Question

A compound microscope has an eyepiece of focal length \[10cm\] and an objective of focal length \[4cm\] . The magnification, if an object is kept at a distance of \[5cm\] from the objective and the final image is formed at the least distance of distinct vision \[\left( {20cm} \right)\] , is:
A. $10$
B. $11$
C. $12$
D. $13$

Answer 1

Hint:The compound microscope is a microscope that uses two lenses to enlarge the image of an object. The lens near the eyes is called eyepiece or ocular and the lens near the object is called the objective lens. The total magnification produced by the compound microscope is the product of magnification produced by the eyepiece and the objective.

The formula used:
$M = \dfrac{{{v_o}}}{{{u_o}}}\left( {1 + \dfrac{D}{{{f_e}}}} \right)$; here, \[D\] is the least distance of distinct vision, ${v_o}$is the distance of the image from the objective, ${u_o}$ is the distance of the object from the objective, and ${f_e}$ is the focal length of the eyepiece.

Complete step by step solution:
Given in the question:
${f_o} = 4cm$, ${f_e} = 10cm$, ${u_o} = 5cm$, $D = 20cm$
We have to find the magnification.
Let us first find the image distance from the objective, using the following formula for the objective.
$\dfrac{1}{{{f_o}}} = \dfrac{1}{{{v_o}}} - \dfrac{1}{{{u_o}}}$
Let us substitute the values.
$\dfrac{1}{4} = \dfrac{1}{{{v_o}}} - \dfrac{1}{{ - 5}} \Rightarrow \dfrac{1}{{{v_o}}} = \dfrac{1}{4} - \dfrac{1}{5} = \dfrac{1}{{20}}$
$ \Rightarrow {v_o} = 20cm$
Hence, the image distance from the objective is \[20cm\].
Now, let us calculate the magnification using the following formula.

Let us substitute the values.
$M = \dfrac{{20}}{5}\left( {1 + \dfrac{{20}}{{10}}} \right) \Rightarrow M = 12$
Therefore, magnification is $12$.
Hence, option (C) $12$ is correct.

Note:
Always substitute the values according to the sign convention.
Magnification is a unitless quantity.
When the final image is formed at infinity, the following formula is used to calculate the magnification.
$M = \dfrac{{{v_o}}}{{{u_o}}}\left( {\dfrac{D}{{{f_e}}}} \right)$
Here, ${v_o}$ is the image distance from the objective, ${u_o}$ is the object distance from the objective, D is the least distance of the distinct vision, and ${f_e}$ is the focal length of the eyepiece.