An object is placed in front of a convex mirror nbsp-at a distance of 60-m from its pole- nbsp-Object starts moving towards the mirror at 20-ms-1- nbsp-Find the average speed of the image for a time interval of 2- s if the nbsp-focal length of the mirror is 20-m-0-568-ms-1-1-ms-1-3 nbsp-ms-1-2-5 nbsp-ms-1-

The correct option is D 2.5 nbsp; ms^ 1 Given: nbsp; Initial object distance, u = 60 m Speed of object, v_o = 20 ms^ 1 Focal length, f = 20 m Time interva ...

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Byju's Answer

Standard IX

Physics

Image Formation in Convex Mirrors

An object is ...

Question

An object is placed in front of a convex mirror at a distance of $60 m$ from its pole. Object starts moving towards the mirror at $20 m s^{- 1}$ . Find the average speed of the image for a time interval of $2 s$ if the focal length of the mirror is $20 m$ .

$0.568 m s^{- 1}$

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$1 m s^{- 1}$

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$3 m s^{- 1}$

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$2.5 m s^{- 1}$

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Solution

The correct option is D
$2.5 m s^{- 1}$

Given:
Initial object distance, $u = - 60 m$
Speed of object, $v_{o} = 20 m s^{- 1}$
Focal length, $f = 20 m$
Time interval, $t = 2 s$

Applying mirror formula to find the initial position of the image $v$ :
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} + \frac{1}{- 60} = \frac{1}{20}$
$v = 15 m$

Distance travelled by the object is the product of speed and time.
$s_{o} = v_{o} t$
$s_{o} = 20 \times 2 = 40 m$

Therefore, now the object distance:
$u_{f} = u + s_{o} = - 60 + 40 = - 20 m$

Applying mirror formula to find the final position of the image $v_{f}$ :
$\frac{1}{v_{f}} + \frac{1}{u_{f}} = \frac{1}{f}$
$\frac{1}{v_{f}} + \frac{1}{- 20} = \frac{1}{20}$
$v_{f} = 10 m$

Average speed is the ratio of total distance travelled to the total time taken.
Average speed of the image:
$v_{i} = \frac{v - v_{f}}{t}$
$v_{i} = \frac{15 - 10}{2} = 2.5 m s^{- 1}$

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