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# A convex lens forms a real image 3 times larger than the object on a screen. Object and screen are moved until the image becomes twice the size of object. If the shift of object is $6\;cm$ The shift of screen is :

$(a)\;12\;cm \\ (b)\;72\;cm \\ (c)\;18\;cm \\ (d)\;36\;cm$

$\large\frac{1}{3x}+\frac{1}{x} =\frac{1}{f}$
or $\large\frac{4}{3x} =\frac{1}{f}$ -----(i)
$\large\frac{1}{2(6+x)}+\frac {1}{6+x} =\frac{1}{f}$
ot $\large\frac{3}{2(6+x)} =\frac {1}{f}$-----(ii)
from equation (i) and (ii) , it gives
$\large\frac{4}{3x} =\frac{3}{2(6+x)}$
$\therefore 9x= 48+8x$
$x=48 \;cm$
Shift of screen $=3x-2(6+X)$
$\qquad= 3 \times 48 -2 (6+ 48)$
$\qquad=36 \;cm$
Hence d is the correct answer.