In a Ramsden eyepiece,two planoconvex lenses, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In a Ramsden eyepiece,the two planoconvex lenses have the same focal length $f$ and are separated by a distance $d = \frac{2f}{3}$.Given the separ

Vedclass · Accepted Answer

$13.5$

Vedclass · Answer

(C) In a Ramsden eyepiece,the two planoconvex lenses have the same focal length $f$ and are separated by a distance $d = \frac{2f}{3}$.Given the separation distance $d = 12 \ cm$,we can find $f$:$d = \frac{2f}{3} \implies 12 = \frac{2f}{3} \implies f = \frac{12 	imes 3}{2} = 18 \ cm$.The equivalent focal length $f_{eq}$ of a combination of two lenses with focal lengths $f_1$ and $f_2$ separated by distance $d$ is given by:$\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$.For a Ramsden eyepiece,$f_1 = f_2 = f$ and $d = \frac{2f}{3}$.Substituting these values:$\frac{1}{f_{eq}} = \frac{1}{f} + \frac{1}{f} - \frac{2f/3}{f^2} = \frac{2}{f} - \frac{2}{3f} = \frac{6-2}{3f} = \frac{4}{3f}$.Therefore,$f_{eq} = \frac{3f}{4}$.Substituting $f = 18 \ cm$:$f_{eq} = \frac{3 	imes 18}{4} = \frac{54}{4} = 13.5 \ cm$.

In a Ramsden eyepiece,two planoconvex lenses,each of focal length $f$,are separated by a distance of $12 \ cm$. The equivalent focal length (in $cm$) of the eyepiece is:

Similar Questions

$A$ compound microscope has a magnifying power of $30$. The focal length of its eyepiece is $5 \, cm$. Assuming the final image is formed at the least distance of distinct vision,the magnification produced by the objective is:

What is the magnifying power of a simple microscope of focal length $5 \ cm$,if the image is formed at the distance of distinct vision?

What are the objective and the eye-piece in an optical instrument?

The objective of a compound microscope is essentially

Vedclass Test Series

Exam Paper Generator

Online Exam Module