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A plane convex lens of refractive index $1.5$ and radius of curvature $30 \mathrm{~cm}$ is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of the size of the object?
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The correct answer is:
$20 \mathrm{~cm}$
$20 \mathrm{~cm}$
$$
\begin{aligned}
& \frac{1}{F}=\frac{2}{f_1}+\frac{1}{f_m} \\
& \text { and } \frac{1}{f_1}=(1.5-1)\left(\frac{1}{\infty}-\frac{1}{-30}\right)=\frac{1}{60} \\
& \text { and } f_m=15 \mathrm{~cm} . \\
& \therefore F=10 \mathrm{~cm} .
\end{aligned}
$$
Object should be placed at $20 \mathrm{~cm}$ from the lens.
\begin{aligned}
& \frac{1}{F}=\frac{2}{f_1}+\frac{1}{f_m} \\
& \text { and } \frac{1}{f_1}=(1.5-1)\left(\frac{1}{\infty}-\frac{1}{-30}\right)=\frac{1}{60} \\
& \text { and } f_m=15 \mathrm{~cm} . \\
& \therefore F=10 \mathrm{~cm} .
\end{aligned}
$$
Object should be placed at $20 \mathrm{~cm}$ from the lens.
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