Question

An object is placed at a distance of 10cm. from a convex mirror of focal length 20cm. Find the position and nature of the image.

Answer 1

Given:-

$u = - 10cm \\ f = 20cm$

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To find:-

$v \: and \: Nature \: of \: the \: image$

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Formula used:-

$Mirror \: formula \\ \frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

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$\frac{1}{20 } = \frac{1}{v} + (\frac{1}{ - 10} ) \\ \frac{1}{20 } = \frac{1}{v} - \frac{1}{ 10} \\ \frac{1}{20 } + \frac{1}{10} = \frac{1}{v} \\ \frac{1 + 2}{20} = \frac{3}{20} \\ v = \frac{20}{3} = 6.67cm$

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Position of image = 6.67cm in front of mirror

Nature of image = virtual, erect and diminished

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Answer 2

$\red{\bf {Question}}$

An object is placed at a distance of 10cm. from a convex mirror of focal length 20cm. Find the position and nature of the image.

$\green{\bf {Answer}}$

$u = - 10cm$(because object is in front of the mirror)

$f = 20cm$

$Using \:Mirror \: formula:- \\ \\ \frac{1}{f} = \frac{1}{v} + \frac{1}{u}\\ \\</p><p>↦\frac{1}{20 } = \frac{1}{v} + (\frac{1}{ - 10} ) \\ \\ ↦ \frac{1}{20 } = \frac{1}{v} - \frac{1}{ 10} \\ \\ ↦\frac{1}{20 } + \frac{1}{10} = \frac{1}{v} \\ \\↦ \frac{1 + 2}{20} = \frac{3}{20} \\ \\ ↦v = \frac{20}{3} =6.67cm$

Position of image(v) = 6.67cm in front of mirror

Nature of image = virtual, erect and diminished

(because it is a convex mirror which makes virtual,erect

and diminished image in all

positions of the object)