A eyepiece of a microscope have 1.25 cm diameter. What is the magnification
Answers
Answer:Focal length of the objective lense (f0bj) = 1.25 cm
Focal length of the eyepiece (feye) = 5 cm
Distance of vision (d) = 25 cm
Angular magnification of the compound microscope = 30X
Total magnification power of the compound microscope (m)= 30
Angular magnification of the eyepiece
meye = (1+d/feye) = (1+25/5) = 6
Angular magnification of the object (mobj) is related to meye
Thus, mobj = m/ meye = 30/6 = 5
Also, mobj = image distance for the objective lens (vobj)/ Objective distance for the objective lens (-uobj)
5 = vobj /-uobj
vobj = -uobj x 5 (i)
Using lens formula
1/ vobj - 1/ uobj = 1/fobj
1/1.25= 1/-5 uobj - 1/ uobj
1/1.25= -6/ uobj
uobj = -6/5 x 1.25 = -1.5 cm
vobj = -5 uobj = -5 x (-1.5) = 7.5 cm
The object should be placed 1.5 cm away from the objective lens.
Using lens formula
1/ veye - 1/ ueye = 1/feye
Where, veye = image distance for the eyepiece = -d = -25 cm
ueye = Object distance for the eyepiece
1/ ueye = 1/ veye - 1/feye = -1/25 - 1/5 = -6/25
ueye = 4.17 cm
Separation between the objective lens and eyepiece = ueye + vobj = 4.17 + 7.5 = 11.67 cm