Question

The length of a compound microscope is $14$ cm and its magnifying power when the final image is formed at a near point is $25$.If the focal length of the eyepiece is $5$ cm.The distance of object from the objective and the focal length of objective lens are :
A. $\dfrac{59}{25}\,cm,\dfrac{69}{31}\,cm$
B. $\dfrac{59}{31}\,cm,\dfrac{69}{25}\,cm$
C. $3\,cm,2\,cm$
D. $4\,cm,1\,cm$

Answer 1

Compound microscope is used to form a real image and the eyepiece forms an enlarged virtual image that can be viewed by the observer. The equation of magnifying power includes the focal length of both eyepiece and the objective, the distance between the two lenses. From the formula of magnifying power, we can easily calculate the focal length of the objective.Ray diagram of compound microscope so that you can get the idea

Formula used:
$\dfrac{1}{{{u}_{0}}}+\dfrac{1}{{{v}_{0}}}=\dfrac{1}{{{f}_{0}}}$
$m=\dfrac{{{v}_{0}}}{{{u}_{0}}}\left( \dfrac{D}{{{f}_{e}}} \right)$

Complete step by step answer:
Given, focal length of eyepiece = ${{f}_{e}}=5\,cm$
Length of the microscope = L = $14\,cm$
Magnifying power = m =$25$

In the above ray diagram, ${{V}_{0}}$ is the distance of the image formed due to the objective lens and${{u}_{0}}$is the distance due to the object for the objective lens.From the figure, we can clearly say that,
${{v}_{0}}+{{f}_{e}}=L$
$\Rightarrow {{V}_{0}}=L-{{f}_{e}}$
$\Rightarrow {{v}_{0}}=14-5=9cm$
Magnifying power, $m=\dfrac{{{v}_{0}}}{{{u}_{0}}}\left( \dfrac{D}{{{f}_{e}}} \right)$
Substituting the values, we get
${{u}_{0}}=\dfrac{9}{25}\times \dfrac{25}{5} \\\ \ \therefore {{u}_{0}}=\dfrac{9}{5}=1.8cm\approx 2cm \\\$
To calculate the focal length of the objective lens, we use lens formula $\dfrac{1}{{{u}_{0}}}+\dfrac{1}{{{v}_{0}}}=\dfrac{1}{{{f}_{0}}}$
$\Rightarrow \dfrac{1}{\dfrac{9}{5}}+\dfrac{1}{9}=\dfrac{1}{{{f}_{e}}} \\\ \Rightarrow \dfrac{5}{9}+\dfrac{1}{9}=\dfrac{1}{{{f}_{0}}} \\\ \Rightarrow \dfrac{6}{9}=\dfrac{1}{{{f}_{0}}} \\\ \therefore \dfrac{1}{{{f}_{0}}}=3\,cm \\\$
Hence, we get ${{f}_{0}}=3\,cm$ and ${{u}_{0}}=2\,cm$

Hence, the correct answer is option C.

Note: Microscopes are used to magnify the objects. The compound microscope has two convex lenses, one is objective and has a very small focal length with a short aperture and other is eyepiece, eyepiece is of moderate focal length and large aperture.
Magnifying power, M = ${{m}_{e}}\times {{m}_{o}}$
Where, ${{m}_{e}}$= magnifying power of eye lens or eyepiece and ${{m}_{o}}$=magnifying power of objective lens.