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(D) Let the distance of the candle from the lens be $u_1$ and $u_2$ in the two cases. The distance of the wall from the lens is $v_1 = 10 \,cm$ and $v

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$10 \,cm$ towards the wall.

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(D) Let the distance of the candle from the lens be $u_1$ and $u_2$ in the two cases. The distance of the wall from the lens is $v_1 = 10 \,cm$ and $v_2 = 30 \,cm$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:Case $1$: $\frac{1}{10} - \frac{1}{-u_1} = \frac{1}{f} \implies \frac{1}{10} + \frac{1}{u_1} = \frac{1}{f}$Case $2$: $\frac{1}{30} - \frac{1}{-u_2} = \frac{1}{f} \implies \frac{1}{30} + \frac{1}{u_2} = \frac{1}{f}$Equating the two: $\frac{1}{10} + \frac{1}{u_1} = \frac{1}{30} + \frac{1}{u_2} \implies \frac{1}{u_1} - \frac{1}{u_2} = \frac{1}{30} - \frac{1}{10} = -\frac{2}{30} = -\frac{1}{15}$.Also,the distance between the candle and the wall is $D = u_1 + 10 = u_2 + 30$,so $u_2 = u_1 - 20$.Substituting $u_2$: $\frac{1}{u_1} - \frac{1}{u_1 - 20} = -\frac{1}{15} \implies \frac{u_1 - 20 - u_1}{u_1(u_1 - 20)} = -\frac{1}{15} \implies \frac{-20}{u_1^2 - 20u_1} = -\frac{1}{15} \implies u_1^2 - 20u_1 - 300 = 0$.Solving the quadratic: $(u_1 - 30)(u_1 + 10) = 0$. Since $u_1 > 0$,$u_1 = 30 \,cm$.Then $f = \frac{1}{\frac{1}{10} + \frac{1}{30}} = \frac{30}{4} = 7.5 \,cm$.For unit magnification $(m = -1)$,the object distance must be $2f$ and image distance must be $2f$. Thus,$u = 15 \,cm$ and $v = 15 \,cm$. The total distance between the candle and the wall is $D = u + v = 30 \,cm$.Initially,the candle was at $u_1 = 30 \,cm$ from the lens,and the lens was $10 \,cm$ from the wall,so the candle was $40 \,cm$ from the wall.To make the distance $30 \,cm$,the candle must be moved $10 \,cm$ towards the wall.

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