$(a)\;1.0\;m \\ (b)\;0.5\;m \\ (c)\;0.6\;m \\ (d)\;0.2 \;m $

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Since object is at 1 m in front of lens of focal length 0.5 m it is at 2 f from lens 10 image $I_1$ is formed at 1m on the other side of lens.

$I_1$ forms object for mirror which is 1 m in front .

$\therefore I_2 $ is formed 1 m behind mirror.

Now final image $I_3$ is formed

Using lens formula

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

$u= 3m$

$f= 0.5 m$

$\large\frac{1}{v} -\frac{1}{-3}=\frac{1}{.5}$

$\large\frac{1}{v} =\frac{1}{.5} -\frac{1}{3}$

$\large\frac{1}{v} $$=2 -\large\frac{1}{3}$

$\large\frac{1}{v} =\frac{5}{3}$

$v= \large\frac{3}{5}$

$\quad= 0.6 \;m$ from lens

Hence c is the correct answer.

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