Question

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

B) A compound microscope consists of an objective lens of focal length $2.0\text{cm}$ and an eyepiece of focal length $6.25\text{cm}$ separated by a distance of $15\text{cm}$. How far from the objective should an object be placed in order to obtain the final image at infinity? What is the magnifying power of the microscope in this case?

Answer 1

A) Step 1: Draw diagram of compound microscope.

Step 2: Find the object distance for the eyepiece.
Given: Focal length of the objective lens, $f_0=2.0\text{cm}$
Focal length of the eyepiece, $f_e=6.25\text{cm}$
Distance between the objective lens and the eyepiece, $d=15\text{cm}$
Using lens formula for the eyepiece,
$\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}$
$\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}$
$\frac{1}{u_e}=\frac{1}{-25}-\frac{1}{6.25}$$\frac{1}{u_e}=\frac{-1-4}{25}=\frac{-1}{5}\text{cm}\Rightarrow u_e=-5\text{cm}$

Step 3: Find the image distance for the objective lens.
Now the image distance for the objective lens is,
$v_o=d-|u_e|$
$v_o=15-5$
$v_o=10\text{cm}$

Step 4: Find the object distance for the objective lens.
Using lens formula for the eyepiece,
$\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_0}$
$\frac{1}{u_o}=\frac{1}{v_o}-\frac{1}{f_0}$
$\frac{1}{u_o}=\frac{1}{10}-\frac{1}{2}$
$\frac{1}{u_o}=\frac{1-5}{10}=\frac{-4}{10}$
$u_o=-2.5\text{cm}$

Step5:Find the Magnification power of microscope.
Magnification of microscope at the least distance of distinct vision is given by,
$m=\frac{v_o}{|u_o|}(1+\frac{D}{f_e})$
$m=\frac{10}{2.5}(1+\frac{25}{6.25})$
$m=4\times 5$
$m=20$
Final Answer: $u_e=-5\text{cm}$
$v_o=10\text{cm}$
$u_o=-2.t\text{cm}$
$m=20$.

B) Step 1: Draw diagram of compound microscope.

Step 2: Find the object distance for the eyepiece.
Given:
Focal length of the objective lens, $f_0=2.0\text{cm}$
Focal length of the eyepiece, $f_e=6.25\text{cm}$
Distance between the objective lens and the eyepiece, $d=15\text{cm}$
Using lens formula for the eyepiece,
$\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}$
$\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}$
$\frac{1}{u_e}=\frac{1}{\infty }-\frac{1}{6.25}$
$u_e=-6.25\text{cm}$

Step 3: Find the image distance for the objective lens.
Now the image distance for the objective lens is,
$v_o=d-|u_e|$
$v_o=15-6.25$
$v_o=8.75\text{cm}$

Step 4: Find the object distance for the objective lens.
Using lens formula for the objective,
$\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o}$
$\frac{1}{u_o}=\frac{1}{v_o}-\frac{1}{f_o}$
$\frac{1}{u_o}=\frac{1}{8.75}-\frac{1}{2}$
$\frac{1}{u_o}=\frac{2-8.75}{17.5}=\frac{-6.75}{17.5}$
$u_o=-2.59\text{cm}$

Step5: Find the Magnification power of microscope.
Magnification of microscope at the least distance of distinct vision is given by,
$m=\frac{v_o}{|u_o|}\left(\frac{D}{f_e}\right)$
$m=\frac{8.75}{2.59}\left(\frac{25}{6.25}\right)$
$m=13.51$
Final Answer:
$u_e=-2.59\text{cm}$
$m=13.51$.