Question

A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image at(a)the least distance of distinct vision (25cm), and(b)at infinity? What is the magnifying power of the microscope in each case?

Answer 1

The focal length of the objective lens,f₁=2.0cm
The focal length of the eyepiece,f₂=6.25cm
Distance between the objective lens and the eyepiece,d=15cm
(a)Least distance of distinct vision,d'=25cm
Image distance for the eyepiece,v₂=-25cm
Object distance for the eyepiece=u₂
According to the lens formula,we have the relation:

\(\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2}\)
\(\frac{1}{u_2}=\frac{1}{v_2}-\frac{1}{f_2}\)

=\(\frac{1}{-25}-\frac{1}{6.25}\)

=\(-\frac{1-4}{25}=-\frac{5}{25}\)
Image distance for the objective lens,v₁=d+u₂=15-5=10cm
Object distance for the objective lens=u1
According to the lens formula, we have the relation:\(\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}\)
\(\frac{1}{u_1}=\frac{1}{v_1}-\frac{1}{f_1}\)
=\(\frac{1}{10}-\frac{1}{2}\)

=\(\frac{1-5}{10}\)

=\(-\frac{4}{10}\)

∴u=-2.5cm
The magnitude of the object distance,|u₁|=2.5cm
The magnifying power of a compound microscope is given by the relation:m=\(\frac{v_1}{|u_1|}\)(\(1+\frac{d'}{f_2}\))=\(\frac{10}{2.5}(1+\frac{25}{6.25})\)=4(1+4)=20
Hence, the magnifying power of a microscope is 20.
(b)The final image is formed at infinity.
Image distance for the eyepiece,v2=∞
Object distance for the eyepiece=u₂
According to the lens formula, we have the relation:\(\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f}\)
\(\frac{1}{\infty}-\frac{1}{u_2}=\frac{1}{6.25}\) ∴u₂=-6.25cm
Image distance for the objective lens,v₁=d+u₂=15-6.25=8.75cm
Object distance for the objective lens=u₁
According to the lens formula, we have the relation:\(\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}\)
\(\frac{1}{u_1}=\frac{1}{v_1}-\frac{1}{f_1}\)=\(\frac{1}{8.75}-\frac{1}{2.0}\)=2-\(\frac{8.75}{17.5}\) ∴u₁=-17.5/6.75=-2.59cm
The magnitude of the object distance,|u₁|=2.59cm
The magnifying power of a compound microscope is given by the relation:m=\(\frac{v_1}{|u_1|}(\frac{d'}{|u_2|})\)=\(\frac{8.75}{2.59}\)×\(\frac{25}{6.25}\)=13.51
Hence, the magnifying power of the microscope is 13.51.