Physics, asked by kamal3225, 1 year ago

Compound microscope and astronomical telescope definition and magnifying power

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Answered by RakeshPateL555
6
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\huge\mathfrak\purple{simple\:microscope}

(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

\huge\mathfrak\purple{Compound\:Microscope}

(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

\huge\mathfrak\purple{Astronomical\:microscope}

(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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