An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25cm and an eyepiece of focal length 5cm. What is the length of a compound microscope to have a final image at a near point of the normal eye?
A. 11.67 cm
B. 3.23 cm
C. 7.5 cm
D. 4.17 cm

Hint: In this question, we are given the angular magnification of the compound microscope, the focal length of the eyepiece as $5cm$ and the objective as \[1.25cm\]. We need to find the length of the compound microscope, so as to set up one with the given quantities.

Complete step by step answer:
Magnifying power of a compound microscope $m={{m}_{o}}{{m}_{e}}$, where $m_0$ is the magnification of the objective and $m_e$ is the magnification of the eyepiece.
The angular magnification of the eyepiece is given by the relation ${{m}_{e}}=1+\dfrac{d}{{{f}_{e}}}$, where me is the magnification of the eyepiece of the microscope, d is the least distance of vision and fe is the focal length of the eyepiece.
The angular magnification of the objective, \[{{m}_{o}}=\dfrac{-{{v}_{o}}}{{{u}_{o}}}\], where $v_0$ is the image distance, $u_0$ is the object distance.
Focal length of the objective lens of the compound microscope, ${{f}_{0}}=1.25cm$ $.......(1)$
Focal length of the eyepiece of the compound microscope, ${{f}_{e}}=5cm$ $.......(2)$
We know that the least distance of distinct vision, $d=25cm$ $.......(3)$
Given, angular magnification of the compound microscope, $m=30x$
 So, we can understand that the magnifying power of the compound microscope as $m=30$ $......(4)$
We must note that the magnifying power of a compound microscope $m={{m}_{o}}{{m}_{e}}$, $....(5)$ where $m_0$ is the magnification of the objective and me is the magnification of the eyepiece.
So, we have to find the angular magnification $m_e$ of the eyepiece and that of the objective lens, $m_o$
The angular magnification of the eyepiece is given by the relation ${{m}_{e}}=1+\dfrac{d}{{{f}_{e}}}$ $=1+\dfrac{25}{5}=1+5=6$ $...(6)$
 The angular magnification of the objective \[{{m}_{o}}=\dfrac{m}{{{m}_{e}}}=\dfrac{30}{6}=5\] \[.......(7)\]
Also, \[{{m}_{o}}=\dfrac{-{{v}_{o}}}{{{u}_{o}}}\]
So we get the image distance \[{{v}_{o}}=-5{{u}_{o}}\] \[......(8)\]
Using the lens maker’s formula, $\dfrac{1}{{{v}_{0}}}-\dfrac{1}{{{u}_{0}}}=\dfrac{1}{{{f}_{0}}}$ \[.........(9)\]
Substituting (1) and (8) and solving the above equation, we get the values of $u_o$ = -1.5 cm and $v_o$ = 7.5 cm
Thus, the object should be placed 1.5 cm away from the objective lens to get the desired magnification.
Also, we know that the image distance for the eyepiece is \[{{v}_{e}}=-d=-25cm\] \[.......(10)\]
Again, using the lens maker’s formula, $\dfrac{1}{{{v}_{e}}}-\dfrac{1}{{{u}_{e}}}=\dfrac{1}{{{f}_{e}}}$ \[.......(11)\]
Substituting (2) and (11), we get the value of \[{{u}_{e}}=-4.17cm\].
Also, the separation between the objective lens and the eyepiece, \[{{u}_{e}}+{{v}_{o}}=11.67cm\].
Hence, the separation between the objective lens and the eyepiece must be \[11.67cm\].

So, the correct answer is “Option A”.

Note:
We must note that the magnifying power (m) is negative, the image seen in a microscope is always inverted, which is upside down and left turned right. For large magnifying power $f_o$ and $f_e$ both have to be small. Also, $f_o$ is considered smaller than that of $f_e$, so that the field of view may get increased.