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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.

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Last updated date: 13th Jun 2024
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Answer
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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.

Complete step by step answer:
The origin of simple microscopes using a single eyepiece dates back to the 13th century. It is amusing to note that the origin of the compound microscope is unknown. Although we know that the first noted compound microscope came into picture in around the 1620s somewhere in Europe.The working of a compound microscope is as follows: The object is placed in front of the objective lens which makes an interim image in between the objective and eyepiece. This image is the object for eyepiece which is further refracted and a final image is formed beyond the objective eyepiece. The final image is virtual, erect and magnified.The magnification of the whole is the product of the magnification of eyepiece and objective lens which is given as:
\[\Rightarrow m = \dfrac{{{f_o}}}{{{u_o} - {f_o}}} \times \left( {1 + \dfrac{D}{{{f_e}}}} \right)\]
On substituting the data from the ques:
\[
\Rightarrow m = \dfrac{{10}}{{15 - 10}} \times \left( {1 + \dfrac{{25}}{{20}}} \right) \\
\Rightarrow m = 2\left( {\dfrac{{45}}{{20}}} \right) \\
\Rightarrow m = 4.5 \\
\]
Therefore, the magnifying power is 4.5.

Note: Here it is not given whether the instrument is a microscope or telescope. So, there are two ways to solve:
-You can calculate the image distance of the objective lens. Then, again using the lens formula, find object distance because final image distance is D (=25cm). So, magnification becomes:
\[\Rightarrow m = {m_o} \times {m_e} = \dfrac{{{v_o}}}{{{u_o}}} \times \dfrac{D}{{{u_e}}}\]
-Use directly the formula as we know for microscopes the focal length of eyepiece is greater than focal length of objective lens. It is opposite in case of telescopes.
It is advisable to use the first method if the nature of the instrument is not mentioned. The answer will be the same. Left as an exercise for students.