Question

In a compound microscope, the focal length of the objective is 3.71 cm and that of the eyepiece is 6.60 cm. The distance between the lenses is 17.6 cm. (a) What is the angular magnification of the microscope if the person using it has a near point of 32.1 cm? (b) If, as normal, the first image lies just inside the focal point of the eyepiece, as in the drawing, how far is the object from the objective? (c) What is the magnification (not the angular magnification) of the objective?

Answer 1

(a)

f_o = focal length of objective = 3.71 cm

f_e = focal length of eyepiece = 6.60 cm

L = distance between the lenses = 17.6 cm

N = near point = 32.1 cm

m = angular magnification

Angular magnification is given as

m = - (L - f_e) N/ (f_o f_e)

inserting the values

m = - (17.6 - 6.60) (32.1)/ ((3.71) (6.60))

m = - 14.4

b)

di = image distance

do = object distance

Image distance is given as

d_i = L - f_e

d_i = 17.6 - 6.60

d_i = 11 cm

Using lens equation

1/d_i + 1/d_o = 1/f_o

1/11 + 1/d_o = 1/3.71

d_o = 5.6 cm

magnification is given as

m = - d_i/d_o = - 11/5.6 = - 1.96