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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Aicharya · Accepted Answer

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ or $ \frac{u}{v}-1=\frac{u}{f} $ $\begin{array}{rlrl} & 	ext { or } & \frac{u}{v} & =\left(\frac{u+f}{f}ight) \ 	herefore & m=\frac{v}{u} & =\left(\frac{f}{u+f}ight) \ \frac{m_{25}}{m_{50}} & =\frac{\left(\frac{20}{-25+20}ight)}{\left(\frac{20}{-50+20}ight)}=6\end{array}$ $	herefore$ answer is 6 . !

The focal length of a thin biconvex lens is $20 \mathrm{~cm}$. When an object is moved from a distance of…

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