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In a compound microscope the focal length of objective lens is $1.2 \mathrm{~cm}$ and focal length of eye piece is $3.0 \mathrm{~cm}$. When object is kept at $1.25 \mathrm{~cm}$ in front of objective, final image is formed at infinity. Magnifying power of the compound microscope should be:
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Verified Answer
The correct answer is:
200
200
Given : $f_0=1.2 \mathrm{~cm} ; f_e=3.0 \mathrm{~cm}$ $u_0=1.25 \mathrm{~cm} ; M_{\infty}=$ ?
From $\frac{1}{f_0}=\frac{1}{v_0}-\frac{1}{u_0}$
$$
\Rightarrow \quad \frac{1}{1.2}=\frac{1}{v_0}-\frac{1}{(-1.25)}
$$
$$
\begin{aligned}
&\Rightarrow \quad \frac{1}{v_0}=\frac{1}{1.2}-\frac{1}{1.25} \\
&\Rightarrow v_0=30 \mathrm{~cm}
\end{aligned}
$$
Magnification at infinity,
$$
\begin{aligned}
&M_{\infty}=-\frac{v_0}{u_0} \times \frac{D}{f_e} \\
&=\frac{30}{1.25} \times \frac{25}{3} \quad(\because D=25 \mathrm{~cm} \text { least distance }
\end{aligned}
$$
of distinct vision)
$$
=200
$$
Hence the magnifying power of the
compound microscope is 200
From $\frac{1}{f_0}=\frac{1}{v_0}-\frac{1}{u_0}$
$$
\Rightarrow \quad \frac{1}{1.2}=\frac{1}{v_0}-\frac{1}{(-1.25)}
$$
$$
\begin{aligned}
&\Rightarrow \quad \frac{1}{v_0}=\frac{1}{1.2}-\frac{1}{1.25} \\
&\Rightarrow v_0=30 \mathrm{~cm}
\end{aligned}
$$
Magnification at infinity,
$$
\begin{aligned}
&M_{\infty}=-\frac{v_0}{u_0} \times \frac{D}{f_e} \\
&=\frac{30}{1.25} \times \frac{25}{3} \quad(\because D=25 \mathrm{~cm} \text { least distance }
\end{aligned}
$$
of distinct vision)
$$
=200
$$
Hence the magnifying power of the
compound microscope is 200
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