Two convex lenses of focal lengths f_1 and f_ | vedclass

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(B) The linear magnification $m$ of a lens in terms of focal length $f$ and object distance $u$ is given by $m = \frac{f}{f+u}$.For the first lens, $m

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$\frac{m_1(1-m_2)}{m_2(1-m_1)}$

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(B) The linear magnification $m$ of a lens in terms of focal length $f$ and object distance $u$ is given by $m = \frac{f}{f+u}$.For the first lens, $m_1 = \frac{f_1}{f_1+u}$, which implies $f_1 + u = \frac{f_1}{m_1}$, so $u = f_1(\frac{1}{m_1} - 1) = f_1(\frac{1-m_1}{m_1})$.For the second lens, $m_2 = \frac{f_2}{f_2+u}$, which implies $u = f_2(\frac{1-m_2}{m_2})$.Since the object distance $u$ is the same for both lenses, we have $f_1(\frac{1-m_1}{m_1}) = f_2(\frac{1-m_2}{m_2})$.Rearranging the terms to find the ratio $f_1/f_2$, we get $\frac{f_1}{f_2} = \frac{m_1(1-m_2)}{m_2(1-m_1)}$.

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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