Question

A compound microscope has objective and eyepiece lenses of focal lengths 0.80 cm and 4.0 cm, respectively. If the microscope length is 15 cm, what is the approximate magnification of the microscope?

Answer 1

M = Mo x Me

Mo is the magnification of the objective lens
Me is the magnification of the eyepiece lens

Mo = D/fo

where D is the distance between objective back focal plane (OBF) and the focal plane of the eyepiece (called the tube length)

OBF = fe(d-fo) / [d-(fo+fe)] = (4)(15-0.8) / (15-4.8) = 5.568 cm

D = OBF + fe = 9.568 cm

Mo = D/fo = 11.96
Me = 25/fe - 1 = 5.25

M = Mo x Me = 62.79

Answer 2

M = Mo x Me

Mo is the magnification of the objective lens
Me is the magnification of the eyepiece lens

Mo = D/fo

where D is the distance between objective back focal plane (OBF) and the focal plane of the eyepiece (called the tube length)

OBF = fe(d-fo) / [d-(fo+fe)] = (4)(15-0.8) / (15-4.8) = 5.568 cm

D = OBF + fe = 9.568 cm

Mo = D/fo = 11.96
Me = 25/fe + 1 = 7.25

M = Mo x Me = 86.71

Answer 3

f_o = 0.80cm

f_e = 4.0 cm

L = 15 cm

m = M_Im_e = (L/f_o)(25cm/f_e)

m = (15/0.80cm)(25cm/4.0cm) = 117.2 $\approx$ 120