The size of the images of an object, formed b | vedclass

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(B) For a thin lens, the magnification $m$ is given by $m = \frac{f}{f+u}$.Since the sizes of the images are equal, the magnitude of magnification mus

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$16$

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(B) For a thin lens, the magnification $m$ is given by $m = \frac{f}{f+u}$.Since the sizes of the images are equal, the magnitude of magnification must be the same, but one image is real (inverted) and the other is virtual (erect).Thus, $m_1 = -m_2$.Let the two positions be $u_1 = -8 \ cm$ and $u_2 = -24 \ cm$.Substituting these into the magnification formula: $\frac{f}{f-8} = -\frac{f}{f-24}$.Canceling $f$ from both sides (assuming $f 
eq 0$): $\frac{1}{f-8} = -\frac{1}{f-24}$.Cross-multiplying gives: $f - 24 = -(f - 8)$.$f - 24 = -f + 8$.$2f = 32$.$f = 16 \ cm$.

The size of the images of an object, formed by a thin lens, are equal when the object is placed at two different positions $8 \ cm$ and $24 \ cm$ from the lens. The focal length of the lens is . . . . . . $cm$.

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