Given: Assume the near point (of the eye) is 25 cm. The distance between eyepiece and objec- tive lens in a certain compound microscope is 1 = 25.8 cm. The focal length of the eyepiece is fe = 1.26 cm, and that of the objective lens is fo = 0.177 cm. What is the overall magnification of the microscope?
Given: Assume the near point (of the eye) is 25 cm. The distance between eyepiece and objec- tive lens in a certain compound microscope is 1 = 25.8 cm. The focal length of the eyepiece is fe = 1.26 cm, and that of the objective lens is fo = 0.177 cm. What is the overall magnification of the microscope?
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25 cm.
The distance between eyepiece and objec-
tive lens in a certain compound microscope is
1 = 25.8 cm. The focal length of the eyepiece
is fe = 1.26 cm, and that of the objective lens
is fo = 0.177 cm.
What is the overall magnification of the
microscope?
Caution: a negative quantity this is. Use
the approximation l – fe l and object dis-
tance d, is approximately the focal length fo.
%3D"
Transcribed Image Text:Given: Assume the near point (of the eye) is
25 cm.
The distance between eyepiece and objec-
tive lens in a certain compound microscope is
1 = 25.8 cm. The focal length of the eyepiece
is fe = 1.26 cm, and that of the objective lens
is fo = 0.177 cm.
What is the overall magnification of the
microscope?
Caution: a negative quantity this is. Use
the approximation l – fe l and object dis-
tance d, is approximately the focal length fo.
%3D
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