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1. The focal lengths of the objective and the eyepiece of a compound microscope are $2 \mathrm{~cm}$ and $3 \mathrm{~cm}$ respectively and the distance between them is $15 \mathrm{~cm}$. The final image formed by the eyepiece is at infinity. The distances of the object and the image produced by the object from the objective lens are respectively
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$2.4 \mathrm{~cm}, 12 \mathrm{~cm}$
Given $\mathrm{f}_0=2 \mathrm{~cm}, \mathrm{f}_{\mathrm{e}}=3 \mathrm{~cm}$ and $\mathrm{f}_0+\mathrm{f}_{\mathrm{e}}=15 \mathrm{~cm}$. Final image is formed at infinity, so object for eye-piece i.e., image formed by objective lens at $\mathrm{f}=3 \mathrm{~cm}$ from it
$\therefore$ Image distance formed by the objective lens from the objective lens $=15-3=12 \mathrm{~cm}$.
Using lens formula, $\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \Rightarrow \frac{1}{2}=\frac{1}{12}-\frac{1}{\mathrm{x}}$
$\therefore \mathrm{x}=-2.4 \mathrm{~cm}$.
$\therefore$ Image distance formed by the objective lens from the objective lens $=15-3=12 \mathrm{~cm}$.
Using lens formula, $\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \Rightarrow \frac{1}{2}=\frac{1}{12}-\frac{1}{\mathrm{x}}$
$\therefore \mathrm{x}=-2.4 \mathrm{~cm}$.
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