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Two convex lenses of focal lengths $f_1$ and $f_2$ form images with magnification $m_1$ and $m_2$, when used individually for an object kept at the same distance from the lenses. Then, $f_1 / f_2$ is
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Verified Answer
The correct answer is:
$\frac{m_1\left(1-m_2\right)}{m_2\left(1-m_1\right)}$
The linear magnification of lens in terms of focal length $f$ and $u$.
$$
m=\frac{f}{f+u}
$$
For first lens,
$$
u_1=\left(\frac{1-m_1}{m_1}\right) f_1
$$
For second lens,
$$
u_2=\left(\frac{1-m_2}{m_2}\right) f_2
$$
According to the question,
$$
\begin{aligned}
u_1 & =u_2 \\
\therefore \quad \frac{\left(m_1-1\right)}{m_1} f_1 & =\frac{\left(m_2-1\right)}{m_2} f_2 \\
\frac{f_1}{f_2} & =\frac{m_1}{m_2} \cdot \frac{\left(m_2-1\right)}{\left(m_1-1\right)} \\
& =\frac{m_1\left(m_2-1\right)}{m_2\left(m_1-1\right)}
\end{aligned}
$$
$$
m=\frac{f}{f+u}
$$
For first lens,
$$
u_1=\left(\frac{1-m_1}{m_1}\right) f_1
$$
For second lens,
$$
u_2=\left(\frac{1-m_2}{m_2}\right) f_2
$$
According to the question,
$$
\begin{aligned}
u_1 & =u_2 \\
\therefore \quad \frac{\left(m_1-1\right)}{m_1} f_1 & =\frac{\left(m_2-1\right)}{m_2} f_2 \\
\frac{f_1}{f_2} & =\frac{m_1}{m_2} \cdot \frac{\left(m_2-1\right)}{\left(m_1-1\right)} \\
& =\frac{m_1\left(m_2-1\right)}{m_2\left(m_1-1\right)}
\end{aligned}
$$
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