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

Only convex lens forms both real and virtual image.

Let f be focal length and n be magnification.

For real image, magnification $=-n$

$u= -12$

$v=+12$

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

$\large\frac{1}{12n} -\frac{1}{-12} =\frac{1}{f}$

$1+ \large\frac{1}{n}=\frac{12}{f}$----(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.

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