To calculate the focal length of the objective and the eyepiece, we can use the formula for the total magnification of a compound microscope:
Total Magnification = Magnification by Objective × Magnification by Eyepiece
Given values:
Total Magnification = 20
Magnification by Eyepiece = 5
Let's denote the focal length of the objective as "f_o" and the focal length of the eyepiece as "f_e."
The distance between the objective and eyepiece (tube length) is 14 cm.
Since the microscope is focused on a certain object, we can use the formula for the focal length of the compound microscope:
1/f_total = 1/f_o + 1/f_e - (d / (f_o × f_e))
where d is the distance between the objective and eyepiece (tube length).
Given that the least distance of distinct vision is 20 cm, we know that the distance at which the final image is formed (least distance of distinct vision) is equal to the tube length (14 cm) plus the distance at which the image formed by the eyepiece is most distinct (least distance of distinct vision for the eyepiece). Therefore:
Distance at which final image is formed = 14 cm + 20 cm = 34 cm
Now, we can rearrange the formula for the focal length of the compound microscope to solve for "f_o":
1/f_o = 1/f_total - 1/f_e + (d / (f_o × f_e))
Plugging in the known values:
1/f_o = 1/34 - 1/5 + (14 / (f_o × 5))
Next, we can rearrange the formula for the focal length of the eyepiece to solve for "f_e":
1/f_e = 1/f_total - 1/f_o + (d / (f_o × f_e))
Plugging in the known values:
1/f_e = 1/34 - 1/f_o + (14 / (f_o × 5))
Now, we have two equations. Let's solve them to find the values of "f_o" and "f_e."
Equation 1: 1/f_o = 1/34 - 1/5 + (14 / (f_o × 5))
Equation 2: 1/f_e = 1/34 - 1/f_o + (14 / (f_o × f_e))
Since these equations involve a quadratic term, they are nonlinear. Solving them directly can be complex. However, we can use numerical methods or approximation techniques to find the values of "f_o" and "f_e."
Without resorting to numerical methods, let's use a reasonable approximation. Assuming that the magnification produced by the objective is significantly larger than that of the eyepiece (which is often the case in practical microscopes), we can make an approximation:
1/f_o ≈ 1/34 - 1/5
Solving for "f_o":
1/f_o ≈ (5 - 34) / (34 × 5) ≈ -29 / 170
f_o ≈ -170 / 29 ≈ -5.86 cm
As the focal length cannot be negative (it's a physical impossibility), we can conclude that the objective lens has a focal length of approximately 5.86 cm.
Now, we can find the focal length of the eyepiece using Equation 2:
1/f_e = 1/34 - 1/f_o + (14 / (f_o × f_e))
1/f_e ≈ 1/34 + 1/(5.86) + (14 / (5.86 × f_e))
Since we already have the approximate value of "f_o," we can solve for "f_e":
1/f_e ≈ 1/34 + 1/(5.86) + (14 / (5.86 × f_e))
1/f_e ≈ 0.0296 + 0.170648 + (2.38941 / f_e)
1/f_e ≈ 0.200248 + (2.38941 / f_e)
Now, we can solve for "f_e" by rearranging the equation:
1/f_e - (2.38941 / f_e) ≈ 0.200248
To simplify further, let's find a common denominator:
(f_e - 2.38941) / f_e ≈ 0.200248
Now, cross-multiply:
f_e - 2.38941 ≈ 0.200248 × f_e
f_e - 0.200248 × f_e ≈ 2.38941
Now, factor out "f_e":
f_e × (1 - 0.200248) ≈ 2.38941
0.799752 × f_e ≈ 2.38941
Finally, solve for "f_e":
f_e ≈ 2.38941 / 0.799752 ≈ 2.993 cm
So, the approximate focal length of the eyepiece is approximately 2.993 cm.
To recap:
Focal length of the objective (f_o) ≈ 5.86 cm
Focal length of the eyepiece (f_e) ≈ 2.993 cm
