Find the magnifying power of a compound microscope whose objective and eyepiece are of focal lengths 4.0 cm and 6.0 cm respectively and the object is placed 5.0 cm beyond the objective. Assume that the final image is formed at the least distance of distinct vision (25 cm)
Question
Find the magnifying power of a compound microscope whose objective and eyepiece are of focal lengths 4.0 cm and 6.0 cm respectively and the object is placed 5.0 cm beyond the objective. Assume that the final image is formed at the least distance of distinct vision (25 cm)
Solution
The magnifying power (M) of a compound microscope is given by the formula:
M = 1 + D/f0
where: D is the near point (assumed to be 25 cm in normal vision) f0 is the focal length of the eyepiece
First, we need to find the image distance (v) for the objective lens using the lens formula:
1/v - 1/u = 1/f
where: u is the object distance f is the focal length of the lens
Given that the object is placed 5.0 cm beyond the objective, u = -5.0 cm (the negative sign indicates that the object is on the same side of the lens as the light source). The focal length of the objective lens is given as 4.0 cm.
Substituting these values into the lens formula gives:
1/v - 1/(-5) = 1/4
Solving for v gives v = 6.67 cm.
The magnification (m) of the objective lens is then given by m = -v/u = -6.67/(-5) = 1.33.
The magnification of the eyepiece is given by M = 1 + D/f0 = 1 + 25/6 = 5.17.
The total magnifying power of the compound microscope is then given by the product of the magnifications of the objective and eyepiece, which is 1.33 * 5.17 = 6.87.
Therefore, the magnifying power of the compound microscope is approximately 6.87.
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