A compound microscope is an optical instrument used for observing highly magnified images of tiny objects.
Construction: a compound microscope consists of two converging lenses (or lens systems); an objective lens O of very small focal length and short aperture and eye piece E of moderate focal length and large aperture. The two lenses are held co-axially at the free ends of two coaxial tubes, at a suitable fixed distance from each other. The distance of the objective lens from the object can be adjusted by rack and pinion arrangement.
AB is a tiny object held perpendicular to the common principle axis, in front of the objective lens beyond its principle focus F0. A real, inverted and enlarged image A’B’ of this object is formed by the objective lens. Now A’B’ acts as an object for the eye lens, whose position is so adjusted that A’B’ lies between optical centre C2 of eye lens E and its principle focus Fe. A virtual and magnified image A’’B’’ is formed by the eye lens. This image is erect w.r.t. A’B’ but inverted w.r.t. AB. the final image A’’B’’ is seen by the eye held close to eye lens E. the adjustments are so made that A”B” is at the least distance of distinct vision (d) from the eye i.e. C2B” = d.
Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the object, when both the final image and the object are situated at the least distance of distinct vision from the eye.
In fig. C2B” = d. imagine the object AB to be shifted to A1B” so that it is at a distance d from the eye. If∠A”C2B” =
And ∠A1”C2B” =, then by definition,
Magnifying power, m = / (1)
For small angles expressed in radians,
tan θ = θ
Therefore, ≈ tan and ≈ tan
From (5), m= tan/ tan (2)
in Δ A”B”C2 tan = A”B”/C2B”
in A”B”C2, tan = A1B”/C2B” = AB/C2B”
Putting in (6), we get
m = A”B”/C2B” × C2B”/AB
= A”B”/AB = A”B”/A’B’ × A’B’/AB
m = me × m0, where me = A”B”/A’B’ (3)
= magnification produced by eye lens, and m0 = A’B’/AB
= magnification produced by objective lens.
Now, me = magnification produced by eye lens (E) is
me = (1 + d)/ƒe (proved in case of simple microscope)
Where d is C2B” = least distance of distinct vision ƒe is focal length of the eye lens.
m0 = A’B’/AB
= distance of image A’B’ from C1/distance of object AB from C1
= C1B’/C1B = v0/-u0
Putting these values in (3), we get
m = v0/-u0 [(1 + d)/ƒe] = v0/|u0| [(1 + d)/ƒe] (4)
As the object AB lies very close to F0, the focus of lens O, therefore,
u0 = C1B ≈ C1F0 = ƒe
= focal length of objective lens.
As A’B’ is formed very close to eye lens, whose focal length is also short, therefore,
v0 = C1B’ ≈ C1C2 = L
= length of microscope tube.
Putting in (4), we get
m = L/-ƒ0 (1 + d/ƒe) = L/|ƒ0| (1 + ƒe) (5)
Discussion: (i) as magnifying power (m) is negative, the image seen in a microscope is always inverted i.e. upside down and left turned right.
(ii) As intermediate image is between the two lenses, a cross wire (or a measuring scale) can be used.
(iii) If final image in a microscope were at infinity (normal setting), eqn. (5) would reduce to
(iv) For large magnifying power, ƒ0 and ƒe both have to be small. Also ƒ0 is taken to the smaller than ƒeso that field of view may be increased.
(v) As aperture of both the lenses in a microscope is small, the defects of images particularly, spherical aberration is minimized.
(vi) In a compound microscope of good quality objective lens and eye piece, both are formed of a number of lenses. This is done to avoid defects of images formed by a single lens.
(vii) Various other factors such as illumination of the object contribute to the quality and visibility of the image.
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