A converging lens forms an image of an object | vedclass

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(C) Let the object distance be $u$ and the image distance be $v$. Given that the image is real and twice the size of the object,the magnification $m =

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half the object size

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(C) Let the object distance be $u$ and the image distance be $v$. Given that the image is real and twice the size of the object,the magnification $m = -v/u = -2$. Thus,$v = 2u$.According to the lens formula,$1/f = 1/v - 1/u$. Substituting $v = 2u$,we get $1/f = 1/(2u) - 1/u = -1/(2u)$,which implies $u = -2f$ and $v = 4f$.The distance between the object and the screen is $D = |u| + v = 2f + 4f = 6f$.When the positions of the object and the screen are interchanged,the new object distance becomes $u' = -4f$ and the new image distance becomes $v' = 2f$.The new magnification $m' = -v'/u' = -(2f)/(-4f) = 1/2$.Therefore,the new image size is half the size of the object.

$A$ converging lens forms an image of an object on a screen. The image is real and twice the size of the object. If the positions of the screen and the object are interchanged,leaving the lens in the original position,the new image size on the screen is

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