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The focal length of a thin biconvex lens is $20 \mathrm{~cm}$. When an object is moved from a distance of $25 \mathrm{~cm}$ in front of it to $50 \mathrm{~cm}$, the magnification of its image changes from $m_{25}$ to $m_{50}$. The ratio $\frac{m_{25}}{m_{50}}$ is
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Verified Answer
The correct answer is:
6
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ or
$$
\frac{u}{v}-1=\frac{u}{f}
$$
$\begin{array}{rlrl} & \text { or } & \frac{u}{v} & =\left(\frac{u+f}{f}\right) \\ \therefore & m=\frac{v}{u} & =\left(\frac{f}{u+f}\right) \\ \frac{m_{25}}{m_{50}} & =\frac{\left(\frac{20}{-25+20}\right)}{\left(\frac{20}{-50+20}\right)}=6\end{array}$
$\therefore$ answer is 6 .
$$
\frac{u}{v}-1=\frac{u}{f}
$$
$\begin{array}{rlrl} & \text { or } & \frac{u}{v} & =\left(\frac{u+f}{f}\right) \\ \therefore & m=\frac{v}{u} & =\left(\frac{f}{u+f}\right) \\ \frac{m_{25}}{m_{50}} & =\frac{\left(\frac{20}{-25+20}\right)}{\left(\frac{20}{-50+20}\right)}=6\end{array}$
$\therefore$ answer is 6 .
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