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(A) Let the object distance be $u$ and the image distance be $v$. Since the image is real and inverted,$v = -3u$ (where $M = v/u = -3$ for real images

Vedclass · Accepted Answer

$15$

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(A) Let the object distance be $u$ and the image distance be $v$. Since the image is real and inverted,$v = -3u$ (where $M = v/u = -3$ for real images,but here magnification magnitude is $3$,so $v/u = -3$ implies $v = -3u$).Given that the distance between the object and the image is $80 \, cm$,we have $v - u = 80 \, cm$.Substituting $v = -3u$ into the equation: $-3u - u = 80 \, cm$.$-4u = 80 \, cm$,which gives $u = -20 \, cm$.Then,$v = -3(-20) = 60 \, cm$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{f} = \frac{1}{60} - \frac{1}{-20} = \frac{1}{60} + \frac{3}{60} = \frac{4}{60} = \frac{1}{15}$.Therefore,$f = 15 \, cm$.

An object is placed on the axis of a convex lens. Its image is formed $80 \, cm$ from the object. The magnification is $3$. The focal length of the lens is.....$cm$

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