At what distance should an object be placed in front of a convex lens of focal length $10$ cm so to obtain an image twice the size of object ?

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Solution

Given focal length $f = 10 c m$ also magnification $m = 2$
We know magnification $m = \frac{- v}{u} => v = - 2 u$ ------(1)
From lens formula we get
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
putting the value found in 1 we get,
$\frac{1}{- 2 u} - \frac{1}{- u} = \frac{1}{- f}$ negative sign given as per sign convention.
$\frac{1}{2 u} = \frac{1}{- 10}$
$u = - 5 c m$

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At what distance should an object be placed in front of a convex lens of focal length 10 cm so to obtain an image twice the size of object ?

Given focal length f=10cm also magnification m=2We know magnificationm=−vu=>v=−2u------(1)From lens formula we get 1v−1u=1fputting the value found in 1 we get,1−2u−1−u=1−f negative sign given as per sign convention.12u=1−10u=−5cm

At what distance should an object be placed in front of a convex lens of focal length $10$ cm so to obtain an image twice the size of object ?