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# : If the distance of object from objective lens is 15 cm and the focal length of objective lens is 10 cm, then find magnifying power, if focal length of eyepiece is 20 cm.

Last updated date: 13th Jun 2024
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Hint:The magnifying power of any lens is given as:
$\Rightarrow m = - \dfrac{v}{u}$
But here two lenses are involved, objective and eyepiece lens. So, the magnifying power will be multiplied i.e. magnifying power of objective x magnifying power of eyepiece which is given as:
$\Rightarrow m = {m_o} \times {m_e} \\ \Rightarrow m = \dfrac{{{f_o}}}{{{u_o} - {f_o}}} \times \left( {1 + \dfrac{D}{{{f_e}}}} \right) \\$
Where,
${f_o}$ is the focal length of objective lens
${f_e}$ is the focal length of eyepiece lens
${u_o}$ is the object distance from objective lens
D is the minimum distance of clear vision i.e. 25 cm
The focal length of an eyepiece is greater than the focal length of an objective lens, so this is the case of compound microscopes.

$\Rightarrow m = \dfrac{{{f_o}}}{{{u_o} - {f_o}}} \times \left( {1 + \dfrac{D}{{{f_e}}}} \right)$
$\Rightarrow m = \dfrac{{10}}{{15 - 10}} \times \left( {1 + \dfrac{{25}}{{20}}} \right) \\ \Rightarrow m = 2\left( {\dfrac{{45}}{{20}}} \right) \\ \Rightarrow m = 4.5 \\$
$\Rightarrow m = {m_o} \times {m_e} = \dfrac{{{v_o}}}{{{u_o}}} \times \dfrac{D}{{{u_e}}}$