pls explain normal adjustment in compound microscope and astronomical telescope.
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Compound microscope Astronomical telescope Objective lens has smaller focal length, than the eyepiece Objective lens has larger focal length than the eyepiece Distance between the objective lens and the eyepiece is greater than f0+fe Distance between the objective lens and the eyepiece is equal to f0+fe It is used to see very small objects It is used to see distant astronomical objects
may be this help u
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may be this help u
i m not sure...
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(See fig 1st)
It is used for observing magnified images of objects. It is consists of a converging lens of small focal length.
Magnifying Power
(i) When final image is formed at least distance of distinct vision (D), then M=1+d/f
where, f= focal length of the lens.
(ii) When final image is formed at infinity, then M = D/f
(See fig. 2nd)
It is a combination of two convex lenses called objective lens and eye piece separated by a distance. Both lenses are of small focal lengths but fo < fe, where fo and feare focal lengths of objective lens and eye piece respectively
Magnifying Power
M = vo / uo {1 + (D/fo)
Where vo= distance of image, formed by objective lens and
uo = distance of object from the objective
(ii) When final image is formed at infinity, then
M = vo/uo . D/fe
(See fig. 3rd)
It is also a combination of two lenses, called objective lens and eye piece, separated by a distance. It is used for observing distinct images of heavenly bodies like stars, planets etc
Magnifying Power
(i) When final image is formed at least distance of distinct vision (D), then M = fo/fe {1+ (D/fe)} where foand fe are focal lengths of objective and eyepiece respectively.
Length of the telescope (L) = (fo + ue)
where, ue = distance of object from the eyepiece.
(ii) When final image is formed at infinity, then M = fo/fe
Length of the telescope (L) = fo + fe
For large magnifying power of a telescope fo should be large and feshould be small.
For large magnifying power of a microscope; fo < fe should be small.
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