A point object is placed at a distance of 12 | vedclass

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(A) Given: Object distance $u = -12 \, cm$,Image distance $v = 18 \, cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substitutin

Vedclass · Accepted Answer

Focal length = $7.2 \, cm$,Yes,magnified

Vedclass · Answer

(A) Given: Object distance $u = -12 \, cm$,Image distance $v = 18 \, cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting the values: $\frac{1}{f} = \frac{1}{18} - \frac{1}{-12} = \frac{1}{18} + \frac{1}{12}$.Taking the $LCM$ of $18$ and $12$ as $36$: $\frac{1}{f} = \frac{2+3}{36} = \frac{5}{36}$.Therefore,$f = \frac{36}{5} = 7.2 \, cm$.The magnification $m$ is given by $m = \frac{v}{u} = \frac{18}{-12} = -1.5$.Since the magnitude of magnification $|m| = 1.5 > 1$,the image is magnified.

$A$ point object is placed at a distance of $12 \, cm$ from a convex lens on its principal axis. Its image is formed on the other side of the lens at a distance of $18 \, cm$ from the lens. Find the focal length of the lens. Is the image magnified? Justify your answer.

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