The distance between an object and its 3 time | vedclass

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(C) For a virtual image formed by a convex lens, the magnification $m = +3$.Since $m = \frac{v}{u}$, we have $v = 3u$.Both the object and the virtual

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$15$

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(C) For a virtual image formed by a convex lens, the magnification $m = +3$.Since $m = \frac{v}{u}$, we have $v = 3u$.Both the object and the virtual image are on the same side of the lens. Let the object distance be $u$ (where $u$ is negative, so $u = -x$) and the image distance be $v$ (where $v = -3x$).The distance between the object and the image is $|v - u| = 20 \,cm$.$|-3x - (-x)| = 20 \,cm$$|-2x| = 20 \,cm \implies 2x = 20 \,cm \implies x = 10 \,cm$.Thus, $u = -10 \,cm$ and $v = -30 \,cm$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{f} = \frac{1}{-30} - \frac{1}{-10} = \frac{-1 + 3}{30} = \frac{2}{30} = \frac{1}{15}$.Therefore, $f = 15 \,cm$.

The distance between an object and its $3$ times magnified virtual image produced by a convex lens is $20 \,cm$. The focal length of the lens used is: (in $\,cm$)

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