Optical instruments
An optical instrument is a device which is constructed by a suitable combination of mirrors, prisms and lenses. The principle of working of an optical instrument is based on the law of reflection and refraction of light.
The common type of optical instruments are :
(i) Projection instruments. These are used to project on the screen a real, inverted and magnified image of an opaque or transparent object so as to be viewed by a large audience. The object is, however, so fitted that its image is seen in erect form.
An eye, a photographic camera, a projection lantern, an episcope, an epidiascope, an overhead projector, a film projector, etc., are examples of projection instruments.
(ii) Microscopes. These are used to see very small objects in magnified form which otherwise cannot be seen distinctly when placed close to the naked eye. Examples : a simple microscope and a compound microscope.
(iii) Telescopes. These are used to see astronomical and distant objects in magnified form which otherwise cannot be seen clearly with the naked eye. Examples : an astronomical telescope, a terrestrial telescope, a Galilean telescope, a reflecting telescope, etc.
Astronomical Telescope
It is an optical instrument used to have clear and detailed view of heavenly objects. I'm simple form, it consists of two convex lenses separated some distance apart and mounted in two metal tubes so as to have a common principal axis. The lens O which faces the object is known as objective and has a large aperture and a large focal length 'fo'. The lens E which faces the eye is known as eyepiece and has a small aperture and a smaller focal length 'fe'. The tube carrying eyepiece can be slide into the tube carrying objective by means of rack and pinion arrangement.
A parallel beam of light coming from a distant object suffers refraction through objective and produces a real, inverted and diminished image at a distance 'fo' from O₁. This image then acts as an object for the eyepiece E and so the final magnified image is obtained after refraction through the eyepiece.
An astronomical telescope can be used in the following two adjustments.
(i) Normal adjustment. An astronomical telescope is said to be in normal adjustment if the final image formed after refraction through both the lenses is situated at infinity. The course of rays, in such an adjustment is shown in Fig. 1.
A₁B₁ is the image of distant object situated at infinity (not shown in fig 1). If A₁B₁ happens to be situated at the principal focus of the lens E, the rays will emerge in parallel direction and a final virtual but magnified image is formed at infinity.
Fig. 1. Astronomical telescope in normal setting. |
Magnification 'M' in normal adjustment is defined as the ratio of the angle ፀi subtended by the final image at the eye as seen through the telescope to the angle ፀo subtended by the object at the unaided eye when both the image and the object are situated at infinity.
Thus, M = ፀi/ፀo = tan ፀi/tan ፀo
(∵ ㄥፀo and ㄥፀi are small)
Now, in triangle A₁O₁B₁,
tan ፀo = A₁B₁/O₁B₁
and in triangle A₁O₂B₁,
tan ፀi = A₁B₁/O₂B₁
∴ M = A₁B₁/O₂B₁ × O₁B₁/A₁B₁
= O₁B₁/O₂B₁
But O₁B₁ = fo = focal length of the objective and O₂B₁ = fe = focal length of eyepiece.
M = fo/fe
Thus, the magnification of astronomical telescope in normal adjustment is equal to ratio of the focal length of objective to the focal length of eyepiece. The distance between the two lenses in this case will be (fo + fe). The strain on the eye will be least if the object is viewed for a long period.
Definition of M in terms of entrance and exit pupil
Fig. 2 shows the co-axial lens system of objective and eye lens of astronomical telescope for a beam of light coming parallel to principal axis Do and De are the diameters of exit pupil and entrance pupil, respectively.
Fig. 2. Entrance and exit pupil of astronomical telescope. |
Triangles A₁O₁I and B₂O₂I are similar
∴ O₁I/O₂I = A₁O₁/B₂O₂
Since O₁I/O₂I = fo/fe = M
∴ M = A₁O₁/B₂O₂ = Do/De
= diameter of objective/diameter of eye lens
So, magnification of astronomical telescope in normal adjustment can also be defined as the ratio between diameter of objective to the diameter of eye lens.
(ii) When the final image is formed at the distance of distinct vision. The course of rays in such an adjustment is shown in (Fig. 3). A parall beam of light coming from an object situated at infinity (not shown in fig. 3) gets refracted through the objective Of and produces a real, inverted and diminished image A₁B₁ at a distance fo from O₁. If A₁B₁ happens to be within the focal length 'fe' from the eyepiece, a final virtual but magnified image A₂B₂ is observed. The position of eyepiece E is so adjusted that final image is obtained at a distance of distinct vision D from the eye.
Fig. 3. Astronomical telescope in setting for distance of distinct vision. |
Magnification in this case is defined as the ratio of the angle 'ፀi' subtended at the eye by the final image formed at the distance of distinct vision to the angle 'ፀo' subtended at the unaided eye by the object situated at infinity.
Thus, M = ፀi/ፀo = tan ፀi/tan ፀo (∵ ㄥs ፀi and ፀo are small)
Now, in triangle A₁B₁O₁,
tan ፀo = A₁B₁/O₁B₁
and in triangle A₁B₁O₂,
tan ፀi = A₁B₁/O₂B₁
M = A₁B₁/O₂B₁ × O₁B₁/A₁B₁
= O₁B₁/O₂B₁
= fo/-u ... (1)
where u is the distance of A₁B₁ from the lens E. Considering refraction through lens E,
1/fe = 1/v - 1/u
Here, v = -D, - (1/u) = 1/fe + 1/D
= fe + D/fe × D
We get u = - (fe × D/fe + D)
Substituting for u in equation (1), we get
M = fo (fe + D/fe × D)
or M = fo/fe (fe + D/D) ... (2)
Out of the two adjustments discussed above the second adjustment gives a higher magnification since the factor (fe + D/D) > 1. Equations (1) and (2) also show that for greater magnification.
(i) the focal length 'fo' of the objective should be large, and
(ii) the focal length 'fe' of the eyepiece should be small.
The objective of an astronomical telescope should have a large aperture so as to allow a large number of rays to all on it. In that case, the intensity of the image is large. This becomes even more essential in the case of heavenly bodies which are situated at large distances. A powerful astronomical telescope placed at Yerke's observatory at Lake Geneva has an objective of aperture about one meter and focal length about 18 meter.
Terrestrial Telescope
An astronomical telescope is used to view heavenly objects since the inversion of their image does not produce any complication whole viewing earthly objects we would prefer to have their images erect. Therefore, astronomical telescope is not suitable in such cases. By using an additional convex lens O in between O₁ and O₂ of an astronomical telescope, we can have the final erect image. The lens O is called erecting lens, while the improved version of the telescope is called terrestrial telescope.
Rays from the distant object get refracted through the objective O₁, giving a real inverted image A₁B₁ [Fig. 4].
Fig. 4. Normal setting of terrestrial telescope. |
The erecting lens O' is so adjusted that its distance from A₁B₁ is equal to twice its (erecting lens) focal length. An image A₂B₂ having small size as that of A₁B₁, inverted w.r.t. A₁B₁and hence erect w.r.t. the object is obtained at a distance 2f on other side of O. A₂B₂ acts as an object for lens at O₂ and final erect and magnified image is obtained after refraction through O₂. If the distance O₂B₂ is equal to focal length 'fe' of the eye lens O₂, final image is formed at infinity [Fig. 4] and the telescope is said to be an in normal adjustment. If the distance O₂B₂ is less than 'fe' them corresponding to a certain value of this distance, a virtual and magnified image is obtained at the distance of distinct vision as shown in [Fig. 5].
Fig. 5. Terrestrial telescope. |
Since the sizes of A₂B₂ and A₁B₁ are same, introduction of the erecting lens O has not produced any change in its magnifying power but has helped in getting the final image erect only. It may also be noted that the use of erecting lens O results in a slight increase (equal to four times the focal length of erecting lens) in the length of the tune of telescope.
Length of the tube,
= O₁B₁ + B₁O + OB₂ + O₂B₂
= fo + 4f + fe
where f = focal length of the erecting lens.
Galileo's Telescope
Instead of using a combination of two lens O₁ and O₂ for getting an erect image, Galileo used only one concave lenses to get the final erect image.
Parallel beam of incident rays from infinity are focussed by the objective O₁. An inverted image A₁B₁ would have been formed after refraction through O₁. Before the rays meet at A₁, a concave lens (at O₂) intercept them [Fig. 6].
Case (i) Final image at infinity
If the position of concave lens is such that
O₂B₁ = fe (focal length of eye lens)
Rays shall enter the eye in a parallel direction [Fig. 6] and appear to come from infinity. Magnification,
M = ፀi/ፀo = tan ፀi/tan ፀo
= (A₁B₁/O₂B₁)/(A₁B₁/B₁O₁)
= B₁O₁/O₂B₁
= fo/fe ... (3)
or M = Focal length of objective/focal length of eye lens
Fig. 6. Final image at infinity. |
Case (ii) Final image at distance of distinct vision
In normal adjustment, distance between O₁ and O₂ is equal to (fo - fe). Let the lens O₂ be moved a small distance towards O₁, so that O₁O₂ is less than (fo - fe). For one particular position of O₂, final erect image A₂B₂ shall be obtained a distance of distinct vision D [Fig. 7].
Let O₂B₁ = u
Fig. 7. Final image at distance of distinct vision. |
Magnification,
M = ፀi/ፀo = tan ፀi/tan ፀo
or M = (A₁B₁/O₂B₁)/(A₁B₁/B₁O₁)
= B₁O₁/O₂B₁
= fo/u ... (4)
Considering refraction through concave lens only
O₂B₁ = +u, O₂B₂ = -D, f = -fe
Since 1/f = 1/v - 1/u
= 1/O₂B₂ - 1/O₂B₁
1/-fe = 1/(-D) - 1/u
or 1/u = 1/fe - 1/D = D - fe/Dfe
or u = feD/D - fe
Substituting for u in equation (4)
M = fo/(feD/D - fe)
= fo/fe(D - fe/D)
= fo/fe (1 - (fe/D)
Resolving power of an optical instrument
Angle subtended by two objects at the eye depends upon the separation between them and at their distances from eye.
The angle may be small in two cases :
(i) When the two objects are situated very close to each other.
(ii) When the two objects (may be widely separated) are situated at large distance away, e.g., stars etc.
It is a limitation of human eye that if this angle is less than one minute, the two objects cannot be distinguished as separate. Therefore, in order to see objects as separate, we make use of optical instruments.
A lens system (telescope or microscope) is used to resolve two point objects, while prism and grating are used to resolve two very closely situated spectral lines.
The process of separation of such closely situated objects is called resolution and the ability of an optical instrument to produce their images as distinctly separate is called the resolving power.
Rayleigh's criterion of resolving power
According to the wave-nature of light, the law of geometric optics and the rectilinear propagation of light are approximate. Even if the lena system is free from aberrations, we do not get a point image of a point object. Instead of that, we get a diffraction pattern which consists of a central maximum surrounded by a series of secondary minimas on either side. If we observe two very closely situated objects, we shall have two such diffraction patterns overlapping each other. Lord Raleigh suggested a criterion for the objects to be just resolved. It can be started as follows :
Two objects are said to be just resolved if their diffraction patterns overlap each other in such a way that the central maximum of one may coincide with the first minimum of the order.
[Fig. 8(i)] shows that position of two diffraction patterns in which central maximum of one is removed away from the first maximum of the other. The objects producing these patterns are said to be widely separated. [Fig. 8(ii)] shows the coincidence of Central maximum of one with the first minimum of the other. In this case, the two objects are just resolved. [Fig. 8(iii)] shows the central maximum of one lies within the separation between central maxima and the first minima of the other. The two objects in this case are said to not resolved. The intensity distribution in each of these cases is shown by thick line. In the first case, there is a prominent dip in between two maximas. There is a small dip in case (ii) while there is no dip in case (iii).
Fig. 8. Rayleigh's criterion of resolving power. |
Resolving Power of a Telescope
Angular separation dθ between two objects which can be just resolved as separate by a telescope is
dθ = 1.22 λ/D₀
Smaller the value of dθ, greater is the resolving power.
(i) Wavelength (λ) of light. Smaller the value of λ greater is the resolving power of telescope. This factor is beyond our control since a telescope is used to see distant objects which emit light of their own.
(ii) Diameter (D₀) of objective. Greater value of D₀ means smaller value of dθ and hence greater resolving power. Greater diameter of objective shall have an additional advantage that it shall allow greater bundle of rays thus, making the image brighter.
Characteristics Of a Good Telescope
A good telescope must have the following characteristics :
(i) High resolving power. Resolving power of a telescope is its ability to see two closely situated objects just separate. The angular limit of resolution of a telescope is given by
∆θ = 1.22 λ/D
where λ = wavelength of light
and D = diameter of the objective.
Smaller the value of '∆θ', greater is the resolving power of the telescope. To have a high resolving power, the objective of the telescope should be of large diameter and the light should of shorter wavelength.
(ii) High magnifying power. To obtain the finer details of the object it is essential to see a magnified view. But higher magnification is useful only if the objects to be viewed are already resolved. Magnifying power M of the telescope in its normal setting is
M = fo/fe
where fo = focal length of the objective
and fe = focal length of the eye piece.
Thus, a greater magnification can be achieved by having an objective of larger focal length and an eye piece of smaller focal length.
(iii) Higher brightness ratio. Brightness ratio of the image obtained through a telescope is defined as the ratio between amount of light entering the objective of the telescope to that entering the pupil of unaided eye.
The amount of light entering any aperture is proportional to the area of the aperture.
∴ Brightness ratio = area of objective/area of eye pupil
= (πD²/4)/(πd²/4)
∴ Brightness ratio = D²/d²
where D = diameter of the objective
and d = diameter of the eye pupil
Thus to have a brighter image, objectives of larger diameter should be used.
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