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When $I_{1}$ and $I_{2}$ are sizes of two images of an object of size I in displacement method of determining the focal length of a convex lens, then size of object I is given by

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${(I_{1} / I_{2})}^{1 / 2}$

${(I_{2} / I_{1})}^{1 / 2}$

${(I_{1} I_{2})}^{1 / 2}$

$I_{1} I_{2}$

answer is D.

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Detailed Solution

$m_{1} = \frac{I_{1}}{O}$

$m_{2} = \frac{I_{2}}{O}$

For lens displacement method

$m_{1} m_{2} = 1$

$\frac{I_{1} I_{2}}{O^{2}} = 1$

$O = \sqrt{I_{1} I_{2}}$

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