In displacement method, the distance between object and screen is 96 cm. The ratio of length of two images formed by a convex lens placed between them is 4.84.

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In the displacement method the two magnifications are reciprocal, so the image-size ratio $m_{1}^{2} = 4.84$ gives $m_{1} = 2.2 = \frac{11}{5}$ - (A) correct.

$m_{1} = \frac{D + d}{D - d} \Rightarrow 2.2 = \frac{96 + d}{96 - d} \Rightarrow d = 36 cm$ - (B) correct.

$f = \frac{D ^{2} - d ^{2}}{4 D} = \frac{9216 - 1296}{384} = 20.625 cm \neq = 22.5$ - (C) wrong.

Lens positions are $\frac{D \mp d}{2} = 30, 66 cm$ ; the shorter image forms at $v = 30 cm$ - (D) correct. Answer: A, B, D.

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