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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Aicharya · Accepted Answer

$d=\frac{2 f}{3}$ $f=\frac{3 d}{2}=\frac{3 	imes 12}{2}=18 \mathrm{~cm}$ Equivalent focal length is $\begin{aligned} f^{\prime} & =\frac{f_1 f_2}{f_1+f_2}+\frac{f}{4}=\frac{18 	imes 18}{18+18}+\frac{18}{4} \ & =9+4.5 \ & =13.5 \mathrm{~cm}\end{aligned}$

In Ramsden eyepiece, the two planoconvex lenses each of focal length $f$ are separated by a distance $12…

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