A compound microscope consists of an objectiv | vedclass

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Question

(A) For the eyepiece:Given $v_e = -25 \ cm$ (final image at least distance of distinct vision),$f_e = 6.25 \ cm$.Using the lens formula $\frac{1}{v_e}

Vedclass · Accepted Answer

$-2.5$

Vedclass · Answer

(A) For the eyepiece:Given $v_e = -25 \ cm$ (final image at least distance of distinct vision),$f_e = 6.25 \ cm$.Using the lens formula $\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$:$\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{6.25} \implies -\frac{1}{u_e} = \frac{1}{6.25} + \frac{1}{25} = \frac{4+1}{25} = \frac{5}{25} = \frac{1}{5}$.So,$u_e = -5 \ cm$.The distance between the lenses is $L = 15 \ cm$,so the image distance for the objective lens is $v_o = L - |u_e| = 15 - 5 = 10 \ cm$.For the objective lens:$v_o = 10 \ cm$,$f_o = 2 \ cm$.Using the lens formula $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$:$\frac{1}{10} - \frac{1}{u_o} = \frac{1}{2} \implies -\frac{1}{u_o} = \frac{1}{2} - \frac{1}{10} = \frac{5-1}{10} = \frac{4}{10} = \frac{2}{5}$.Therefore,$u_o = -2.5 \ cm$.The object should be placed $2.5 \ cm$ in front of the objective lens.

Similar Questions

$A$ card sheet divided into squares each of size $1\,mm^2$ is being viewed at a distance of $9\,cm$ through a magnifying glass (a converging lens of focal length $9\,cm$) held close to the eye. What is the angular magnification (magnifying power) of the lens?

In which of the following is the final image erect?

How should people wearing spectacles work with a microscope?

The magnifying power of a simple microscope is $6$. The focal length of its lens in metres will be,if the least distance of distinct vision is $25\,cm$.

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