In a Ramsden eyepiece, two plano-convex lense | vedclass

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(C) For a Ramsden eyepiece, the separation distance $d$ between the two plano-convex lenses is given by $d = \frac{2f}{3}$.Given that $d = 12 \ cm$, w

Vedclass · Accepted Answer

$13.5$

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(C) For a Ramsden eyepiece, the separation distance $d$ between the two plano-convex lenses is given by $d = \frac{2f}{3}$.Given that $d = 12 \ cm$, we can find the focal length $f$ of each lens:$12 = \frac{2f}{3} \implies f = \frac{12 	imes 3}{2} = 18 \ cm$.The equivalent focal length $f_{eq}$ of a Ramsden eyepiece is given by the formula:$f_{eq} = \frac{f_1 f_2}{f_1 + f_2 - d}$.Substituting $f_1 = f_2 = f$ and $d = \frac{2f}{3}$:$f_{eq} = \frac{f^2}{2f - \frac{2f}{3}} = \frac{f^2}{\frac{4f}{3}} = \frac{3f}{4}$.Substituting $f = 18 \ cm$:$f_{eq} = \frac{3 	imes 18}{4} = \frac{54}{4} = 13.5 \ cm$.

In a Ramsden eyepiece, two plano-convex lenses, each of focal length $f$, are separated by a distance of $d = 2f/3$. If the separation distance is $12 \ cm$, what is the equivalent focal length (in $cm$) of the eyepiece?

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