Answer:
According to the question, we have ā
Focal length of the given objective lens is f1=2.0 cm
Focal length of the given eyepiece is f2=6.25 cm
Distance between the eyepiece and the objective lens is d =15 cm
(a) Least distance of distinct vision is denoted by dā = 25 cm
Image distance for the eyepiece,
$v_{2}$= -25cm
Object distance for the eyepiece= is represented by u2
Using the lens formula, we have the relation:
$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$
$\frac{1}{u_{2}}=\frac{1}{v_{2}}-\frac{1}{f_{2}}$
$\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}=\frac{-5}{25}$
$u_{2}=-5cm$
Now, the image distance for the objective lens becomes ā
$v_{1}=d+u_{2}=15 ā 5=10 cm$
Object distance for the objective lens is denoted by $u_{1}$
$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$
$\frac{1}{u_{1}}=\frac{1}{v_{1}}-\frac{1}{f_{1}}$
$\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=\frac{-4}{10}$
$u_{1}=-2.5cm$
Now for the Magnifying power ā
$m=\frac{v_{0}}{|u_{0}|}(1+\frac{d}{f_{e}}) $
$=\frac{10}{2.5}(1+\frac{25}{6.25}) $
$=20$
Hence, the magnifying power of the microscope is 20.
(b) The final Image formed is at infinity.
Therefore image distance for the eyepiece is
$v_{2}=\infty$
Object distance for the eyepiece is denoted by $ u_{2}$
With respect to the lens formula, we have the relation:
$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$
$\frac{1}{\infty}-\frac{1}{u_{2}}=\frac{1}{6.25}$
$u_{2}=-6.25cm$
Now, image distance for the objective lens becomes ā
$v_{1} = d + u_{2}=15 -6.25 = 8.75 cm$
Object distance for the objective lens=u1
From the lens formula, we have ā
$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$
$\frac{1}{u_{1}}=\frac{1}{v_{1}}-\frac{1}{f_{1}}$
$\frac{1}{8.75}-\frac{1}{2.0}=\frac{2-8.75}{17.5}$
$u_{1}=-\frac{17.5}{6.75}=-2.59cm$
Magnitude of the object distance, $|u_{1}|$=2.59 cm
The magnifying power of a compound microscope is given by the relation ā
m=$\frac{v_{1}}{|u_{1}|}(\frac{dā}{|u_{2}|})$
=$\frac{8.75}{2.59}\times\frac{25}{6.25}$
=13.51
Therefore, 13.51 is the magnifying power of the microscope.