Before discussing the details of simple microscope, first we briefly discuss about visual angle subtended by an object at the eye. This is because in the optical instruments like telescopes and microscopes, we are concerned with the visual angle.
An object NM is placed at some distance from the eye. This object is subtended an angle q at the eye. The length of the image q formed by the eye is proportional to angle q subtended at the eye by the object. This angle is called the visual angle.
Using the relation
Angle = Arc/Radius
q = pΘ
q is directly proportional to Θ (as p is constant)
This shows that the visual angle is directly proportioned to apparent size of the object. Optical Instruments such as telescopes and microscopes are designed to increase the visual angle. The resultant effect of this is that the image of the object formed on the retina becomes much bigger than it is when these instruments are not used to view them i.e., image formed on the retina when these instruments are used become much magnified than when they are not used.
Simple Microscope (in normal use):
The figure given above shows that two objects (of different size) MN and M'N' are subtending same visual angle Θ at the eye, thus seems to be of equal size. But in actual, objects are of different sizes and object MN is bigger in size.
The simple microscope in normal use signifies that image is formed at near point as shown in figure above. Here h is length of object viewed at near point (it means at D). Visual angles subtended are α (in radian) and α' (in radian). α' is increased angle when the simple microscope is utilized to view object given in above second figure.
As you can see, magnified image is attained that is erect and distance of image is equal to D.
Angular magnification is maximum when image is at near point of the eye.
Angular magnification in terms of visual angle is
M = α'/α..................................Eq.1
Now values of α and α' can be attained from figure
α'= h'/d and α = h/d
Therefore Eq.1 becomes
M = (h'/D)(h/D) = h'/h = (v/u)..........Eq.2
Here u is the object distance and v is the image distance.
As you know that
1/u + 1/v = 1/f
On multiplying the above Eq. by v and on rearranging the terms, we get
v/u = v/f - 1
v/u = -D/f -1
Therefore h'/h = v/u = -(D/f + 1)..........Eq.3
Therefore h'/h = v/u = (D/f + 1)
Substituting Eq.3 in Eq.2, we get expression
M = -(D/f + 1)
Numerically, magnification can be written as
M = (D/f + 1)..........Eq.4
Equation 4 provides angular magnification of the microscope in normal use and negative sign is the indication that final image is virtual. Further, it can be seen that for higher angular magnification, the lens of short focal length is required. You know that eye has tendency to focus on the image formed anywhere between near point and infinity by the simple microscope.
Simple Microscope (with Image at Infinity):
Eye has tendency to focus on the image formed anywhere between near point and infinity by the simple microscope. The simple microscope is the instrument that is utilized to see very small objects. When it is in normal use, image is formed at D (least distance of distinct vision)
The simple microscope with image formed at infinity signifies that eye should be accommodated to bring image to infinity as given in figure.
Where f is focal length of the lens. Magnification M can be stated as
M = α'/α = (h/f)/(h/D) = D/f
Thus M = D/f..........Eq.5
Equation 5 provides the angular magnification with the microscope having the single lens. Magnification can be further increased by using one or two additional lenses.
A simple microscope in normal adjustment has its magnification numerically as
M = D/f + 1
A decrease in f implies an increase in angular magnification. But in practice, it is difficult to obtain a very small f. Therefore two lenses can be used to increase angular magnification. This two lens microscope is known as the compound microscope as shown in Figure below.
The compound Microscope in normal use means that final image is formed at the near point. Details of image formation are given below.
Compound microscope basically comprises of two convex lenses of focal length f1 and f2 in which one of the lenses (of focal length f1) is objective lens and second lens (of focal length f2) is eye piece. Objective lens is placed near object being viewed while eyepiece is lens near the eye as shown in (a) figure. fo is focus of objective lens and fe is focus of eyepiece. h1 is height of the image formed by objective and at last we get image h2 by the eyepiece.
Lenses are arranged such that separation is less than f1 + f2. As such image of object formed by the objective lens is situated from second lens at the distance less than focal length of the second lens. Therefore image of first image formed by second lens should be virtual and magnified. As a result final image formed is numerous times larger than object to observer.
The formula of angular magnification for compound microscope is provided by
Therefore M = (D/f2 +1)(v/f1 - 1).............................Eq.6
From Eq.6, it can be noted that M is large for small f1 and f2. It signifies that if focal lengths of objective lens and eyepiece lens are both small, angular magnification will be high.
The angular magnification of a Telescope is defined as the ratio:
M = α'/α..............Eq.7
Where α is the visual angle subtended by the distance object at the unaided eye and α is the angle subtended at the eye by its final image when telescope is used.
The Astronomical Telescope in Normal Adjustment:
Astronomical telescope like compound microscope, consist of two lenses: objective and eyepiece. The objective is of large focal length and eyepiece is of short focal length closer to the image. f1 is the focal length of the objective while f2 is the focal length of eyepiece.
The parallel rays are collected by the objective lens O and an image h is formed. Final image is formed at infinity (∞).
In normal adjustment the two foci f1 and f2 coincides and it therefore implies that the distance between the two lenses is f1 + f2.
If α is the angle subtended by the unaided eye and α¢ is the angle subtended by the aided eye. Then, since α and α¢ are small the angular magnification of the telescope is
M = α'/α = (h/f2)/(h/f1)
M = f1/f2..............Eq.8
So from Eq.8, angular magnification is a ratio of focal length of objective to focal length of eyepiece. For high angular magnification eyepiece must have a small focal length and objective must have high focal length.
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