Lens Separation and Magnifying Power of a Compound Microscope
A person with normal near point (25 cm) uses a compound microscope with:
Objective focal length = 8.0 mm
Eyepiece focal length = 2.5 cm
Object placed at 9.0 mm from objective.
Step 1: Objective Lens Calculation
The objective forms the intermediate image 72 mm from it.
Step 2: Eyepiece Calculation
Near point = 25 cm = 250 mm
Eyepiece focal length = 25 mm
Step 3: Separation Between Lenses
Step 4: Magnifying Power
Final Answers
Thus, the microscope lenses must be separated by approximately 9.5 cm, and the magnifying power of the microscope is about 88.
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Frequently Asked Questions
What is the role of the objective lens in a compound microscope?
The objective lens forms a real, inverted, and magnified intermediate image of the object. This image is then further magnified by the eyepiece.
How is the separation between the objective and eyepiece calculated?
The separation equals the distance of the intermediate image formed by the objective plus the object distance required for the eyepiece to form the final image at the near point.
Why is the final image formed at the near point?
For maximum magnification in normal adjustment, the final image is formed at the least distance of distinct vision (25 cm), allowing comfortable viewing.
How is the magnifying power of a compound microscope calculated?
Magnifying power equals the product of the magnification of the objective and the magnification of the eyepiece. That is, M = mo × me.
What is the magnifying power obtained in this problem?
The magnifying power calculated for this microscope setup is approximately 88.