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  4. An object is placed at a distance of 1.5 m from a screen and a convex lens is interposed between them. The magnification produced is 4. The focal length of the lens is

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Q. An object is placed at a distance of $1.5\, m$ from a screen and a convex lens is interposed between them. The magnification produced is $4$. The focal length of the lens is

Ray Optics and Optical Instruments

Solution:

Here, $m = \frac{v}{u} = -4 $ or $u = \frac{-v}{4}$
Also, $|u|+|v| = 1.5$
$\frac{v}{4} + v = 1.5 $
or $v = 1.2\,m$ and
$u = \frac{-1.2}{4} = -0.3\,m$
$\therefore f = \frac{uv}{u-v} $
$= \frac{-03 \times 1.2}{-0.3 - 1.2} = 0.24\,m$