Question

a lens form a vitual image of twice the size of an object placed 15 cm away from the lens find the distanceof images from thelens and focal length of the lens?


hi guys
for evening

anyone solve this sum and send me the answer guys​

Answer 1

$\sf Given \begin {cases} & \sf { Distance \: of \: object \: = u = \: -15 } \\ & \sf{ Magnification = m = \: +2 } \end {cases}$

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$\sf To \: find \begin {cases} & \sf { Distance \: of \: image \: = v = \: ?? } \\ & \sf{ Focal length = f \: ?? } \end {cases}$

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

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$\star\;{\boxed{\sf{\pink{ Magnification = \dfrac{ Distance_{(image)}} { Distance_{(Object) } } }}}}$

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$:\implies\sf 2 = \dfrac{ Distance_{(image)}} { -15 }$

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$:\implies\sf Distance_{(image)} = 2 \times - 15$

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$: \implies \sf Distance_{(image)} = -30 cm$

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$:\implies{\underline{\boxed{\frak{Image \: distance = -30 \: cm }}}}\;\bigstar$

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

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

Where ;

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• f = focal length

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• u = object distance

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• v = image distance

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$:\implies\sf \dfrac{1}{f} = \dfrac{- 1}{30} - \dfrac{ - 1}{15}$

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$:\implies\sf \dfrac{1}{f} = \dfrac{- 1}{30} + \dfrac{1}{15}$

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$:\implies\sf \dfrac{1}{f} = \dfrac{-1 + 2 }{30}$

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$:\implies\sf \dfrac{1}{f} = \dfrac{1}{30}$

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$:\implies{\underline{\boxed{\frak{\purple{Focal \: length = 30 cm }}}}}\;\bigstar$

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$\therefore\:{\underline{\sf{Focal \;length \: of \: the \:lens \; is \: \bf{ 30\: cm }}}}$

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Answer 2

Answer:

Given :-

• A lens from a virtual image of twice the size of an object placed 15 cm away from the lens.

To Find :-

• What is the distance image from the lens.
• What is the focal length of the lens.

Formula Used :-

$\clubsuit$ Magnification Formula :

$\longmapsto \sf\boxed{\bold{\pink{Magnification =\: \dfrac{Image\: Distance}{Object\: Distance}}}}$

$\clubsuit$ Lens Formula :

$\longmapsto \sf \boxed{\bold{\pink{\dfrac{1}{f} =\: \dfrac{1}{v} - \dfrac{1}{u}}}}$

where,

• v = Image Distance
• u = Object Distance
• f = Focal Length

Solution :-

First, we have to find the image distance :

Given :

• Magnification = 2
• Object Distance = - 15 cm

According to the question by using the formula we get,

$\implies \sf 2 =\: \dfrac{Image\: Distance}{- 15}$

By doing cross multiplication we get,

$\implies \sf Image\: Distance =\: - 15(2)$

$\implies \sf Image\: Distance =\: - 15 \times 2$

$\implies \sf\bold{\red{Image\: Distance =\: - 30\: cm}}$

$\therefore$ The distance image from the lens is - 30 cm .

$\rule{150}{2}$

Now, we have to find the focal length :

Given :

• Image Distance = 30 cm
• Object Distance = 15 cm

According to the question by using the formula we get,

$\implies \sf \dfrac{1}{f} =\: \dfrac{- 1}{30} - \dfrac{- 1}{15}$

$\implies \sf \dfrac{1}{f} =\: \dfrac{- 1 + 2}{30}$

$\implies \sf \dfrac{1}{f} =\: \dfrac{1}{30}$

By doing cross multiplication we get,

$\implies \sf f =\: 30(1)$

$\implies \sf f =\: 30 \times 1$

$\implies \sf\bold{\red{f =\: 30\: cm}}$

$\therefore$ The focal length of the lens is 30 cm .