Question

When an object is placed at a distance of 15 cm from a convex lens , 4 times magnified real image is formed. Where should the object be placed so that 4 times magnified virtual image is formed? ​

Answer 1

$\sf Given\: that \begin{cases} & \sf{Initial \: Distance\: of \: object \: = \frak{- 15\: cm}} \\ & \sf{ Magnification_{(initial)} \: = \bf{ - 4 }}\\ & \sf{ Magnification_{(final)} = \: + 4 } \end{cases}$

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$\sf To \: find \begin {cases} & \sf { Distance \: of \: image \: = \: ?? } \\ & \sf{ Final\: Distance\: of \: object = \: ??? } \end {cases}$

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$\bigstar\:{\underline{\sf From \: magnification \:formulae\::}}$

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$\star\;{\boxed{\sf{\pink{ Magnification_{(initial)} = \dfrac{ Initial \:Distance_{(image)}} { Initial \: Distance_{(Object) } } }}}}$

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$:\implies\sf - 4 = \dfrac{ Distance_{(image)}} { -15 }$

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$:\implies\sf Distance_{(image)} = - 4 \times - 15$

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$: \implies \sf Distance_{(image)} = 60cm$

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$:\implies{\underline{\boxed{\frak{Image \: distance = 60 \: cm }}}}\;\bigstar$

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$\bigstar\:{\underline{\sf Using\:Lens\:Formula\::}}$

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$\star\;{\boxed{\sf{\pink{\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} }}}}$

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Where

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• f = focal length
• u = object distance
• v = image distance

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$:\implies\sf \dfrac{1}{f} = \dfrac{1}{60} - \dfrac{ - 1}{15}$

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$:\implies\sf \dfrac{1}{f} = \dfrac{1}{60} + \dfrac{1}{15}$

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$:\implies\sf \dfrac{1}{f} = \dfrac{1 + 4 }{60}$

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$:\implies\sf \dfrac{1}{f} = \dfrac{5}{60}$

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$: \implies\sf \dfrac{1}{f} = \dfrac{1}{12}$

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$:\implies{\underline{\boxed{\frak{\purple{f = 12cm }}}}}\;\bigstar$

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$\bigstar\:{\underline{\sf From \: magnification \:formulae\::}}$

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$\star\;{\boxed{\sf{\pink{ Magnification_{Final} = \dfrac{ Final \:Distance_{(image)} = v } { Final \: Distance_{(Object) } = u} }}}}$

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$:\implies\sf 4 = \dfrac{ v } { u }$

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$:\implies\sf v = 4u$

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$\bigstar\:{\underline{\sf Using\:Lens\:Formula\::}}$

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$\star\;{\boxed{\sf{\pink{\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}}}}}$

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Where

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• f = focal length
• u = object distance
• v = image distance

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$:\implies\sf \dfrac{1}{12} = \dfrac{1}{4u} - \dfrac{ 1}{u}$

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$:\implies\sf \dfrac{1}{12} = \dfrac{u - 4u }{4u^2}$

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$:\implies\sf \dfrac{1}{12} = \dfrac{-3u}{4u\times u }$

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$: \implies\sf \dfrac{1}{12} = \dfrac{-3}{4u}$

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$: \implies \sf u = \dfrac{-3\times 12}{4}$

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$: \implies \sf u = - 3 \times 3$

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$:\implies \sf u = - 9cm$

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$:\implies{\underline{\boxed{\frak{\purple{Final \:distance _{(object)} = - 9cm }}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Object \;will \: be \: placed \:at \; \bf{9\:cm \: in \: front\: of \: mirror }.}}}$

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Answer 2

Given :-

• Initial Distance of object = -15
• Magnification ( initial ) = - 4
• Magnification ( final ) = + 4

To find :-

• Distance of an image
• Final distance of an object

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using magnification formula

➪Magnification ( initial ) = Initial Distance ( image ) / Initial distance ( Object )

➪ - 4  = Distance ( image ) / - 5

➪ Distance ( image )  = - 4 × - 5

➪ Distance ( image ) = 60 cm

∴ Image distance is 60 cm .

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using lens formula

➪ 1/f = 1/v - 1/u

➪ 1/f  = 1/60  - (-1)/15

➪ 1/f = 1/60 + 1/15

➪ 1/f =  1 + 4 / 60

➪ 1/f  = 5/60

➪ 1/f = 1/12 = 12cm

∴ Focal length is 12cm

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Using magnification formula

➪ Magnification ( final ) = v/u

➪ 4 =v/u

➪ v = 4u

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using lens  formula

➪ 1/f = 1/v - 1/u

➪ 1/12 = 1/4u - 1/u

➪ 1/12 = u - 4u/ 4u²

➪ 1/12 = - 3u/4u × u

➪ 1/12 = -3/4u

➪ u = -3 × 12 / 4

➪ u = -3 ×  3

➪ u = - 9 cm

∴ Final distance ( object ) = -9cm

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Therefore object will be placed at 9cm in the front of mirror