Question #b5c81

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Question #b5c81

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Answer 1 · 2016-08-28T02:55:01

As the lens is being used as magnifying glass the image formed will be virtual and magnified,

The lens formula

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $1/v-1/u=1/f$

where

$v \to Image distance$ $v->"Image distance"$

$u \to object distance$ $u->"object distance"$

$f \to focal length$ $f->"focal length"$

The magnification given here is 5 times the size of an object and image is formed 20 m from lens. So

$v \to Image distance = - 20 m (-ve sign for virtual image)$ $v->"Image distance"=-20m(" -ve sign for virtual image")$

$u \to object distance = \frac{Image distance}{magnification} = - \frac{20}{5} = - 4$ $u->"object distance"="Image distance"/"magnification"=-20/5=-4$

Inserting these in lens equation we have

$\frac{1}{- 20} - \frac{1}{- 4} = \frac{1}{f}$ $1/(-20)-1/(-4)=1/f$

$\Rightarrow \frac{1}{f} = \frac{1}{- 20} - \frac{1}{- 4} = \frac{1}{4} - \frac{1}{20} = \frac{5 - 1}{20} = \frac{1}{5}$ $=>1/f=1/(-20)-1/(-4)=1/4-1/20=(5-1)/20=1/5$

$f = 5 m$ $f=5m$

Hence the focal length of the lens is 5m and the object is to be placed at 4m distance to have 5 times magnified virtual image at a distance 20m from the lens.

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Question #b5c81

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