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A microscope has objective of focal $1.5\; cm$ and eye piece of focal length $2.5 \;cm$. If length of tube $25\;cm$ What must be the magnification produced.

$(a)\;100\;times \\ (b)\;120\;times \\ (c)\;140\;times \\ (d)\;75\;times $

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Length of tube $=25\;cm$
$f_e =2.5 cm$
we have $v_0 +f_e =L$
$v_0 =25 -2.5 =22.5 \;cm$
$\large\frac{1}{v_0} -\frac{1}{u_0} =\frac{1}{f_0}$
$\large\frac{1}{22.5} -\frac{1}{40} =\frac{1}{1.5}$
$u_0 = 1.6 \;cm$
magnification =$\large\frac{v_0}{u_0} \times \frac{D}{f_e}$
$\qquad= \large\frac{22.5}{1.6} \times \frac{22}{2.5}$
Hence c is the correct answer
answered Jan 22, 2014 by meena.p

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