Telescope - astronomical, terrestrial and Galileo's, Rayleigh's criterion of resolving power and Characteristics of a good telescope

 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. 

Lens - Types, Formula and Power

 Lens

A portion of refracting material bound between two spherical surfaces is called a lens. 

Types of lens:

There are two types of lenses-

     (i) Converging lens (Convex lens). A lens is said to be converging if the width of a beam decreases after refraction through it. 

     Focal length of a converging lens is taken to be positive. Different types of converging lenses are shown below in Fig. 1.

Fig. 1. Converging lens. 

     Bi-convex lens. When radii of curvature of two sides of the convex lens is same then a double convex lens becomes a bi-convex lens. 

   (ii) Diverging lens (Concave lens). A lens is said to be diverging if the width of the beam increases after refraction through it. 

     Focal length of a diverging lens is taken as negative. Different types of diverging lenses are shown below in Fig. 2.

Fig. 2. Diverging lens. 

     Rule for nomenclature of a lens. While assigning a name to a lens, name of the surface having large radius of curvature is to be written first. 

Some definitions connected with lenses

     (i) Centre of curvature (C₁ and C₂). There are two centres of curvature for a lens, one each belonging to both the surfaces. 

     Centre of curvature of a surface of a lens is defined as the centre of that sphere of which that surface forms a part. 

    (ii) Radii of curvature (R₁ and R₂). There are two radii of curvature of a lens, one each belonging to both the surfaces. 

     Radius of curvature of a surface of a lens is defined as the radius of that sphere of which the surface forms a part. 

   (iii) Optical centre (C). It is a point, laying on the principal axis of the lens, with in it or outside such that a ray of light passing through it goes undeviated. 

     A ray of light AB which coincides with the principal axis goes undeviated and undisplaced after passing the optical centre C [Fig. 3]. 

Fig. 3. Refraction through a convex lens. 

     Any other ray A'CB', which also happens to pass through the optical centre C also goes undeviated but not undisplaced. In this case, the emergent ray EB' will be parallel to incident ray A'D. Position of the optical centre is so chosen that it divides the thickness of the lens in the direct ratio of the radii of curvature of the two surfaces. If the two surfaces are of same radius of curvature optical centre lies exactly in the center of the lens. 

   (iv) Principal axis. A line joining the two centres of curvature and passing through the optical centre is called principal axis. 

    (v) Principal foci. Two principal foci can be defined in case of a lens. 

     (a) First principal focus (F₁). It is a point situated on the principal axis of the lens such that a beam of light starting from it in case of a convex lens, [Fig. 4(i)] or directed towards it in case of a concave lens, [Fig. 4(ii)] becomes parallel to principal axis after refraction through the lens. 

Fig. 4. First principal focus. 

     (b) Second principal focus (F₂). It is a point situated on the principal axis of the lens such that a beam of light coming  parallel to principal axis meets in case of a convex lens, [Fig. 5(i)] or appears to meet in case of a concave lens, [Fig. 5(ii)] after refraction through the lens. 

Fig. 5. Second principal focus. 

    (vi) Focal lengths. Corresponding to the two principal foci, there are two focal lengths in case of lenses. 

     (a) First focal length. Distance of first principal focus from the optical centre of the lens is known as first principal focus length. It is denoted by f₁  [Fig. 4]. 

     (b) Second focal length. Distance of second principal focus from the optical centre of the lens is known as second focal length. It is denoted by f₂ [Fig. 5].

     For beams of light coming parallel to the principal axis, principal foci shall be on the point where principal axis cuts the focal plane. If the incident beam has rays parallel to themselves but not parallel to principal axis, the focus (not principal focus now) shall be on the focal plane, a certain distance below/above the principal axis. Various possible cases for a converging and diverging lens is shown in Fig. 6.

   (vii) Focal planes. Again corresponding to two principal foci, there are two focal planes. 

     (a) First focal plane (P₁). It is a plane passing through the first principal focus and perpendicular to principal axis as shown in [Fig. 6(i) and (ii)].

     (b) Second focal plane (P₂). It is a plane passing through the second principal focus and perpendicular to principal axis as shown in [Fig. 6(i) and (ii)].

Fig. 6. Position of focus. 

Action of lenses

Action of a lens (converging and diverging) can be explained by imagining the lens to be consisting of a number of prisms placed one above the other. 

     (i) Converging lens. A converging lens XY can be supposed to be consisting of a number of prisms placed one above the other while its central region can be treated as a glass slab [Fig. 7].

Fig. 7. Converging action of a convex lens. 

     Refracting edges of these imaginary prisms, in upper half, are directed upward while for those in lower half are directed downwards. Since a prism bends a ray towards the base rays crossing the upper half bend downwards while those crossing the lower half bend upwards. The rays passing through the central region go undeviated (imagined to be passing through a slab). The net effect on the beam is that the emergent beam meets at a point and is said to have converged. 

    (ii) Diverging lens. The refracting edges of the imaginary prisms of a diverging lens, in upper half are downwards while for those in lower half are upwards [Fig. 8].

Fig. 8. Diverging action of a concave lens. 

     Trying to bend the rays towards the base, prism bends the rays in upper half upward while those in lower half are bent downwards. As the result, the beam becomes more divergent and appears to meet at a point when produced backwards. Again in this case, the central region which is equivalent to a glass slab sends the rays undeviated. 

Lens formula (1/f = 1/v - 1/u) 

It is a relation connecting focal length of the lens with the distances of objects and images. 

     Sign conventions. Following sign conventions will be used for refraction through lenses. 

     (i) All the distances will be measured from the pole of the surface. 

    (ii) Distances measured against the incident ray are taken as negative while those measured along the incident ray are taken as positive. 

   (iii) All transverse measurements done above the principal axis are taken as positive while those done below the principal axis are taken as negative.

     Assumptions. All relations derived for refraction through lenses will hold good if the following conditions are satisfied. 

     (i) The source is point one. 

    (ii) The aperture of the lens is small. 

   (iii) Lens is thin. 

   (iv) Rays of light make smaller angles with principal axis. 

     Let 'AB' be an object situated on the principal axis of a lens. Consider two rays AM and AC. First goes parallel to the principal axis while second is heading towards the optical centre. If the object is situated beyond 'F' of a convex lens, a real image A₁B₁ is formed as shown in [Fig. 9(i)]. If the object is situated in between 'C' and 'F' of a convex lens, a virtual image  A₁B₁ is formed as shown in [Fig. 9(ii)]. In case of a concave lens also, a virtual image A₁B₁ is formed as shown in [Fig. 9(iii)].

Fig. 9. Refraction through a lens. 

     Triangles ABC and A₁B₁C₁ are similar, 

     ∴            AB/A₁B₁ = BC/B₁C                    ... (1) 

     Triangles CMF and A₁B₁F are also similar, 

     ∴          CM/A₁B₁ = CF/B₁F

     Since           CM = AB

     ∴          AB/A₁B₁ = CF/B₁F                     ... (2) 

     From equations (1) and (2), we get

                 BC/B₁C = CF/B₁F                       ... (3) 

     Case (i) Convex lens producing a real image

     From [Fig. 9(i)]. B₁F = B₁C - CF

     Substituting for B₁F in equation (3), we get

               BC/B₁C = CF/B₁C - CF                ... (4) 

     According to sign convention, 

             BC = -u, B₁C = +v, CF = +f

     Substituting these un equation (4), we get

                   -u/v = f/v - f

or              -u (v - f) = vf

or             -uv + uf = vf

or              uf - vf = uv

     Dividing throughout by uvf, we get

            uf/uvf - vf/uvf = uv/uvf

                                1/f = 1/v - 1/u

     Case (ii) Convex lens producing a virtual image

     From [Fig. 9(ii)], B₁F = B₁C + CF

     Substituting for B₁F in equation (3), we get

            BC/B₁C = CF/B₁C + CF                     ... (5) 

     According to sign convention, 

             BC = -u, B₁C = -v, CF = +f

     Substituting these in equation (5), we get

              -u/-v = f/-v + f

       -u (-v + f) = -vf, +uv - uf = -vf, uf - vf = uv

     Dividing throughout by uvf, we get

            uf/uvf - vf/uvf = uv/uvf

or                            1/f = 1/v - 1/u

     Case (iii) Concave lens producing a virtual image

     From [Fig. 9(iii)], B₁F = CF - B₁C

     Substituting for B₁F in equation (3), we get

               BC/B₁C = CF/CF - B₁C                   ... (6) 

     According to sign convention, 

            BC = -u, B₁C = -v, CF = -f

     Substituting these in equation (6), we get

                -u/-v = -f/-f - (-v) 

or           -u/-v = -f/-f +v

or          -u (-f + v) = vf

or              +if - uv = vf

or                uf - vf = uv

      Dividing throughout by uvf, we get

              uf/uvf - vf/uvf = uv/uvf

or                             1/f = 1/v - 1/u

     Thus, it is clear that whatever the type of lena may be and the image produced by it may be real or virtual, u, v and f are connected together by the relation

                1/f = 1/v - 1/u                               ... (7) 

     Equation (7) is called lens formula. 

Linear magnification

It is the ratio between the size of the image to the size of the object. 

     Let the size of the image and the size of the object be denoted by 'I' and 'O' respectively. 

                      m = I/O                                 ... (8) 

(a) Magnification produced by a convex lens

     (i) Real image

     Triangles ABC and A₁B₁C [Fig. 9(i)] are similar, 

                 A₁B₁/AB = CB₁/CB

     According to sign convention, 

         A₁B₁ = -I   (Measurement below principal axis) 

          AB = +O   (Measurement above principal axis) 

         CB₁ = +v   (Along incident ray) 

          CB = -u   (Against incident ray) 

     Making the substitution

              -I/O = +v/-u

     ∴       I/O = v/u                                      ... (9) 

     Using equation (8) and (9), 

                 m = v/u

    (ii) Virtual image

     Triangles ABC and A₁B₁C [in Fig. 9(ii)] are similar

            A₁B₁/AB = CB₁/CB

     Here

           A₁B₁ = +I     (Measurement above the principal axis) 

             AB = +O    (Measurement above the principal axis) 

            CB₁ = -v     (Against incident ray) 

            CB = -u      (Against incident ray) 

     Making substitution, 

               +I/+O = -v/-u

or               I/O = v/u                                 ... (10) 

     From equation (8) and (9) 

                     m = v/u

(b) Magnification produced by a concave lens

     Triangles A₁B₁C and ABC [in Fig. 9(iii)] are similar

     ∴          A₁B₁/AB = CB₁/CB

     Applying sign conventions, 

         A₁B₁ = +I    (Measurement above the principal axis) 

           AB = +O   (Measurement above the principal axis) 

          CB₁ = -v     (Against incident ray) 

           CB = -u     (Against incident ray) 

     Making the substitution

              +I/+O = -v/-u

or              I/O = v/u                                   ... (11) 

     Using equation (8) and (9), 

                    m = v/u

     Thus, it is clear that the linear magnification produced by a lens is 

                     m = v/u

     This relation true both for convex and concave lenses. This relation is also independent of the nature of image (real or virtual). 

Expression for m in terms of u, v and f

     (i) In terms of v and f

     For a lens,         1/f = 1/v - 1/u

     Multiplying throughout by v, 

                  v/f = 1- (v/u) = 1 - m

     ∴             m = 1 - (v/f) 

     ∴             m = f - v/f

    (ii) In terms of u and f

     Again, for a lens     1/f = 1/v - 1/u

     Multiplying throughout by u, 

               u/f = u/v - (1) = 1/m - (1) 

     ∴        1/m = 1 + (u/f) = f + u/f

     ∴           m = f/f + u

     Thus, linear magnification for a lens can be expressed as

               m = I/O = v/u = f + v/f = f/f + u

Position of the image as the objective is gradually moved from infinity to the pole of the lens

As the object is shifted gradually from infinity to the pole of the lens, the image produced by the lens undergoes a variation not only in nature and size but its position is also affected. We discuss below the different cases of relative change in position, nature and size of the image as the object is shifted. 

(i) Object being at infinity

     In this case,          u = ∞

     ∴                    1/f = 1/v - 1/∞

or                       1/f = 1/v    or      v = f

     Therefore, the image 'A₁B₁' is obtained at a distance 'f' from the lens, i.e., on the principal focus 'F' of the lens. Formation of the image by a convex lens is shown in (Fig. 10). Magnification in this case is extremely small and the image is said to be real and inverted. 

Fig. 10. Object at infinity. 

(ii) Object lying beyond '2f'

     According to lens formula,  

                        1/f = 1/v - 1/u

     ∴                1/v = 1/f + 1/u

     If u > - 2f, 2f > v > -f, i.e., the image lies between f and 2f, m(= v/u) always is less than one. 

     Thus, as object moves from infinity to the position 2f, position of image changes from 'f' to '2f' distance away from lens. Formation of image is shown in (Fig. 11). The image is diminished in size and is real and inverted. 

Fig. 11. Object beyond 2f.

(iii) Object at 2f [Fig. 12]

                        u = -2f

     ∴               1/f = 1/v - (1/-2f) = 1/v + 1/2f

or                  1/v = 1/f - 1/2f = 1/2f

     ∴                  v = 2f

     ∴                 m = v/u = 2f/-2f = -1

Fig. 12. Object at 2f.

     Thus, the image of an object situated at '2f' is also obtained at the same distance on the other side of the lens. The formation of the image is shown in (Fig. 12). The size of the image is same as that of the object and is real and inverted. 

(iv) Object lying between 'f' and '2f'

     According to lens formula, 

                          1/f = 1/v - 1/u

     ∴                  1/v = 1/f + 1/u

     If -2f > u > -f, i.e., can be senn from above relation that, v > 2f.

     Thus, the image lies at a distance greater than 2f on the other side of the lens, 

     ∴    m (= v/u) is greater than one. 

Fig. 13. Object between f and 2f.

     Thus, if an object is moved from '2f' to 'f' the image goes beyond '2f'. Formation of image is shown in (Fig. 13). The image is magnified in size and is real and inverted. 

(v) Object at f

     In this case,         u = -f

     ∴                          1/f = 1/v - 1/u

     ∴                         1/v = 1/f + 1/u = 1/f - (1/-f) 

     ∴                            v = ∞

     Since u = -f and v = ∞

     ∴    m (= v/u) is infinite. 

Fig. 14. Object at f. 

     Therefore, the image of an object situated at a distance 'f' from the lens, i.e., on the principal focus is obtained at infinity. The formation of the image is shown in (Fig. 14). The image is infinitely large in size and is real and inverted. 

(vi) Object lying between 'f' and optical centre 'C'

     At f,              u = -f

     ∴                 1/f = 1/v - (1/-f) = 1/v + 1/f

or                   1/v = 1/f - 1/f = 0

     ∴                   v = ∞

     At C,            u = 0.

     ∴               1/v = 1/f + 1/0 = ∞     

     ∴                  v = ∞

     Thus, as the object is moved from principal focus 'F' to the optical centre 'C' of the lens, its image moves from infinity to the optical centre and lies on the same side of the lens. Formation of the image is shown in (Fig. 15). The image is magnified, virtual and erect. 

Fig. 15. Object between F and optical center. 

     

     Conclusion. A change in nature, size and position of the image due to a change in the position of object can be summed up in the form of following table. 

S. No.	Position of object	Position of image	Relative size of image	Nature of image
1.	At infinity	At f	Very small	Real, inverted
2.	Beyond 2f	Between f and 2f	Diminished	Real, inverted
3.	At 2f	At 2f	Same size	Real, inverted
4.	Between f and 2f	Beyond 2f	Large	Real, inverted
5.	At f	At infinity	Very large	Real, inverted
6.	Between f and pole	On same side as the object, away from lens	Large	Virtual, erect

Power of a lens

The function of lens is to converge a beam of light (in the case of convex lens) or to diverge it (in the case of concave lens). Since a convex lens of shorter focal length produces a greater convergence, therefore, power of a lens varies inversely as its focal length. 

     The reciprocal of the focal length of a lens, expressed in metre, is called its power. 

     Representing focal length of a lens by 'f' in metre and power by 'P', we can write

                    P = 1/f(m) 

    The unit of power is dioptre is defined as the power of a lens of focal length one metre. 

     Thus,     P (dioptre) = 1/f(metre) 

                                        = 100/f(cm) 

     Thus, the power of a convex lens of focal length 25 cm is 4 dioptre. Opticians specify lenses by their powers. The power of a convex lens is taken as positive and that of a concave lens negative. 

Key formulae

1. Lens formula : 1/f = 1/v - 1/u

2. Magnification (linear) : 

            m = I/O = v/u = f - v/f = f/f + u

3. Lens maker's formula :

     (i) When the lens (μ₂)  has same medium (μ₁) on its two sides

               1/f = (1μ₂ - 1) (1/R1 - 1/R2)

    (ii) When medium situated on the two sides of the lens is different. 

             1/f = μ - μ1/μ3R1 + μ3 - μ₂/μ3R2 

4. Power of a lens :

             P (dioptre) = 1/f(metre) = 100/f(cm)