A thin converging lens of focal length 25 ,cm | vedclass

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Question

(D) Given: Focal length $f = 25 \,cm$. Initial image distance $v_1 = 75 \,cm$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$, we fin

Vedclass · Accepted Answer

$12.5 \,cm$ away from the lens

Vedclass · Answer

(D) Given: Focal length $f = 25 \,cm$. Initial image distance $v_1 = 75 \,cm$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$, we find the initial object distance $u_1$:$\frac{1}{25} = \frac{1}{75} - \frac{1}{u_1} \Rightarrow \frac{1}{u_1} = \frac{1}{75} - \frac{1}{25} = \frac{1-3}{75} = -\frac{2}{75}$.So, $u_1 = -37.5 \,cm$ (object is $37.5 \,cm$ in front of the lens).Now, the screen is moved $25 \,cm$ closer to the lens, so the new image distance is $v_2 = 75 - 25 = 50 \,cm$.For the new image to be sharp, we find the new object distance $u_2$:$\frac{1}{25} = \frac{1}{50} - \frac{1}{u_2} \Rightarrow \frac{1}{u_2} = \frac{1}{50} - \frac{1}{25} = \frac{1-2}{50} = -\frac{1}{50}$.So, $u_2 = -50 \,cm$ (object is $50 \,cm$ in front of the lens).The shift in the object position is $|u_2| - |u_1| = 50 \,cm - 37.5 \,cm = 12.5 \,cm$.Since the object distance from the lens must increase from $37.5 \,cm$ to $50 \,cm$, the object must be shifted $12.5 \,cm$ away from the lens.

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