An object placed at a distance of $16 c m$ from a convex lens produces an image of magnification $m (m > 1)$ . If the object is moved towards the lens by $8 c m$ then again an image of magnification $m$ is obtained. Find the numerical value of focal length of the lens.

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Solution

Since, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{u}{v} - 1 = \frac{1}{f}$
$m = \frac{v}{u} = \frac{f}{f + u}$
given, $u_{1} = - 16 m_{1} = m$
$u_{2} = - 8 m_{2} = - m$
$\frac{f}{f - 16} = \frac{- f}{f - 8}$
$f^{2} - 8 f = {- f}^{2} + 16 f$
${2 f}^{2} - 24 f = 0$
$f^{2} - 12 f = 0$
So $f = 12 c m$

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An object placed at a distance of 16cm from a convex lens produces an image of magnification m(m>1). If the object is moved towards the lens by 8cm then again an image of magnification m is obtained. Find the numerical value of focal length of the lens.

Since, 1v−1u=1f uv−1=1f m=vu=ff+ugiven, u1=−16m1=m u2=−8m2=−m ff−16=−ff−8 f2−8f=−f2+16f 2f2−24f=0 f2−12f=0So f=12cm

An object placed at a distance of $16 c m$ from a convex lens produces an image of magnification $m (m > 1)$ . If the object is moved towards the lens by $8 c m$ then again an image of magnification $m$ is obtained. Find the numerical value of focal length of the lens.