A thin converging lens of focal length f=25 , | vedclass

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(C) According to the first condition,for a converging lens,the lens formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Given $f = 25 \,cm$ and $v =

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$12.5$

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(C) According to the first condition,for a converging lens,the lens formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Given $f = 25 \,cm$ and $v = 75 \,cm$,we have:$\frac{1}{25} = \frac{1}{75} - \frac{1}{u}$$\frac{1}{u} = \frac{1}{75} - \frac{1}{25} = \frac{1-3}{75} = -\frac{2}{75}$$u = -37.5 \,cm$.So,the object is placed at a distance of $37.5 \,cm$ from the lens.According to the second condition,the screen is moved closer to the lens by $25 \,cm$,so the new image distance is $v_1 = 75 \,cm - 25 \,cm = 50 \,cm$.Using the lens formula again:$\frac{1}{25} = \frac{1}{50} - \frac{1}{u_1}$$\frac{1}{u_1} = \frac{1}{50} - \frac{1}{25} = \frac{1-2}{50} = -\frac{1}{50}$$u_1 = -50 \,cm$.So,the new object distance is $50 \,cm$.The distance through which the object has to be shifted is $\Delta u = |u_1| - |u| = 50 \,cm - 37.5 \,cm = 12.5 \,cm$.

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