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(C) Let the distance between the two parallel walls be $D$. The bulb is on the first wall and the image is formed on the second wall.For a convex lens

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only $d/2$

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(C) Let the distance between the two parallel walls be $D$. The bulb is on the first wall and the image is formed on the second wall.For a convex lens to form a real image of equal size (magnification $m = -1$),the object distance $u$ and image distance $v$ must satisfy $u = v = 2f$.The total distance between the object and the screen is $D = u + v = 2f + 2f = 4f$.In this problem,the lens is placed at a distance $d$ from the second wall,which means the image distance $v = d$.Since the image is of equal size,the object distance $u$ must also be equal to $d$ (because $u = v$ for $m = -1$).Therefore,the total distance between the walls is $D = u + v = d + d = 2d$.Using the relation $D = 4f$,we get $2d = 4f$.Solving for $f$,we find $f = 2d / 4 = d / 2$.

$A$ bulb is located on a wall. Its image of equal size is to be obtained on a parallel wall with the help of a convex lens. The lens is placed at a distance $d$ ahead of the second wall. The required focal length will be:

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