Question

A person with a normal near point(25cm)using a compound microscope with an objective of focal length 8.0mm and an eyepiece of focal length 2.5cm can bring an object placed at 9.0mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.

Answer 1

Focal length of the objective lens,f₀=8mm=0.8cm
Focal length of the eyepiece,fe=2.5cm
Object distance for the objective lens,u_o=-9.0mm=-0.9cm
Least distance of the eyepiece,ve=-d=-25cm
Object distance for the eyepiece=ue
Using the lens formula, we can obtain the value of ue as:\(\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}\)
\(\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}\)

=\(\frac{1}{-25}-\frac{1}{2.5}\)

=-\(1-\frac{10}{25}\)

=\(-\frac{11}{25}\)

 ∴\(u_e=-\frac{25}{11}\)=-2.27cm
We can also obtain the value of the image distance for the objective lens(vº)using the lens formula.
\(\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}\)
=\(\frac{1}{0.8}-\frac{1}{0.9}\)

=\(\frac{0.9-0.8}{0.72}\)

=\(\frac{0.1}{0.72}\)
The distance between the objective lens and the eyepiece=|u_e|+vº=2.27+7.2=9.47cm
The magnifying power of the microscope is calculated as 

\(\frac{v_e}{|u_e|}(1+\frac{d}{f_e})\)=\(\frac{7.2}{0.9}(1+\frac{25}{2.5})\)=8(1+10)=88
Hence, the magnifying power of the microscope is 88.