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When the distance between the object and the screen is more than 4 \mathrm{f}, we can obtain the image of the object on the screen for the two positions of a convex lens of focal length \mathrm{f}. It is called displacement method. In one case the image is magnified. If \mathrm{I_{1}} and \mathrm{I_{2}} be the sizes of the two images, then the size of the object is:

Option: 1

\left(I_{1}+I_{2}\right) / 2


Option: 2

I_{1}-I_{2}


Option: 3

\sqrt{ }\left(I_{1} I_{2}\right)


Option: 4

\sqrt{ }\left(I_{1} / I_{2}\right)


Answers (1)

best_answer

\text { If } \mathrm{I}_{1}=\mathrm{m} \mathrm{I}_{0} \text { where } \mathrm{m} \text { is magnification and } \mathrm{I}_{0} \text { is actual size }

\mathrm{I}_{2}=\mathrm{I}_{0} / \mathrm{m}

\therefore \mathrm{I}_{0}=\sqrt{\mathrm{I}_{1} \mathrm{I}_{2}}.

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chirag

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