$(a)\;15\;cm \\ (b)\;25\;cm \\ (c)\;35\;cm \\ (d)\;20\;cm $

For real image

$u=u_1,v=-2u_1,f=20\;cm$

Substituting in $\large\frac{1}{v} +\frac{1}{u} =\frac{1}{f}$

Then, we get,

$\large\frac{1}{-2u}-\frac{1}{u_1}=\frac{-1}{20}$

or $u_1=30\;cm$

For virtual image $u=-u_2,v=2u_2$

$f= -20\;cm$

$u_2= 10\;cm$

$\therefore$ Distance between two position of object are $u_1-u_2$

or $30-10=20\;cm$

Hence d is the correct answer.

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