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Q.
An object has image thrice of its original size when kept
at $8\, cm$ and $16 \,cm$ from a convex lens. Focal length of the
lens is
Ray Optics and Optical Instruments
Solution:
$m=\pm 3$, using $m=\frac{f}{f+u}$
For virtual image $3=\frac{f}{f-8}$ $\ldots $(i)
For real image $-3=\frac{f}{f-16}$ ....(ii)
Solving (i) and (ii) we get $f=12 \,cm$