Question

The focal length of the objective and eye lens of a compound microscope are 2cm and 6.25 cm respectively. The distance between the lenses is 15 cm. How far the objective lens will the object be kept so as to obtain the final imageat the near point of the eye. Also calculate its magnifying power

Answer 1

 length of the objective lens (f1) = 2.0 cm

Focal length of the eyepiece (f2) = 6.25 cm

Distance between the objective lens and the eyepiece (d)= 15 cm

(a) Least distance of distinct vision, (d1) = 25 cm

Therefore, Image distance for the eyepiece (v2) = – 25 cm

Object distance for the eyepiece =  u2

According to the lens formula,

1/ v2 - 1/u2 = 1/ f2

1/u2 = 1/ v2 -1/ f2  = 1/-25 - 1/6.25 = -1-4/ 25 = -5/25

 u2 = -5 cm

Image distance for the objective lens, (v1) =- d + u2= 15-5 = 10 cm

Object distance for the objective lens = u2

According to the lens formula,

1/ v1 - 1/u1 = 1/f1

1/u1 = 1/ v1 - 1/f1

        = 1/10 - 1/2 = 1-5/10 = -4/10

u1 = -2.5 cm

Magnitude of the object distance (u1) = 2.5 cm