(a) Draw a labeled ray diagram showing the formation of a final image by a compound microscope at least distance of distinct vision. (b) The total magnification produced by a compound microscope is 20. The magnification produced by the eyepiece is 5. The microscope is focused on a certain object. The distance between the objective and eyepiece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal length of the objective and the eyepiece.

where, AB = object, A'B' = image formed by objective and A"B" = image formed by eyepiece
fo= focal length of objective,
uo= object distance from objective
vo= image distance from objective
D= distance of least distinct vision
l= length of the microscope
(b). For the least distance of clear vision, the total magnification is given by:
m = −L/fo(1+D/fe) = mo.me.. (1)
n2
where, L is the separation between the eyepiece and the objective
fo is the focal length of the objective
fe is the focal length of the eyepiece
D is the least distance for clear vision
Also, the given magnification for the eyepiece:
me=5=(1+D/fe)
5=1+20/fe => fe=5cm
1/ue = −1/vo = −ue/vo
ue = −vo
L=vo+|ue|
14=vo+u => vo=10 => uo=vo/m = 10/4 = 2.5
Hence
f0 = −uovo/5 = 4×2.5/5 = 2cm