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)