac{1}{u} = \frac{ - 1}{50} \\ u = - 50$

Now let us find the final position of the object :

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \frac{7}{50} + \frac{1}{u} = \frac{1}{10} \\ \frac{1}{u} = \frac{ - 2}{50} \\ u = - 25$

Therefore the shift in object is : 50-25 = 25 m

The time taken for this shift = 30 s

$Speed = \frac{25 \: m}{30 \: s} = \frac{3 \: km}{hr}$

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