Question

A person with near point 25cm using a compound microscope having focal length 8mm and 2.5 cm can bring an object placed at 9mm from the objective in sharp focus.what is the separation between the lenses?calculate the magnifying power of the compound microscope.

Answer 1

Focal length of the objective lense (f_0bj) = 8 mm = 0.8 cm

Focal length of the eyepiece (f_eye) = 2.5 cm

Object distance for the objective lens (U_obj) = -.9 mm = -0.9 cm

Distance of vision (d) = 25 cm

Image distance for the eyepiece (v_eye) = -d = -25 cm

Object distance for the eyepiece (u_eye) = ?

Using lens formula

1/ v_eye - 1/ u_eye = 1/f_eye

1/ u_eye =1/ v_eye -1/f_eye

1/ u_eye = 1/-25 - 1/2,5 = -1-10/25 = -11/25

u_eye = -25/11 = -2.27 cm

Value of the image distance for objective lens (f_0bj) can be obtained by using lens formula

1/ v_obj - 1/u_obj = 1/f_obj

1/ v_obj =1/f_obj + 1/ u_obj = 1/0.8 - 1/0.9 = 0.9-0.8/0.72 = 0.1/0.72

v_obj = 7.2 cm

So, sum of the distance for objective lens and the eyepiece = u_eye + v_obj = 2.27 + 7.2 = 9.47 cm

The magnifying power of the microscope is = v_obj (1+ d/ f_obj)/u_obj

= 7.2(1 + 25/2.5)/0.9 = 1(1+10) = 88