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  • The focal lengths of the objective and eye-piece of a microscope are <img alt="1 \mathrm{~cm} \: \& \: 5$ $\mathrm{cm}" src="https://entrancecorner.oncodecogs.com/gif.latex?1%20%5Cmathrm%7B%7Ec

The focal lengths of the objective and eye-piece of a microscope are 1 \mathrm{~cm} \: \& \: 5$ $\mathrm{cm} respectively. If the magnifying power for the normal adjustment is -45 , the length of the tube is ( Least distance of clear vision =25 \mathrm{~cm} )

Option: 1

23 \mathrm{~cm}


Option: 2

25 \mathrm{~cm}


Option: 3

15 \mathrm{~cm}


Option: 4

13 \mathrm{~cm}


Answers (1)

For a microscope,

\mathrm{m}=\frac{\mathrm{v}_{0}}{\mathrm{u}_{0}} \frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}=-45

\Rightarrow \frac{\mathrm{v}_{0}}{\mathrm{u}_{0}}=-9
\text { also } \frac{1}{\mathrm{v}_{0}}-\frac{1}{\mathrm{u}_{0}}=\frac{1}{\mathrm{f}_{0}}

\Rightarrow \mathrm{v}_{0}=10 \mathrm{~cm}.

\therefore \text { Length of tube }=\mathrm{v}_{0}+\mathrm{f}_{\mathrm{e}}=15 \mathrm{~cm}.

Posted by

Kshitij

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