Question

The medium-power objective lens in a laboratory microscope has a focal length fobjective = 3.75 mm. If this lens produces a lateral magnification of -53.0, what is its "working distance"; that is, what is the distance from the object to the objective lens? What is the focal length of an eyepiece lens that will provide an overall magnification of -130? Assume student's near-point distance is N=25cm.

Answer 1

f_e = focal length of the eyepiece = ?

N = near point distance = 25 cm

L = lateral magnification = - 53

M = overall magnification = - 130

M = overall magnification = (L) (N/f_e)

inserting the values

- 130 = (- 53) (25/f_e)

f_e = 10.2 cm

d_o = distance of object

d_i = image distance

f = focal length of objective lens

using the equation

1/d_i + 1/d_o = 1/f

multiplying both side by "d_o"

d_o/d_i + 1 = d_o/f

we know that : - di/do = - 53

hence d_i/d_o = 53

1/53 + 1 = d_o/3.75

d_o = 3.82 mm