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A microscope consists of an objective of focal length $1.9 \mathrm{~cm}$ and eye piece of focal length $5 \mathrm{~cm}$. The two lenses are kept at a distance of $10.5 \mathrm{~cm}$. If the image is to be formed at the least distance of distinct vision, the distance at which the object is to be placed before the objective is (least distance of distinct vision is $25 \mathrm{~cm}$ )
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Verified Answer
The correct answer is:
$2.7 \mathrm{~cm}$
For eye piece
$$
\begin{array}{cc}
& V_e=-25 \mathrm{~cm}, \quad f_e=5 \mathrm{~cm} \\
\Rightarrow & \frac{1}{-25}-\frac{1}{u_e}=\frac{1}{5} \\
\Rightarrow \quad & u_e=-\frac{25}{6} \mathrm{~cm} \\
& v_0=L-\left|u_e\right|=10.5-\frac{25}{6}=\frac{38}{6} \mathrm{~cm}
\end{array}
$$
For objective
$$
\begin{array}{ll}
& \frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0} \\
& \frac{1}{38 / 6}-\frac{1}{u_0}=\frac{1}{1.9} \\
\Rightarrow \quad & \frac{1}{u_0}=\frac{6}{38}-\frac{1}{1.9} \\
& u_0=2.7 \mathrm{~cm}
\end{array}
$$
$$
\begin{array}{cc}
& V_e=-25 \mathrm{~cm}, \quad f_e=5 \mathrm{~cm} \\
\Rightarrow & \frac{1}{-25}-\frac{1}{u_e}=\frac{1}{5} \\
\Rightarrow \quad & u_e=-\frac{25}{6} \mathrm{~cm} \\
& v_0=L-\left|u_e\right|=10.5-\frac{25}{6}=\frac{38}{6} \mathrm{~cm}
\end{array}
$$
For objective
$$
\begin{array}{ll}
& \frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0} \\
& \frac{1}{38 / 6}-\frac{1}{u_0}=\frac{1}{1.9} \\
\Rightarrow \quad & \frac{1}{u_0}=\frac{6}{38}-\frac{1}{1.9} \\
& u_0=2.7 \mathrm{~cm}
\end{array}
$$
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