Question

An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25cm and an eyepiece of focal length 5cm. How will you set up the compound microscope?

Answer 1

Focal length of the objective lens, ƒ_º = 1.25cm
Focal length of the eyepiece, f_e = 5cm
Least distance of distinct vision, d = 25cm
Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microscope, m = 30
The angular magnification of the eyepiece is given by the relation: \(me = (1 + \frac{d }{ ƒ_e}) = (1 + \frac{25}{5}) = 6\)
The angular magnification of the objective lens (m_º) is related to me as: m_º m_e = m
\(m_{º} = \frac{m}{m_e} =\frac{ 30 }{ 6} = 5\)
We also have the relation:\( m_{º} = \frac{Image \space distance \space for \space the \space objective \space lens \space (v_{º}) }{ Object \space distance \space for \space the \space objective \space lens (-u_{º})}\)
\(5 =\frac{ v_{º} }{ -u_{º}} ∴ v_{º} = -5u_{º}...(1)\)
Applying the lens formula for the objective lens: \(\frac{1 }{ ƒ_{º}} = \frac{1 }{ v_{º}} - \frac{1 }{ u_{º}}\)
\(\frac{1 }{ 1.25}\) = \(\frac{1 }{ -5u_{º}} - \frac{1 }{ u_{º}}\) = \(\frac{-6 }{ 5u_{º}}\) 
∴ \(u_{º} = \frac{-6 }{ 5} \times 1.25 = -1.5cm\) 
And \( v_{º} = -5u_{º} \)
= -5 × (-1.5) = 7.5cm
The object should be placed 1.5cm away from the objective lens to obtain the desired magnification. Applying the lens formula for the eyepiece: Where,

\( \frac{1 }{ v_{e}} - \frac{1 }{ u_e} = \frac{1 }{ ƒe}\)

 v_e = Image distance for the eyepiece = -d = -25cm 

u_e = Object distance for the eyepiece
\(\frac{1 }{ u_e} = \frac{1 }{ v_e} - \frac{1 }{ ƒ_e} \)
= \(\frac{-1 }{ 25} - \frac{1 }{ 5} \)
= \(\frac{-6 }{ 25} \)
∴ u_e = -4.17cm
Seperation between the objective lens and the eyepiece = |u_e| + |v_º| = 4.17 + 7.5 = 11.67cm
Therefore, the separation between the objective lens and the eyepiece should be 11.67cm.