Question

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

• A. $\frac{m_{1}\left(1-m_{1}\right)}{m_{2}\left(1-m_{2}\right)}$
• B. $\frac{m_{1}\left(1-m_{2}\right)}{m_{2}\left(1-m_{1}\right)}$
• C. $\frac{m_{2}\left(1-m_{1}\right)}{m_{1}\left(1-m_{2}\right)}$
• D. $\frac{m_{2}\left(1-m_{2}\right)}{m_{1}\left(1-m_{1}\right)}$

Answer 1

Answer: (B)

The linear magnification of lens in terms of focal length $f$ and $u$.
$m=\frac{f}{f+u}$
For first lens,
$u_{1}=\left(\frac{1-m_{1}}{m_{1}}\right) t_{1}$
For second lens,
$u_{2}=\left(\frac{1-m_{2}}{m_{2}}\right) f_{2}$
According to the question,
$u_{1} =u_{2} $
$ \therefore \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)} $
$=\frac{m_{1}\left(m_{2}-1\right)}{m_{2}\left(m_{1}-1\right)}$