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In a compound microscope, let ' $u_0$ ' and ' $v_0$ ' be the object distance and image distance respectively. The objective of focal length ' $f_0$ ' magnifies a tiny object into a real, inverted image. The linear magnification of the objective is
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Verified Answer
The correct answer is:
$\frac{f_0}{f_0+u_0}$
Using lens formula:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
And definition of magnification,
$m=\frac{v}{u}$
Multiply lens formula with $u$,
$\left(\frac{u}{v}\right)-1=\frac{u}{f}$
$\left(\frac{1}{m}\right)-1=\frac{u}{f}$
$m=\frac{f}{(u+f)}$
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
And definition of magnification,
$m=\frac{v}{u}$
Multiply lens formula with $u$,
$\left(\frac{u}{v}\right)-1=\frac{u}{f}$
$\left(\frac{1}{m}\right)-1=\frac{u}{f}$
$m=\frac{f}{(u+f)}$
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