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

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(C) According to the first condition:$f = 25 \ cm, v = 75 \ cm$Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{25} = \frac{

Vedclass · Accepted Answer

$12.5$

Vedclass · Answer

(C) According to the first condition:$f = 25 \ cm, v = 75 \ cm$Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\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$According to the second condition,the screen is moved closer by $25 \ cm$,so the new image distance is $v_1 = 75 - 25 = 50 \ cm$.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$The distance through which the object has to be shifted is:$\Delta u = |u_1| - |u| = 50 \ cm - 37.5 \ cm = 12.5 \ cm$.

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