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An object is kept at a distance of 12 cm from a thin lens and image formed is real. If the object is kept at a distance of $8\;cm$ from the same lens image formed is virtual. If the size of image formed is equal length of lens will be :

$(a)\;15\;cm \\ (b)\;17\;cm \\ (c)\;10\;cm \\ (d)\;8\;cm $

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Only convex lens forms both real and virtual image.
Let f be focal length and n be magnification.
For real image, magnification $=-n$
$u= -12$
$\large\frac{1}{u} -\frac{1}{u}=\frac{1}{f}$
$\large\frac{1}{12n} -\frac{1}{-12} =\frac{1}{f}$
$1+ \large\frac{1}{n}=\frac{12}{f}$----(1)
When virtual image of same size informed magnification is $-n$
$\large\frac{1}{-8 n} +\frac{1}{8}=\frac{1}{f}$
$1- \large\frac{1}{n} =\frac{8}{f}$----(2)
Solving (1) and (2)we get,
$f=10 cm$
Hence c is the correct answer.


answered Jan 22, 2014 by meena.p
edited Jul 15, 2014 by thagee.vedartham

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