Question

The focal length of the eyepiece of a certain microscope is 18.0 mm. The focal length of the objective is 9.00 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

a)

Since the image formed from the eyepiece is at infinity, the
image formed by the objective lens (which is used as the object for
the eyepiece) must be at located at the focal point of the
eyepiece."

So, the distance from the objective to the image is:

19.7 cm - 18.0 mm = 197 mm - 18.0 mm = 179 mm

formula 1/s + 1/s' = 1/f, so, for the objective:

1/s + 1/179 = 1/9

1/s = 1/9 - 1/179

1/s = 0.111- 0.005=0.105

s = 9.486 mm= 0.9486 cm

Ans- distance from the objective to the object being viewed is 0.9486 cm

b)

m=-s'/s

Linear magnification= - 179/9.486

= - 18.869

Ans- the magnitude of the linear magnification produced by the objective is -18.869

c)

Angular magnification of the microscope= 25s'/f1f2

= 25x 179/ 18x 9

= 4475/ 162

= 27.62

Ans- the overall angular magnification of the microscope is
27.62

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