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Question

The focal length of a thin biconvex lens is $$20\ cm$$. When an object is moved from a distance of $$25\ cm$$ in front of it to $$50\ cm$$, the magnification of its image changes from $$m_{25}$$ to $$m_{60}$$. The ratio $$m_{25}/ m_{50}$$ is


Solution

$$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$$

$$\implies \dfrac{u}{f}=\dfrac{u}{v}-1 $$     and $$m=\dfrac{v}{u}$$

$$\implies \dfrac{u}{f}=\dfrac{1}{m}-1$$

$$\implies m=\dfrac{f}{f+u}$$

Now, $$m_{25}=\dfrac{20}{20-25}=-4$$

And, $$m_{50}=\dfrac{20}{20-50}=\dfrac{-2}{3}$$

$$\implies \dfrac{m_{25}}{m_{50}}=\dfrac{-4}{-2/3}=6$$

Physics

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