Question #cf067

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Answer 1 · 2016-12-12T06:02:52

We know the lens formula as follows

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

where

$u \to object distance = - v e$ $u->"object distance"=-ve$

$v \to imsge distance = + v e$ $v->"imsge distance"=+ve$

$m \to magnification = - 1$ $m->"magnification"=-1$

$since real image is formed$ $color(red)"since real image is formed"$

$Now \frac{v}{u} = m = 1 \Rightarrow v = - u =$ $"Now " v/u=m=1=>v=-u=$

$f \to focal length = f for convex lens$ $f->"focal length"=f" for convex lens"$

So the above equation becomes

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

$\Rightarrow u = - 2 f$ $=>u=-2f$

Hence it is possible when

$object distance (u) = 2 \times focal length (f)$ $color(blue)("object distance"(u) = 2xx"focal length"(f))$