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

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