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In Ramsden eyepiece, the two planoconvex lenses each of focal length $f$ are separated by a distance $12 \mathrm{~cm}$. The equivalent focal length (in $\mathrm{cm}$ ) of the eyepiece is
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The correct answer is:
$13.5$
$d=\frac{2 f}{3}$
$f=\frac{3 d}{2}=\frac{3 \times 12}{2}=18 \mathrm{~cm}$
Equivalent focal length is
$\begin{aligned} f^{\prime} & =\frac{f_1 f_2}{f_1+f_2}+\frac{f}{4}=\frac{18 \times 18}{18+18}+\frac{18}{4} \\ & =9+4.5 \\ & =13.5 \mathrm{~cm}\end{aligned}$
$f=\frac{3 d}{2}=\frac{3 \times 12}{2}=18 \mathrm{~cm}$
Equivalent focal length is
$\begin{aligned} f^{\prime} & =\frac{f_1 f_2}{f_1+f_2}+\frac{f}{4}=\frac{18 \times 18}{18+18}+\frac{18}{4} \\ & =9+4.5 \\ & =13.5 \mathrm{~cm}\end{aligned}$
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