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(D) Let the focal length of the lens be $f$. The object distance is $u = -30\, cm$.Using the lens formula,$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we

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$20$

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(D) Let the focal length of the lens be $f$. The object distance is $u = -30\, cm$.Using the lens formula,$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we get $\frac{1}{v} = \frac{1}{f} + \frac{1}{-30} = \frac{30-f}{30f}$,so $v = \frac{30f}{30-f}$.The magnification is $m = \frac{v}{u} = \frac{30f/(30-f)}{-30} = \frac{f}{f-30}$.Since the lens is cut and separated by $h$,each half is shifted by $h/2$ from the original principal axis. The image formed by each half is shifted by $m 	imes (h/2)$ from the new principal axis of that half.The distance of each image from the original central axis is $\frac{h}{2} + |m| \frac{h}{2} = \frac{h}{2}(1 + |m|)$.The total distance $d$ between the two images is $2 	imes \frac{h}{2}(1 + |m|) = h(1 + |m|)$.From the graph,the slope is $\frac{d}{h} = \frac{6\, cm}{2\, cm} = 3$.Thus,$1 + |m| = 3$,which means $|m| = 2$.Since the images are real,the magnification $m$ must be negative,so $m = -2$.Substituting $m = \frac{f}{f-30} = -2$,we get $f = -2f + 60$,so $3f = 60$,which gives $f = 20\, cm$.

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