A thin converging lens of focal length $25 \mathrm{~cm}$ forms a sharp image of an object on a screen placed at a distance of $75 \mathrm{~cm}$ from the lens. Later the screen is moved closer to the lens by a distance $25 \mathrm{~cm}$. The distance through which the object is to be shifted…

Question

A thin converging lens of focal length $25 \mathrm{~cm}$ forms a sharp image of an object on a screen placed at a distance of $75 \mathrm{~cm}$ from the lens. Later the screen is moved closer to the lens by a distance $25 \mathrm{~cm}$. The distance through which the object is to be shifted so that its image on the screen is sharp again is

Aicharya · Accepted Answer

According to the question,

Given, focal length, $f=25 \mathrm{~cm}$ and distance between image of an object and screen, $v=75 \mathrm{~cm}$ Now, By lens formula, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ [ $\because$ Because screen is moved closer to the lens.] $\frac{1}{u}=\frac{1}{f}-\frac{1}{v}$ $\begin{aligned} \frac{1}{u} & =\frac{1}{25}-\frac{1}{75} \\ \Rightarrow \quad u & =\frac{75 \times 25}{50}=\frac{75}{2} \mathrm{~cm} \end{aligned}$ When the screen shift upto by $25 \mathrm{~cm}$, then the screen will be at $2 f$. $\therefore$ For to get sharp image, object has to be at $2 f$. So, the distances is $v-u=f$. $50-\frac{75}{2}=\frac{25}{2}=12.5 \mathrm{~cm}$ $\left[\begin{array}{l} \text {at } 2 f \\ v=50 \mathrm{~cm} \end{array}\right]$ This is the distance through which the object shifted from the lens.

A thin converging lens of focal length \(25 \mathrm{~cm}\) forms a sharp image of an object on a screen…

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