Question

The focal length of the eyepiece of a certain microscope is 20.0 mm. The focal length of the objective is 5.00 mm. The distance between the objective and eyepiece is 19.7 cm. The final image formed by the eyepiece is at infinity. Treat all lenses as thin.

(a) What is the distance from the objective to the object being
viewed?

________cm

(b) What is the magnitude of the linear magnification produced by
the objective?

_________?

(c) What is the overall angular magnification of the
microscope?

__________?

Answer #1

for the eyepiece ::

f_{e} = focal length = 20 mm

d_{i} = image distance = infinite

d_{o} = object distance = ?

using the lens equation ::

**1/d _{o} + 1/d_{i} =
1/f_{e}**

1/d_{o} + 1/infinte = 1/20 mm

**d _{o} = 20 mm**

distance between the lenses = d = 19.7 cm = 197 mm

distance of image formed by the objective lens = d_{oi}
= 197 - 20 = 177 mm

focal length of objective lens = f_{o} = 5 mm

object distance from objective lens = d_{oo} = ?

using the lens equation ::

**1/d _{oo} + 1/d_{oi} =
1/f_{o}**

**1/d _{oo} + 1/177 = 1/5**

**d _{o} = 5.16 mm**

**d _{o} = 0.516 cm**

**b)**

magnification is given as ::

m = -d_{i}/d_{o} = -177/5.16 = -34.3

magnification is negative since image is inverted.

c)

overall angular magnification is given as

m = - 250 L/(f_{o} f_{e})

m = - 250 x 197 / (20 x 5)

m = -492.5

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