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(D) Let $h_o$ be the height of the object,$h_1 = 9 \, cm$ be the height of the first image,and $h_2 = 4 \, cm$ be the height of the second image.Using

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All of the above

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(D) Let $h_o$ be the height of the object,$h_1 = 9 \, cm$ be the height of the first image,and $h_2 = 4 \, cm$ be the height of the second image.Using the property of displacement method,$h_o = \sqrt{h_1 	imes h_2} = \sqrt{9 	imes 4} = \sqrt{36} = 6 \, cm$.For the first position,magnification $m_1 = \frac{h_1}{h_o} = \frac{9}{6} = 1.5$. Since $m = \frac{v}{u}$,we have $v = 1.5u$.Given the total distance $D = u + v = 90 \, cm$,substituting $v = 1.5u$ gives $u + 1.5u = 90 \implies 2.5u = 90 \implies u = 36 \, cm$.Thus,$v = 90 - 36 = 54 \, cm$.Using the lens formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ (with sign convention $\frac{1}{f} = \frac{1}{v} - \frac{1}{-u}$),we get $\frac{1}{f} = \frac{1}{54} + \frac{1}{36} = \frac{2+3}{108} = \frac{5}{108}$.$f = \frac{108}{5} = 21.6 \, cm$.Since all statements are correct,the answer is $D$.

$A$ convex lens forms an image of an object on a screen. The height of the image is $9 \, cm$. The lens is now displaced until an image is again obtained on the screen. The height of this image is $4 \, cm$. The distance between the object and the screen is $90 \, cm$.

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