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Optic axis of a thin equiconvex lens is the x-axis. The co-ordinates of a point object and its image are $(-40\;cm,1\;cm)$ and $(50\;cm,-2\;cm)$ respectively . lens is located at

$(a)\;x=15\;cm \\ (b)\;x=10\;cm \\ (c)\;x=20\;cm \\ (d)\;x=-25\;cm$

Linear magnification $m= \large\frac{-2 \;cm}{1\;cm}$$=-2$
ie image is real and inverted $|v|=2 |u|$
Let $|u|=x$ then $|v|=2x$
Now $|u|+|v|=50-(-40)=90$
$\therefore x+2x=90\;or\;x=30\;cm$
Hence distance of object from the lens is $30\;cm$ and of object is $60\;cm$.
$\therefore$ lens should be located at $x=-10\;cm$ as shown.
Hence b is the correct answer.