In a compound microscope,the focal lengths of | vedclass

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(D) Given: Focal length of objective $f_{o} = 1.5 \ cm$,focal length of eyepiece $f_{e} = 6.25 \ cm$,object distance $u_{o} = -2 \ cm$,and final image

Vedclass · Accepted Answer

$11$

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(D) Given: Focal length of objective $f_{o} = 1.5 \ cm$,focal length of eyepiece $f_{e} = 6.25 \ cm$,object distance $u_{o} = -2 \ cm$,and final image distance $v_{e} = -25 \ cm$.For the objective lens,using the lens formula $\frac{1}{f_{o}} = \frac{1}{v_{o}} - \frac{1}{u_{o}}$:$\frac{1}{1.5} = \frac{1}{v_{o}} - \frac{1}{-2} \Rightarrow \frac{2}{3} = \frac{1}{v_{o}} + \frac{1}{2} \Rightarrow \frac{1}{v_{o}} = \frac{2}{3} - \frac{1}{2} = \frac{4-3}{6} = \frac{1}{6}$.Thus,$v_{o} = 6 \ cm$.For the eyepiece,using the lens formula $\frac{1}{f_{e}} = \frac{1}{v_{e}} - \frac{1}{u_{e}}$:$\frac{1}{6.25} = \frac{1}{-25} - \frac{1}{u_{e}} \Rightarrow \frac{1}{u_{e}} = -\frac{1}{25} - \frac{1}{6.25} = -\frac{1}{25} - \frac{4}{25} = -\frac{5}{25} = -\frac{1}{5}$.Thus,$u_{e} = -5 \ cm$.The distance between the two lenses (tube length) is $L = v_{o} + |u_{e}| = 6 \ cm + 5 \ cm = 11 \ cm$.

In a compound microscope,the focal lengths of two lenses are $1.5 \ cm$ and $6.25 \ cm$. An object is placed at $2 \ cm$ from the objective lens and the final image is formed at $25 \ cm$ from the eye lens. The distance between the two lenses is $.... \ cm$.

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