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Magnification produced by astronomical telescope for normal adjustment is 20 and length of telescope is $1.05\;m$ The magnification when image is formed at least distance of district vision $(D= 25 cm)$ is

$(a)\;6 \\ (b)\;10 \\ (c)\;14 \\ (d)\;24 $

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$M_1 =20 =\large\frac{f_0}{f_e}$----(1)
Also (1) and (2) $f_0 +f_e =1.05 \;m$ ----(2)
Solving (1) and (2) we get,
$f_0 =100\;cm$ and $ f_e =5 \;cm$
Then $M_2 =f_0 \bigg[ \large\frac{1}{D}+\frac{1}{f_e}\bigg]$
$\qquad= 100 \bigg [\large\frac{1}{25}+\frac{1}{5} \bigg]$
$\qquad= 100 \times \large\frac{6}{25}$
$\qquad= 24$
Hence d is the correct answer.
answered Jan 22, 2014 by meena.p

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