Question

The focal length of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively find the position of the object related to the objective in order to obtain an angular magnification of 30 in normal adjustment

Answer 1

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

Answer 2

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