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Three thin lenses are combined by placing them in contact with each other to get more magnification in an optical instrument. Each lens has a focal length of $3 \mathrm{~cm}$. If the least distance of distinct vision is taken as $25 \mathrm{~cm}$, the total magnification of the lens combination in normal adjustment is
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Verified Answer
The correct answer is:
26
Combined focal length $\left(\frac{1}{F}\right)=\frac{1}{F_1}+\frac{1}{F_2}+\frac{1}{F_3}$
$$
\begin{aligned}
& \Rightarrow \quad \frac{1}{F}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3} \\
& \Rightarrow \quad F=1 \mathrm{~cm} \\
&
\end{aligned}
$$
Magnification of the lens combination in normal adjustment is
$$
M=1+\frac{D}{F}=1+\frac{25}{1}=26
$$
$$
\begin{aligned}
& \Rightarrow \quad \frac{1}{F}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3} \\
& \Rightarrow \quad F=1 \mathrm{~cm} \\
&
\end{aligned}
$$
Magnification of the lens combination in normal adjustment is
$$
M=1+\frac{D}{F}=1+\frac{25}{1}=26
$$
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