Optics Past Paper PDF

9. Optics Can you recall? 1. What are laws of reflection and refraction? 4. What is total internal reflection? 2. What is dispersion of light? 5. How does light refract at a curved surface? 3. What is refractive...

9. Optics Can you recall? 1. What are laws of reflection and refraction? 4. What is total internal reflection? 2. What is dispersion of light? 5. How does light refract at a curved surface? 3. What is refractive index? 6. How does a rainbow form? 9.1 Introduction: c = 299792458 m s-1 According to Einstein’s “See it to believe it” is a popular saying. special theory of relativity, this is the maximum In order to see, we need light. What exactly possible speed for any object. For practical is light and how are we able to see anything? purposes we write it as c = 3×108 m s-1. We will explore it in this and next standard. Commonly observed phenomena We know that acoustics is the term used for concerning light can be broadly split into three science of sound. Similarly, optics is the term categories. used for science of light. There is a difference (I) Ray optics or geometrical optics: A in the nature of sound waves and light waves particular direction of propagation of which you have seen in chapter 8 and will learn energy from a source of light is called in chapter 13. a ray of light. We use ray optics for 9.2 Nature of light: understanding phenomena like reflection, Earlier, light was considered to be that form refraction, double refraction, total of radiant energy which makes objects visible internal reflection, etc. due to stimulation of retina of the eye. It is a (II) Wave optics or physical optics: For form of energy that propagates in the presence explaining phenomena like interference, or absence of a medium, which we now call diffraction, polarization, Doppler effect, waves. At the beginning of the 20th century, it etc., we consider light energy to be in the was proved that these are electromagnetic (EM) form of EM waves. Wave theory will be waves. Later, using quantum theory, particle further discussed in XIIth standard. nature of light was established. According to (III) Particle nature of light: Phenomena like this, photons are energy carrier particles. By an photoelectric effect, emission of spectral experiment using countable number of photons, lines, Compton effect, etc. cannot be it is now an established fact that light possesses explained by using classical wave theory. dual nature. In simple words we can say that These involve the interaction of light with light consists of energy carrier photons guided matter. For such phenomena we have to by the rules of EM waves. In vacuum, these use quantum nature of light. Quantum waves (or photons) travel with a speed of nature of light will be discussed in XIIth standard. In a material medium, the speed of EM 9.3 Ray optics or geometrical optics: waves is given by , In geometrical optics, we mainly study image formation by mirrors, lenses and prisms. where permittivity ε and permeability µ It is based on four fundamental laws/ principles are constants which depend on the electric which you have learnt in earlier classes. and magnetic properties of the medium. (i) Light travels in a straight line in a c homogeneous and isotropic medium. The ratio n = is called the absolute Homogeneous means that the properties v refractive index and is the property of the of the medium are same every where in medium. the medium and isotropic means that the 159  properties are the same in all directions. Solution: (ii) Two or more rays can intersect at a point Speed of light in vacuum, c = 3×108m/s without affecting their paths beyond that nglass = 10.5 point. ∴ Speed of light in glass = (iii) Laws of reflection: c 3  108 (a) Reflected ray lies in the plane formed by   2  108 m / s nglass 1.5 incident ray and the normal drawn at the point of incidence; and the two rays are on Distance to be travelled by light in glass, either side of the normal. s = 2 mm = 2×10-3 m (b) Angles of incidence and reflection are ∴Time t required by light to travel this distance, equal. s 2  10 3 (iv) Laws of refraction: These apply at the t   10 11 s boundary between two media v glass 2  10 8 (a) Refracted ray lies in the plane formed by Most convenient prefix to express this small incident ray and the normal drawn at the time is pico (p) = 10-12 point of incidence; and the two rays are on ∴ t = 10 × 10-12 = 10 ps either side of the normal. 9.3.1 Cartesian sign convention: (b) Angle of incidence ( θ 1 in a medium of While using geometrical optics it is refractive index n 1) and angle of refraction necessary to use some sign convention. The ( θ 2 in medium of refractive index n 2) are relation between only the numerical values of related by Snell’s law, given by u, v and f for a spherical mirror (or for a lens) will be different for different positions of the ( n 1)sin θ 1 = ( n 2)sin θ 2. object and the type of mirror. Here u and v are the distances of object and image respectively Do you know ? from the optical center, and f is the focal length. Properly used suitable sign convention Interestingly, all the four laws stated enables us to use the same formula for all above can be derived from a single different particular cases. Thus, while deriving principle called Fermat’s (pronounced a formula and also while using the formula it ''Ferma'') principle. It says that “While is necessary to use the same sign convention. travelling from one point to another by one Most convenient sign convention is Cartesian or more reflections or refractions, a ray of sign convention as it is analogous to coordinate light always chooses the path of least geometry. According to this sign convention, time”. (Fig. 9.1): Ideally it is the path of extreme time, i.e., path of minimum or maximum time. We strongly recommend you to go through a suitable reference book that will give you the proof of i = r during reflection and Snell’s law during refraction using Fermat’s principle. Example 9.1: Thickness of the glass of a Fig. 9.1 Cartesian sign convention. spectacle is 2 mm and refractive index of its i) All distances are measured from the optical glass is 1.5. Calculate time taken by light to center or pole. For most of the optical cross this thickness. Express your answer with objects such as spherical mirrors, thin the most convenient prefix attached to the unit lenses, etc., the optical centers coincides ‘second’. with their geometrical centers. 160  ii) Figures should be drawn in such a way b) If we are standing on the bank of a still that the incident rays travel from left to water body and look for our image formed right. A diverging beam of incident rays by water (or if we are standing on a plane corresponds to a real point object (Fig. 9.2 mirror and look for our image formed by (a)), a converging beam of incident rays the mirror), the image is laterally reversed, corresponds to a virtual object (Fig. 9.2 of the same size and on the other side. (b)) and a parallel beam corresponds to an c) If an object is kept between two plane object at infinity. Thus, a real object should mirrors inclined at an angle θ (like in a be shown to the left of pole (Fig. 9.2 (a)) kaleidoscope), a number of images are and virtual object or image to the right of formed due to multiple reflections from pole. (Fig. 9.2 (b)) both the mirrors. Exact number of images depends upon the angle between the mirrors and where exactly the object is kept. It can be obtained as follows (Table 9.1): 360 Calculate n   Let N be the number of images seen. (I) If n is an even integer, N   n  1 , Fig. 9.2: (a) Diverging beam from a real irrespective of where the object is. object (II) If n is an odd integer and object is exactly on the angle bisector, N   n  1. (III) If n is an odd integer and object is off the angle bisector, N = n (IV) If n is not an integer, N = m, where m is integral part of n. Fig. 9.2: (b) Converging beam towards a Table 9.1 virtual object. Angle 360 Position of iii) x-axis can be conveniently chosen as the n N principal axis with origin at the pole. θ0  the object iv) Distances to the left of the pole are On angle 120 3 2 negative and those to the right of the pole bisector are positive. Off angle 120 3 3 v) Distances above the principal axis (x-axis) bisector are positive while those below it are 110 3.28 Anywhere 3 negative. Unless specially mentioned, we shall always 90 4 Anywhere 3 consider objects to be real for further 80 4.5 Anywhere 4 discussion. On angle 9.4 Reflection: 72 5 4 bisector 9.4.1 Reflection from a plane surface: Off angle 72 5 5 a) If the object is in front of a plane reflecting bisector surface, the image is virtual and laterally 60 6 Anywhere 5 inverted. It is of the same size as that of the object and at the same distance as that of 50 7.2 Anywhere 7 object but on the other side of the reflecting surface. 161 Example 9.2: A small object is kept (concave or convex) mirror is related to object symmetrically between two plane mirrors distance and image distance as inclined at 38°. This angle is now gradually 1 1 1 increased to 41°, the object being symmetrical    --- (9.1) f v u all the time. Determine the number of images visible during the process. Solution: According to the convention used in the table above, 360   380  n   9.47  38 ∴ N = 9. This is valid till the angle is 40° as the object is kept symmetrically Beyond 40°, n < 9 and it decreases upto 360 = 8.78 Fig. 9.3 (a): Parallel rays incident from left 41. appear to be diverging from F, lying on the Hence now onwards there will be 8 images till positive side of origin (pole). 41°. 9.4.2 Reflection from curved mirrors: In order to focus a parallel or divergent beam by reflection, we need curved mirrors. You might have noticed that reflecting mirrors for a torch or headlights, rear view mirrors of vehicles are not plane but concave or convex. Mirrors for a search light are parabolic. We shall restrict ourselves to spherical mirrors only which can be studied using simple mathematics. Such mirrors are parts of a sphere polished from outside (convex) or from inside (concave). Radius of the sphere of which a mirror Fig. 9.3 (b): Parallel rays incident from is a part is called as radius of curvature (R) left appear to converge at F, lying on the of the mirror. Only for spherical mirrors, half negative side of origin (pole). of radius of curvature is focal length of the By a small mirror we mean its aperture  R (diameter) is much smaller (at least one tenth) mirror  f  2 . For a concave mirror it is than the values of u, v and f.   the distance at which parallel incident rays Focal power: Converging or diverging ability converge. For a convex mirror, it is the distance of a lens or of a mirror is defined as its focal from where parallel rays appear to be diverging 1 after reflection. According to sign convention, power. It is measured as P =. f the incident rays are from left to right and they In SI units, it is measured as diopter. should face the polished surface of the mirror. 1dioptre  D   1m1 Thus, focal length of a convex mirror is positive Lateral magnification: Ratio of linear size (Fig 9.3 (a)) while that of a concave mirror is of an image to that of the object, measured negative (Fig. 9.3 (b)). perpendicular to the principal axis, is defined as Relation between f, u and v: v For a point object or for a small finite the lateral magnification m = u object, the focal length of a small spherical For any position of the object, a convex mirror 162 always forms virtual, erect and diminished to a cone of very small angle. image, m < 1. In the case of a concave mirror (iii) If there is a parallel beam of rays, it is it depends upon the position of the object. paraxial, i.e., parallel and close to the principal Following Table 9.2 will help you refresh your axis. knowledge. However, in reality, these assumptions do Table 9.2 not always hold good. This results into distorted Concave mirror (f negative) or defective image. Commonly occurring Real(R) defects are spherical aberration, coma, Lateral Position of Position of or astigmatism, curvature, distortion. Except magnifi- object image Virtual spherical aberration, all the other arise due to -cation (V) beams of rays inclined to principal axis. These u =∞ v= f R m =0 are not discussed here. u>2f 2f >v> f R m u> f v>2f R m >1  R  earlier, the relation  f   giving a single u= f v R m  2 u< f v >u V m >1 point focus is applicable only for small aperture spherical mirrors and for paraxial rays. In reality, Example 9.3: A thin pencil of length 20 cm when the rays are farther from the principal axis, is kept along the principal axis of a concave the focus gradually shifts towards pole (Fig. mirror of curvature 30 cm. Nearest end of the 9.4). This phenomenon (defect) arises due to pencil is 20 cm from the pole of the mirror. spherical shape of the reflecting surface, hence What will be the size of image of the pencil? called as spherical aberration. It results into a Solution: R = 30 cm unsharp (fuzzy) image with unclear boundaries. f = R/2 =-15 cm... (Concave mirror) 1 1 1   f v u For nearest end, u = u1 = - 20 cm. Let the image distance be v1 1 1 1     v1  60 cm 15 v1 20 Nearest end is at 20 cm and pencil itself is 20 Fig. 9.4: Spherical aberration for curved cm long. Hence farthest end is 20 + 20 = 40 mirrors. cm  u2 The distance between FM and FP (Fig. Let the image distance be v2 9.4) is measured as the longitudinal spherical 1 1 1     v 2  24 cm aberration. If there is no spherical aberration, 15 v 2 40 we get a single point image on a screen placed ∴ Length of the image = 60 -24 = 36 cm. perpendicular to the principal axis at that Defects or aberration of images: The theory location, for a beam of incident rays parallel to of image formation by mirrors or lenses, the axis. In the presence of spherical aberration, and the formulae that we have used such as no such point is possible at any position of the R 1 1 1 f = or    etc. screen and the image is always a circle. At a 2 f v u particular location of the screen, the diameter of are based on the following assumptions: (i) this circle is minimum. This is called the circle Objects and images are situated close to the of least confusion. In the figures it is across principal axis. AB. Radius of this circle is transverse spherical (ii) Rays diverging from the objects are confined aberration. 163  In the case of curved mirrors, this defect Absolute refractive index: can be completely eliminated by using a Absolute refractive index of a medium is parabolic mirror. Hence surfaces of mirrors defined as the ratio of speed of light in vacuum used in a search light, torch, headlight of a car, to that in the given medium. telescopes, etc., are parabolic and not spherical. c n = where c and v are respective speeds v Do you know ? of light in vacuum and in the medium. As n is the ratio of same physical quantities, it is a Why does a parabolic mirror not have unitless and dimensionless physical quantity. spherical aberration? Parabola is a geometrical shape drawn For any material medium (including in such a way that every point on it is air) n > 1, i.e., light travels fastest in vacuum equidistant from a straight line and from a than in any material medium. Medium having point. Figure 9.5 shows a parabola. Points A, greater value of n is called optically denser. An B, C, … on it are equidistant from line RS optically denser medium need not be physically (called directrix) and point F (called focus). denser, e.g., many oils are optically denser than Hence A′A = AF, B′B = BF, C′C = CF, …. water but water is physically denser than them. R Relative refractive index: Refractive index of medium 2 with respect to medium 1 is defined as the ratio of speed of light v1 in medium 1 to its speed v2 in medium n2 v1 2. Thus, n= = 1 2 n1 v 2 S Fig. 9.5: Single focus for parabolic mirror. Do you know ? If rays of equal optical path converge at a point, that point is the location of real (a) Logic behind the convention 1n2 : Letter image corresponding to that beam of rays. n is the symbol for refractive index, Paths A″AA′, B″BB′. C″CC′, etc., n2 corresponds to refractive index of are equal paths in the absence of mirror. medium 2 and 1n2 indicates that it is If the parabola ABC… is a mirror then with respect to medium 1. In this case, the respective optical paths will be A″AF, light travels from medium 1 to 2 so we B″BF, C″CF, … and from the definition of need to discuss medium 2 in context to parabola, these are also equal. Thus, F is the medium 1. single point focus for entire beam parallel to (b) Dictionary meaning of the word refract the axis with NO spherical aberration. is to change the path`. However, in context of Physics, we should be more 9.5 Refraction: specific. We use the word deviate for Being an EM wave, the properties of light changing the path. During refraction at (speed, wavelength, direction of propagation, normal incidence, there is no change etc.) depend upon the medium through which in path. Thus, there is refraction but it is traveling. If a ray of light comes to an no deviation. Deviation is associated interface between two media and enters into with refraction only during oblique another medium of different refractive index, incidence. Deviation or changing the it changes itself suitable to that medium. This path or bending is associated with phenomenon is defined as refraction of light. many phenomena such as reflection, The extent to which these properties change is diffraction, scattering, gravitational decided by the index of refraction, 'n'. bending due to a massive object, etc. 164 Illustrations of refraction: 1) When seen from Example 9. 4: A crane flying 6 m above a still, outside, the bottom of a water body appears to clear water lake sees a fish underwater. For the crane, the fish appears to be 6 cm below the be raised. This is due to refraction at the plane surface of water. In this case, water surface. How much deep should the crane Real depth immerse its beak to pick that fish? nwater ≅ apparent depth For the fish, how much above the water surface does the crane appear? Refractive index of This relation holds good for a plane water = 4/3. parallel transparent slab also as shown below. Solution: For crane, apparent depth of the fish Figure 9.6 shows a plane parallel slab of a is 6 cm and real depth is to be determined. transparent medium of refractive index n. A point object O at real depth R appears to be at For fish, real depth (height, in this case) of the I at apparent depth A, when seen from outside crane is 6 m and apparent depth (height) is to (air). Incident rays OA (traveling undeviated) be determined. and OB (deviating along BC) are used to locate the image. For crane, it is water with respect to air as real depth is in water and apparent depth is as seen from air 4 R R  n     R  8cm 3 A 6 For fish, it is air with respect to water as the Fig. 9.6: Real and apparent depth. real height is in air and seen from water. By considering i and r to be small, we can write, 3 R 6  n     A  8m x x 4 A A tan  r    sin  r  and tan  i    sin  i  A R 9.6 Total internal reflection: x sin  r   A  R Realdepth n     sin  i   x  A Apparentdeptth R   2) A stick or pencil kept obliquely in a glass containing water appears broken as its part in water appears to be raised. Small angle approximation: For small angles, Fig. 9.7: Total internal reflection. expressed in radian, sin     tan. For example, for Figure 9.7 shows refraction of light emerging from a denser medium into a rarer medium for various angles of incidence. The angles of refraction in the rarer medium are larger than the corresponding angles of In this case the error is 0.5236  0.5  0.0236 in incidence. At a particular angle of incidence ic 0.5, which is 4.72 %. in the denser medium, the corresponding angle For practical purposes we consider angles less of refraction in the rarer medium is 900. For than 100 where the error in using sin    is angles of incidence greater than ic , the angle less than 0.51 %. (Even for 600, it is still 15.7 %) of refraction become larger than 900 and the ray It is left to you to verify that this is almost does not enter into rarer medium at all but is equally valid for tanθ till 200 only. reflected totally into the denser medium. This is 165 called total internal reflection. In general, there is always partial reflection and partial refraction at the interface. During total internal reflection TIR, it is total reflection and no refraction. The corresponding angle of incidence in the denser medium is greater than or equal to the critical Fig. 9.8 (a): Optical fibre construction. angle. Critical angle for a pair of refracting media can be defined as that angle of incidence in the Fig. 9.8 (b): Optical fibre working. denser medium for which the angle of refraction in the rarer medium is 90°. An optical fibre essentially consists of an extremely thin (slightly thicker than a human Do you know ? hair), transparent, flexible core surrounded by optically rarer (smaller refractive index), In Physics the word critical is used when flexible cover called cladding. This system is certain phenomena are not applicable or coated by a buffer and a jacket for protection. more than one phenomenon are applicable. Entire thickness of the fibre is less than half a Some examples are as follows. mm. (Fig. 9.8(a)). Number of such fibres may (i) In case of total internal reflection, the be packed together in an outer cover. phenomenon of reversibility of light An optical signal (ray) entering the core is not applicable at critical angle and refraction is possible only for angles of suffers multiple total internal reflections (Fig. incidence in the denser medium smaller 9.8 (b)) and emerges after several kilometers than the critical angle. with extremely low loss travelling with highest (ii) At the critical temperature, a substance possible speed in that material ( ~ 2,00,000 coexists into all the three states; km/s for glass). Some of the advantages of solid, liquid and gas. At all the other optic fibre communication are listed below. temperatures, only two states are (a) Broad bandwidth (frequency range): For simultaneously possible. TV signals, a single optical fibre can (iii) For liquids, streamline flow is possible carry over 90000 channels (independent till critical velocity is achieved. signals). At critical velocity it can be either (b) Immune to EM interference: Being streamline or turbulent. electrically non-conductive, it is not able Let µ be the relative refractive index to pick up nearby EM signals. of denser medium with respect to the rarer. (c) Low attenuation loss: The loss is lower Applying Snell’s law at the critical angle of than 0.2 dB/km so that a single long cable 1 incidence, iC , we can write sin ( ic )  as, can be used for several kilometers.  (d) Electrical insulator: No issue with ground (µ)sin (ic) = (1) sin 90° loops of metal wires or lightning. For commonly used glasses of (e) Theft prevention: It is does not use copper µ = 1.5, ic = 41° 49′ ≅ 42° and for water of or other expensive material. 4 (f) Security of information: Internal damage is   , ic = 48° 35′ (Both, with respect to air) 3 most unlikely. 9.6.1 Applications of total internal reflection: (ii) Prism binoculars: Binoculars using (i) Optical fibre: Though little costly for initial only two cylinders have a limitation of field set up, optic fibre communication is undoubtedly of view as the distance between the two the most effective way of telecommunication cylinders can’t be greater than that between by way of EM waves. the two eyes. This limitation can be overcome 166 by using two right angled glass prisms From the dimensions given, ( iC ~ 420 ) used for total internal reflection as 3 3 sin 90 5 shown in the Fig. 9.9. Total internal reflections tan ( ic )   sin ( ic )   nliquid   4 5 sin( ic ) 3 occur inside isosceles, right angled prisms. 9.7 Refraction at a spherical surface and lenses: In the section 9.5 we saw that due to refraction, the bottom of a water body appears Real depth to be raised and nwater = apparent depth. However, this is valid only if we are dealing with refraction at a plane surface. In many cases such as liquid drops, lenses, ellipsoid paper weights, etc, curved surfaces are present and the formula mentioned above may not be true. In Fig. 9.9: Prism binoculars such cases we need to consider refraction at one (iii) Periscope: It is used to see the objects on or more spherical surfaces. This will involve the surface of a water body from inside water. parameters including the curvature such as The rays of light should be reflected twice radius of curvature, in addition to refractive through right angle. Reflections are similar indices. to those in the binoculars (Fig 9.10) and total Lenses: Commonly used lenses can be internal reflections occur inside isosceles, right visualized to be consisting of intersection of angled prisms. two spheres of radii of curvature R1 and R2 or of one sphere and a plane surface (R = ∞). A lens is said to convex if it is thicker in the middle and narrowing towards the periphery. A lens is concave if it is thicker at periphery and narrows down towards center. Convex lens is visualized Fig. 9.10: Periscope. to be internal cross section of two spheres (or Example 9.5: There is a tiny LED bulb at the one sphere and a plane surface) while concave center of the bottom of a cylindrical vessel of lens is their external cross section (Figs. 9.11-a to diameter 6 cm. Height of the vessel is 4 cm. The 9.11-f). Concavo-convex and convexo-concave beaker is filled completely with an optically lenses are commonly used for spectacles of dense liquid. The bulb is visible from any positive and negative numbers, respectively. inclined position but just visible if seen along For lenses of material optically denser than the edge of the beaker. Determine refractive the medium in which those are kept, convex index of the liquid. lenses have positive focal length [according Solution: As seen from the accompanying to Cartesian sign convention] and converge figure, if the bulb is just visible from the edge, the incident beam while concave lenses have angle of incidence in the liquid (at the edge) negative focal length and diverge the incident must be the critical angle of incidence, iC beam. For most of the applications of lenses, maximum thickness of lens is negligible (at least 50 times smaller) compared with all the other distances such as R1 and R2, u, v, f, etc. Such a lens is called as a thin lens and physical 167 center of such a lens can be assumed to be the For lenses, the relations between u, v, R and f common pole (or optical center) for both its depend also upon the refractive index n of the refracting surfaces.  R material of the lens. The relation  f  2  Fig. 9.11 (a): Convex   lens as internal does NOT hold good for lenses. Below we shall cross section of two derive the necessary relation by considering spheres. refraction at the two surfaces of a lens independently. Fig. 9.11 (b): Concave Unless mentioned specifically, we assume lens as external cross lenses to be made up of optically denser section of two spheres. material compared to the medium in which those are kept, e.g., glass lenses in air or in water, etc. As special cases we may consider Fig. 9.11 (c): Plano lenses of rarer medium such as an air lens convex lens in water or inside a glass. A spherical hole inside a glass slab is also a lens of rarer medium. In such case, physically (or geometrically or shape-wise) convex lens Fig. 9.11 (d): Plano diverges the incident beam while concave concave lens lens converges the incident beam. Refraction at a single spherical surface: Consider a spherical surface YPY’ of radius of Fig. 9.11 (e): concave- curvature R, separating two transparent media convex lens of refractive indices n1 and n2 respectively with n1 < n2. P is the pole and X’PX is the principal axis. A point object O is at an object distance Fig. 9.11 (f) convex- -u from the pole, in the medium of refractive concave lens index n1. Convexity or concavity of a surface is always with respect to the incident rays, i.e., 1 1 1 with respect to a real object. Hence in this case For any thin lens,    --- (9.2) the surface is convex (Fig. 9.12). f v u If necessary, we can have a number of thin lenses in contact with each other having common principal axis. Focal power of such combination is given by the algebraic addition (by considering ± signs) of individual focal powers. 1 1 1 1 1 Fig. 9.12: Refraction at a single refracting ∴         f  f i  f1 f 2 f 3 surface.  P1  P2  P3 ..  Pi  P To locate its image and in order to minimize spherical aberration, we consider two paraxial For only two thin lenses, separated in air by rays. The ray OP along the principal axis distance d, travels undeviated along PX. Another ray OA 1 1 1 d strikes the surface at A. CAN is the normal     P1  P2  dP1P2  P f f1 f 2 f 1 f 2 from center of curvature C of the surface at A. Angle of incidence in the medium n1 at A is i. 168 As n1< n2 , the ray deviates towards the normal, travels along AZ and cuts the principal axis at I. Thus, real image of point object O is formed at I. Angle of refraction = in medium n2 is r. n refractive = indexoftheothermedium 1 2 According to Snell’s law, u  4 cm n1 sin  i   n2sin  r  --- (9.3) v=? Let be the angles subtended by R  3cm incident ray, normal and refracted ray with the n  n n n principal axis. 2 1  2 1 R v u  i    andr     1  1.5 1 1.5 1 1 3 For paraxial rays, all these angles are        v  4.8cm 3 v 4 6 v 8 small and PA can be considered as an arc for. In this case apparent depth is NOT less than real depth. This is due to curvature of the refracting surface. arc AP arc AP In this case (Fig. 9.12) we had considered Also,    , PO u the object placed in rarer medium, real image arc AP arc AP   and in denser medium and the surface facing the PC R object to be convex. However, while deriving arc AP arc AP the relation, all the symbolic values (which   PI v could be numeric also) were substituted as  n1i  n2 r per the Cartesian sign convention (e.g. ‘u’ as negative, etc.). Hence the final expression  n1      n2      (Eq. 9.4) is applicable to any surface   n2  n1    n2  n1 separating any two media, and real or virtual image provided you substitute your values Substituting and canceling 'arc AP', (symbolic or numerical) as per Cartesian we get sign convention. The only restriction is that n2  n1 n2 n1 n1 is for medium of real object and n2 is the   --- (9.4) R v u other medium (not necessarily the medium of image). Only in the case of real image, it Example 6: A glass paper-weight (n =1.5) of will be in medium n2. If virtual, it will be in radius 3 cm has a tiny air bubble trapped inside the medium n1 (with image distance negative it. Closest distance of the bubble from the how do you justify this?). surface is 2 cm. Where will it appear when seen We strongly suggest you to do the from the other end (from where it is farthest)? derivations yourself for any other special Solution: Accompanying Figure below case such as object placed in the denser illustrates the location of the bubble. medium, virtual image, concave surface, etc. It must be remembered that in any case you will land up with the same expression as in Eq. (9.4). Lens makers’ equation: Relation between refractive index (n), focal length (f ) and radii of curvature R1 and R2 for a thin lens. Consider a lens of radii of curvature R1 and According to the symbols used in the Eq. R kept in a medium such that n is refractive 2 (9.4), we get, 169 index of material of the lens with respect to the Adding Eq. (9.5) and (9.6), we get, outside medium. Assuming the lens to be thin,  1 1  1 1 P is the common pole for both the surfaces. O is  n  1      a point object on the principal axis at a distance  R1 R2  v u u from P. First refracting surface of the lens For of radius of curvature R1 faces the object (Fig  1 1  1 9.13).    n  1    --- (9.7) f  R1 R2  For preparing spectacles, it is necessary to grind the glass (or acrylic, etc.) for having the desired radii of curvature. Equation (9.7) can be used to calculate the radii of curvature for the lens, hence it is called the lens makers’ equation. (It should be remembered that while Fig. 9.13: Lens maker's equation. solving problems when you are using equations Axial ray OP travels undeviated. Paraxial 9.1, 9.2, 9.4, 9.7, etc., we will be substituting the ray OA deviates towards normal and would values of the corresponding quantities. Hence intersect axis at I1, in the absence of second this time it is algebraic substitution, i.e., with refracting surface. PI1 = v1 is the image distance Special cases: for intermediate image I1. Most popular and most common special Thus, the symbols to be used in Eq. (9.4) are case is the one in which we have a thin, n2 = n, n1 = 1, R = R1 , u = u , v = v1 symmetric, double lens. In this case, R1 and R2 n 1 n 1 are numerically equal. ∴   --- (9.5) (A) Thin, symmetric, double convex lens: R1 R1 v1  u  is positive, R2 is negative and numerically (Not that, in this case, we are not substituting equal. Let R = 1 R= 2 R. the algebraic values but just using different 1  1 1  2  n  1 symbols.)    n  1    f  R  R  R Before reaching I1, the ray PI1 is intercepted at B by the second refracting surface. In this Further, for popular variety of glasses, case, the incident rays AB and OP are in the n ≅ 1.5. In such a case, f = R. medium of refractive index n and converging (B) Thin, symmetric, double concave lens: towards I1. Thus, I1 acts as virtual object for R1 is negative, R2 is positive and numerically second surface of radius of curvature (R2) and equal. Let R = 1 R= 2 R. object distance is  u  v1 . As the incident rays 1  1 1  2  n  1 are in the medium of refractive index n, this    n  1    is the medium of (virtual) object ∴ n1 = n and f  R R  R refractive index of the other medium is n2 = 1. Further if (C) Thin, planoconvex lenses: One radius is After refraction, the ray bends away from 1 n 1 the normal and intersects the principal axis at I R and the other is ∞.   f R which is the real image of object O formed due to the lens. ∴ PI = v. Further if Substituting all these symbols in Eq. (9.4), proper ± sign) we get Example 7: A dense glass double convex lens 1 n n 1 1    n --- (9.6)  n  2  designed to reduce spherical aberration R2  R2 v  v1  has |R1|:|R2|=1:5. If a point object is kept 15 cm in front of this lens, it produces its real image at 170 7.5 cm. Determine R1 and R2. parallel surfaces must be separated over very Solution: u = - 15 cm, v = + 7.5 cm (real image large distance and i should be large. is on opposite side). 1 1 1 1 1 1         f  5cm f v u f 7.5 15 The lens is double convex. Hence, R1 is positive and R2 is negative. Also, R2 = 5 R1 and n = 2.  1 1  1  n  1      R1 R2  f  1 1  1   2  1       R   5 R   5  1 1  Fig. 9.15: Lateral dispersion due to plane  6  1  1     R1  6 cm  R2  30cm parellal slab.  1 5 5 R Example 8: A fine beam of white light is 9.8 Dispersion of light and prisms: incident upon the longer side of a plane parallel The colour of light that we see depends glass slab of breadth 5 cm at angle of incidence upon the frequency of that ray (wave). The 600. Calculate angular deviation of red and refractive index of a material also depends upon violet rays within the slab and lateral dispersion the frequency of the wave and increases with between them as they emerge from the opposite frequency. Obviously refractive index of light side. Refractive indices of the glass for red and is different for different colours. As a result, violet are 1.51 and 1.53 respectively. for an obliquely incident ray, the angles of Solution: As shown in the Fig. 9.15 above, refraction are different for each colour and they VM = LV and RT = LR give respective lateral separate (disperse) as they travel along different deviations for violet and red colours and LVR = directions. This phenomenon is called angular LV - LR is the lateral dispersion between these dispersion Fig 9.14. colours. nR = 1.51, nV = 1.53 and i = 60° sin i sin 600  sin rR    0.5735 nR 1.51 sin i sin 600 sin rV    0.566 nV 1.53  rR  350 and RV  340 28’  RV  rR  rV 32 ’ Fig. 9.14: Angular dispersion at a single   i  rR  250 , i  rV  250 32’ surface. If a polychromatic beam of light (bundle of rays of different colours) is obliquely incident upon a plane parallel transparent slab, emergent beam consists of all component colours parallel to each other and also parallel to initial R   R  separated out. However, in this case all those are L  RT  AR sin i  r  2.58cm direction. This is lateral dispersion which is LV  VM  AV  sin i  rV   2.58cm measured as the perpendicular distance between the direction of incident ray and respective directions of dispersed emergent rays (LR and It shows that the lateral dispersion is too LV) Fig 9.15. For it to be easily detectable, the small to detect. 171  In order to have appreciable and observable reflecting surface AB. Normal passing through dispersion, two parallel surfaces are not useful. the point of incidence Q is MQN. Angle of In such case we use prisms, in which two incidence at Q is i. After refraction at Q, the ray refracting surfaces inclined at an angle are deviates towards the normal and strikes second used. Popular variety of prisms are having refracting surface AC at R which is the point three rectangular surfaces forming a triangle. of emergence. MRN is the normal through R. At a time two of these are taking part in the Angles of refraction at Q and R are r1 and r2 refraction. The one, not involved in refraction is respectively. called base of the prism. Fig 9.16. n Fig. 9.16: Prism consisting of three plane surfaces. Fig. 9.18: Deviation through a prism. Any section of prism perpendicular to the base After R, the ray deviates away from normal and is called principal section of the prism. Usually finally emerges along RS making e as the angle we consider all the rays in this plane. Fig 9.17 a of emergence. Incident ray PQ is extended as and 9.17 b show refraction through a prism for QT. Emergent ray RS meets QT at X if traced monochromatic and white beams respectively. backward. Angle TXS is angle of deviation δ. Angular dispersion is shown for white beam. ∠ AQN = ARN  900 …… (Angles at normal) ∴ From quadrilateral AQNR, A + ∠ QNR = 1800 --- (9.8) From ∆ QNR, r1 + r2 + ∠ QNR = 1800 --- (9.9) ∴ From Eqs. (9.8) and (9.9), A  r1  r2 --- (9.10) Fig. 9.17 (a): Refraction through a prism Angle δ is exterior angle for triangle XQR. (monocromatic light).  XQR  XRQ     i  r1    e  r2       i  e    r1  r2    Hence, using Eq. (9.10),  i  e    A     i  e  A   --- (9.11) Fig. 9.17 (b): Angular dispersion through a Deviation curve, minimum deviation and prism. (white light). prism formula: From the relations (9.10) and Relations between the angles involved: (9.11), it is clear that δ ,e,r1 and r2 depend upon Figure 9.18 shows principal section ABC of a i, A and n. After a certain minimum value of prism of absolute refractive index n kept in air. angle of incidence imin, the emergent ray is Refracting surfaces AB and AC are inclined at possible. This is because of the fact that for angle A, which is refracting angle of prism or i< imin , r2 > ic and there is total internal reflection simply ‘angle of prism’. Surface BC is the base. at the second surface and there is no emergent A monochromatic ray PQ obliquely strikes first ray. This will be shown later. Then onwards, 172  sin i as i increases, r1 increases as sin r = n but r2 (I) Grazing emergence and minimum angle 1 and e decrease. However, variation in δ with of incidence: At the point of emergence, the increasing i is different. It is as plotted in the ray travels form a denser medium into rarer Fig. 9.19. (popular prisms are of denser material, kept 1  1  in rarer). Thus if r2  sin  n  is the critical   angle, the angle of emergence e = 90. This 0 is called grazing emergence or we say that the ray just emerges. Angle of prism A is constant for a given prism and A  r1  r2. Hence the corresponding r1 and i will have their minimum possible values. Fig. 9.19: Deviation curve for a prism. It shows that, with increasing values of i, the angle of deviation δ decreases initially to a certain minimum  m  and then increases. It should also be noted that the curve is not a symmetric parabola, but the slope in the part after is less. It is clear that except at    m , (Angle of minimum deviation) there are two (II) For commonly used glass prisms, values of i for any given δ.By applying the 1  1  principle of reversibility of light to path PQRS n = 1.5, sin 1    sin 1   it is obvious that if one of these values is i, n  1.5  the other must be e and vice versa. Thus at  410 49'   r2 max    m , we have i = e. Also, in this case, r1 = r2 A If prism is symmetric (equilateral), and A = r1 + r2 = 2r   r  A  600  r1  600  410 49’  18011’ 2 Only in this case QR is parallel to base BC and the figure is symmetric.  imin  27055’  280. Using these in Eq. (9.11), we get, (III) For a symmetric (equilateral) prism, i  i  A  m  i  A  m  the prism formula can be written as 2  60   m     According to Snell’s law, sin  sin  30  m    A  m  n  2   2  sin    60  sin  30  n   2  sin    A  2  sin   --- (9.12) 2     2sin  30  m  Equation (9.12) is called prism formula.  2  Example 9.9: For a glass (n =1.5) prism having refracting angle 600, determine the range of (IV) For a prism of denser material, angle of incidence for which emergent ray kept in a rarer medium, the incident ray is possible from the opposite surface and the deviates towards the normal during the corresponding angles of emergence. Also first refraction and away from the normal calculate the angle of incidence for which during second refraction. However, during i = e. How much is the corresponding angle of both the refractions it deviates towards the minimum deviation? base only. 173  Solution: As shown in the box above, separated. This is angular dispersion (Fig. 9.20). imin = 27055'. Angle of emergence for this is emax = 900. From the principle of reversibility of light, =imax 90 = 0 and emin 27055’ Also, from the box above, n  60   m  sin   n  2  Fig. 9.20: Angular dispersion through a  60  prism. sin    2  It is measured for any two component   colours.  sin  30  m   21   2  1  2   m    2sin  30   sin  30   2  Normally we do it for extreme colours. For white light, violet and red are the extreme colours.  VR   V   R Using deviation for thin prism (Eq. 9.13), we can write  21   2  1  A  n2  1  A  n1  1  A  n2  n1  where n1 and n2 are refractive indices for the i  e  A   and i  e for    m two colours. ∴ i + i = 60 + 37° 10′ = 97°10′ ∴ i = 48°35′ Also, Thin prisms: Prisms having refracting angle  VR   V   R  A  nV  1  A  nR  1   less than 100 A  100 are called thin prisms. For such prisms we can comfortably use  A  nV  nR  --- (9.14) sin   . For such prisms to deviate the incident Yellow is practically chosen to be the mean ray towards the base during both refractions, it colour for violet and red. is essential that i should also be less than 100 so This gives mean deviation that all the other angles will also be small.   Thus  VR  V R   Y  A  nY  1 --- (9.15) 2 Do you know ?  i  nr1 and e  nr2 (i) If you see a rainbow widthwise, yellow Using these in Eq. (9.11), we get, appears to be centrally located. Hence i  e  nr1  nr2  n  r1  r2   nA  A   angular deviation of yellow is average   A  n  1 --- (9.13) for the entire colour span. This may be A and n are constant for a given prism. Thus, the reason for choosing yellow as the for a thin prism, for small angles of incidences, mean colour. Remember, red band is angle of deviation is constant (independent of widest and violet is much thinner than blue. angle of incidence). (ii) While obtaining the expression for ω, Angular dispersion and mean deviation: we have used thin prism formula for δ. As discussed earlier, if a polychromatic beam However, the expression for ω (equation is incident upon a prism, the emergent beam 9.16) is valid as well for equilateral consists of all the individual colours angularly prisms or right-angled prisms. 174 Dispersive power: Ability of an optical is called a mirage (Fig. 9.21). material to disperse constituent colours is its dispersive power. It is measured for any two colours as the ratio of angular dispersion to the mean deviation for those two colours. Thus, for the extreme colours of white light, dispersive power is given by   V   R   V  R V  R  Y Fig. 9.21: The Mirage.  2  On a hot day the air in contact with the   road is hottest and as we go up, it gets gradually A  nV  nR  nV  nR   --- (9.16) cooler. The refractive index of air thus increases A  nY  1 nY  1 with height. As shown in the figure, due to this gradual change in the refractive index, the ray of As ω is the ratio of same physical quantities, light coming from the top of an object becomes it is unitless and dimensionless quantity. From more and more horizontal as it almost touches the expression in terms of refractive indices the road. For some reason (mentioned later) it it should be understood that dispersive power bends above. Then onwards, upward bending depends only upon refractive index (hence continues due to denser air. As a result, for an material only) and not upon the dimensions of observer, it appears to be coming from below prism. For commonly used glasses it is around thereby giving an illusion of reflection from an 0.03. (imaginary) water surface. Example 10: For a dense flint glass prism of Rainbow: Undoubtedly, rainbow is an eye- refracting angle 100, obtain angular deviation catching phenomenon occurring due to rains for extreme colours and dispersive power of and Sunlight. It is most popular because it is dense flint glass. ( observable from anywhere on the Earth. A V  A(nV  1)  10(1.792  1)  (7.92) few lucky persons might have observed two  R  A(nR  1)  10(1.712  1)  (7.12)) rainbows simultaneously one above the other. Some might have seen a complete circular  Angulardispersion,VR  V   R   0.8   0 rainbow from an aeroplane (Of course, this time dispersive power, ω = it’s not a bow!). Optical phenomena discussed  R till now are sufficient to explain the formation  V of a rainbow.  V   R   2  The facts to be explained are:   (i) It is seen during rains and on the opposite  7.92  7.12   2  side of the Sun.  7.92  7.12  (ii) It is seen only during mornings and 2  0.8 evenings and not throughout the day.   0.11064 15.04 (iii) In the commonly seen rainbow red arch is (This is much higher than popular crown glass) outside and violet is inside. 9.9 Some natural phenomena due to Sunlight: (iv) In the rarely occurring concentric Mirage: On a hot clear Sunny day, along secondary rainbow, violet arch is outside a level road, a pond of water appears to be and red is inside. there ahead. However, if we physically reach (v) It is in the form of arc of a circle. the spot, there is nothing but the dry road and (vi) Complete circle can be seen from a higher water pond again appears ahead. This illusion altitude, i.e., from an aeroplane. 175  (vii) Total internal reflection is not possible in emerge from V′ and R′ and can be seen by an this case. observer on the ground. For the observer they Conditions necessary for formation of a appear to be coming from opposite side of rainbow: Light shower with relatively large the Sun. Minimum deviation rays of red and raindrops, morning or evening time and enough violet colour are inclined to the ground level at Sunlight. θR = 42.8° ≅ 43° and θV = 40.8 ≅ 41° respectively. Optical phenomena involved: During the As a result, in the ‘bow’ or arch, the red is above formation of a rainbow, the rays of Sunlight or outer and violet is lower or inner. incident on water drops, deviate and disperse during refraction, internally (NOT total A internally) reflect once (for primary rainbow) White or twice (for secondary rainbow) and finally sunlight refract again into air. At all stages there is angular dispersion which leads to clear separation of the colours. Primary rainbow: Figure 9.22 (a) shows the optical phenomena involved in the formation of a primary rainbow due to a spherical water drop. Observer on ground Fig. 9.22 (a): Formation of primary rainbow. Do you know ? Possible reasons for the upward bending at the road during mirage could be: (i) Angle of incidence at the road is glancing. At glancing incidence, the reflection coefficient is very large which causes reflection. (ii) Air almost in contact with the road is not steady. The non-uniform motion of the air bends the ray upwards and once it has bent upwards, it continues to do Fig. 9.22 (b): Formation of secondary so. rainbow. (iii) Using Maxwell’s equations for EM Secondary rainbow: Figure 9.22 (b) shows waves, correct explanation is possible some optical phenomena involved in the for the reflection. formation of a secondary rainbow due to a It may be pointed out that total internal spherical water drop. White ray AB from the reflection is NEVER possible here because Sun strikes from lower portion of a water drop the relative refractive index is just less than 1 at an incident angle i. On entering into water, it and hence the critical angle (discussed in the deviates and disperses into constituent colours. article 9.6) is also approaching 900. Extreme colours violet(V) and red(R) are shown. White ray AB from the Sun strikes from upper Refracted rays BV and BR finally emerge the portion of a water drop at an incident angle i. drop from V' and R' after suffering two internal On entering into water, it deviates and disperses reflections and can be seen by an observer on into constituent colours. Extreme colours the ground. Minimum deviation rays of red and violet(V) and red(R) are shown. Refracted rays violet colour are inclined to the ground level at BV and BR strike the opposite inner surface θR ≅ 51° and θV ≅ 53° respectively. As a result, of water drop and suffer internal (NOT total in the ‘bow’ or arch, the violet is above or outer internal) reflection. These reflected rays finally and red is lower or inner. 176  Do you know ? of i and r. Again, by using Figs. a and b, we (I) Why total internal reflection is not can obtain the corresponding angles θ R and θ V possible during formation of a rainbow? at the horizontal, which is the visible angular Angle of incidence i in air, at the water position for the rainbow. drop, can’t be greater than 90°. As a result, (III) Why is the rainbow a bow or an arch? angle of refraction r in water will always less Can we see a complete circular rainbow? than the critical angle. From Fig a and b and Figure c illustrates formation of primary by simple geometry, it is clear that this r itself and secondary rainbows with their common is the angle centre O is the point where the line joining of incidence the sun and the observer meets the Earth at any point when extended. P is location of the observer. for one or Different colours of rainbows are seen on more internal arches of cones of respective angles described reflections. earlier. Fig. a O b v i o u s l y, total internal reflection is not possible. (II) Why is rainbow seen only for a definite angle range with respect to the ground? For clear visibility we must have a beam of enough intensity. From the deviation curve (Fig 9.19) it is clear that near minimum deviation the curve is almost parallel to x-axis, i.e., for majority of angles of incidence in this range, the angle of deviation is nearly the same Fig. c and those are almost parallel Smallest half angle refers to the cone of forming a beam violet colour of primary rainbow, which is of enough 410. As the Sun rises, the common centre of intensity. Thus, the rainbows moves down. Hence as the Sun Fig. b comes up, smaller and smaller part of the the rays in the near vicinity rainbows will be seen. If the Sun is above of minimum 410, violet arch of primary rainbow cannot deviation are almost parallel to each other. be seen. Obviously beyond 530, nothing will Rays beyond this range suffer wide angular be seen. That is why rainbows are visible dispersion and thus will not have enough only during mornings and evenings. intensity for visibility. However, if observer moves up (may be in By using simple geometry for Figs. an aeroplane), the line PO itself moves up a and b it can be shown that the angle of making lower part of the arches visible. deviation between final emergent ray and the After a certain minimum elevation, entire incident ray is δ = π + 2i - 4r during primary circle for all the cones can be visible. rainbow, and δ = 2π + 2i - 6r during secondary (IV) Size of water drops convenient for rainbow. Using these relations and Snell’s rainbow: Water drops responsible for the law sini = nsinr , we can obtain derivatives formation of a rainbow should not be too of δ. Second derivative of δ comes out small. For too small drops the phenomenon to be negative, which shows that it is the of diffraction (redistribution of energy due minima condition. Equating first derivative to obstacles, discussed in XIIth standard) to zero we can obtain corresponding values dominates and clear rainbow can’t be seen. 177 9.10 Defects of lenses (aberrations of optical images): As mentioned in the section 9.4 for aberration for curved mirrors, while deriving various relations, we assume most of the rays to be paraxial by using lenses of small aperture. In reality, we have objects of finite sizes. Also, Fig. 9.24: Chromatic aberration: (a) we need optical devices of large apertures Convex lens. (lenses and/or mirrors of size few meters for telescopes, etc.). In such cases the beam of rays is no more paraxial, quite often not parallel also. As a result, the spherical oberration discussed for spherical mirrors can occur for lenses also. Only one defect is mentioned corresponding to monochromatic beam of light. Chromatic aberration: In case of mirrors there is no dispersion of light due to refractive index. However, lenses are prepared by using a transparent material medium having different Fig. 9.24: Chromatic aberration: (b) refractive index for different colours. Hence Concave lens angular dispersion is present. A convex lens can Reducing/eliminating chromatic aberration: be approximated to two thin prisms connected base to base and for a concave lens those are Eliminating chromatic aberration vertex to vertex. (Fig. 9.23 (a) and 9.23 (b)) simultaneously for all the colours is impossible. We try to eliminate it for extreme colours which reduces it for other colours. Convenient methods to do it use either a convex and a concave lens in contact or two thin convex lenses with proper separation. Such a combination is called achromatic combination. Achromatic combination of two lenses in contact: Let ω1 and ω2 be the dispersive powers of materials of the two component lenses used Fig. 9.23: (a) Convex lens (b) Concave lens in contact for an achromatic combination. Their focal lengths f for violet, red and yellow If the lens is thick, this will result into (assumed to be the mean colour) are suffixed by notably different foci corresponding to each respective letters V, R and Y. colour for a polychromatic beam, like a white light. This defect is called chromatic  1 1  Also, let K1     for lens 1 and aberration, violet being focused closest to pole  R1 R2 1 as it has maximum deviation. (Fig 9.24 (a) and  1 1  9.24 (b)) Longitudinal chromatic aberration, K 2     for lens 2. transverse chromatic aberration and circle of  R1 R2  2 least confusion are defined in the same manner For two thin lenses in contact, as that of spherical aberration for spherical 1 1 1 ……   mirrors. f f1 f 2 To be used separately for respective colours. For the combination to be achromatic, the 178 resultant focal length of the combination must Example 9.11: After Cataract operation, a be the same for both the colours, i.e., person is recommended with concavo-convex 1 1 spectacles of curvatures 10 cm and 50 cm. = fV f=R or  fV fR Crown glass of refractive indices 1.51 for red 1 1 1 1 and 1.53 for violet colours is used for this.     Calculate the lateral chromatic aberration f1V f 2V f1 R f 2 R occurring due to these glasses. (n1V-1) K1 + (n2V-1) K2 = (n1R-1) K1 + (n2R-1) K2 Solution: For a concavo-concave lens, both …… using lens makers’ Eq. (9.7) the radii of curvature are either positive or both K1 n2V  n2 R negative. If convex shape faces object, both   --- (9.17) will be positive. See the accompanying figure. K 2 n1V  n1R For mean colour yellow, 1 1 1    fY f1Y f 2Y 1 with    n1Y  1 K1  Fig. Concavo-convex f1Y lens with convex face 1 receiving incident rays and   n2Y  1 K 2 f 2Y K1  n2Y  1   f 2Y   R1  10 cmandR2  50 cm     --- (9.18) K 2  n1Y  1   f1Y   1 1   1 1  1 Equating R.H.S. of (9.17) and (9.18) and         0.08cm  1 R R2   10 50  rearranging, we can write f 2Y  n  n2 R   n1R  n1R     2V    f1Y  n2Y  1   n1Y  1    2 --- (9.19) 1 Equation (9.19) is the condition for achromatic 1  1 1  and   nV  1    combination of two lenses, in contact. fV  R1 R2  Dispersive power ω is always positive. Thus,  1.53  1  0.08  0.0424 one of the lenses must be convex and the other concave.  fV  23.58cm  If second lens is concave, f 2Y is negative. ∴ Longitudinal chromatic aberration 1 1 1 = fV - fR=25.51 - 23.58 ∴   = 1.93 cm,... (quite appreciable!) fY f1Y f 2Y For this combination to be converging, fY Verify that you get the same answer even should be positive. if you consider the concave surface facing the Hence, f1Y < f 2Y and 1  2 incident rays. Thus, for an achromatic combination if there Spherical aberration: Longitudinal spherical is a choice between flint glass ( n = 1.655) aberration, transverse spherical aberration and and crown glass ( n = 1.517 ), the convergent circle of least confusion are defined in the same (convex) lens must be of crown glass and the manner as that for spherical mirrors. (Fig 9.25 divergent (concave) lens of flint glass. (a) and 9.25 (b)) 179  9.11 Optical instruments: Introduction: Whether an object appears bigger or not does not necessarily depend upon its own size. Huge mountains far off may appear smaller than a small tree close to us. This is because the angle subtended by the mountain at the eye from that distance (called the visual angle) is smaller than that subtended by the tree Fig. 9.25 (a): Spherical aberration, Convex from its position. Hence, apparent size of an lens. object depends upon the visual angle subtended by the object from its position. Obviously, for an object to appear bigger, we must bring it closer to us or we should go closer to it. However, due to the limitation for focusing the eye lens it is not possible to take an object closer than a certain distance. This distance is called least distance of distance vision D. For a normal, unaided human eye D = 25cm. If an object is brought closer than this, we cannot see it clearly. If an object is too small (like Fig. 9.25 (b): Spherical aberration, Concave the legs of an ant), the corresponding visual lens angle from 25 cm is not enough to see it and Methods to reduce/eliminate spherical if we bring it closer than that, its image on the aberration of lenses: retina is blurred. Also, the visual angle made (i) Cheapest method to reduce the spherical by cosmic objects far away from us (such as aberration is to use a planoconvex or stars) is too small to make out minor details and planoconcave lens with curved side facing we cannot bring those closer. In such cases we the incident rays (real object). Reversing it need optical instruments such as a microscope increases the aberration appreciably. in the former case and a telescope in the latter. (ii) Certain ratio of radii of curvature for a It means that microscopes and telescopes help given refractive index almost eliminates us in increasing the visual angle. This is called the spherical aberration. For n = 1.5, the angular magnification or magnifying power. R1 1 1 Magnifying power: Angular magnification or ratio is = and for n = 2, it is R2 6 5 magnifying power of an optical instrument is (iii) Use of two thin converging lenses defined as the ratio of the visual angle made separated by distance equal to difference by the image formed by that optical instrument between their focal lengths with lens of (β) to the visual angle subtended by the object larger focal length facing the incident rays when kept at the least distance of distinct vision considerably reduces spherical aberration. ( α ). (Figure 9.26 (a) and 9.26 (b)) In the case of telescopes, α is the angle subtended by the (iv) Spherical aberration of a convex lens is object from its own position as it is not possible positive (for real image), while that of a to get it closer. concave lens is negative. Thus, a suitable combination of them (preferably a double Simple microscope or a reading glass: In convex lens of smaller focal length and order to read very small letters in a newspaper, a planoconcave lens of greater focal sometimes we use a convex lens. You might length) can completely eliminate spherical have seen watch-makers using a special type aberration. of small convex lens while looking at very tiny 180 parts of a wrist watch. Convex lens used for D D this purpose is a simple microscope.  M max   1 u f (ii) For minimum magnifying power, v  , i.e., u = f (numerically) D D Fig. 9.26: (a) Visual Angle α.  M min   u f Thus the angular magnification by a lens of D  D focal length f is between  f  and  1 f      only. For common human eyesight, D = 25 cm. Thus, if f = 5 cm, Fig. 9.26: (b) Visual Angle β. Figure 9.26 (a) shows visual angle α made D  D M min     5and M max   1    6 by an object, when kept at the least distance of  f   f . distinct vision D. Without an optical instrument Hence image appears to be only 5 to 6 times this is the greatest possible visual angle as we bigger for a lens of focal length 5 cm. cannot get the object closer than this. Figure D v 9.26 (b) shows a convex lens forming erect, For M min   f   5 , v  . ∴ m  u  .   virtual and magnified image of the same object, Thus, the image size is infinite times that of the when placed within the focus. The visual angle object, but appears only 5 times bigger. β of the object and the image in this case are For the same. However, this time the viewer is D looking at the image which is not closer than M max  1     6 , D. Hence the same object is now at a distance  f  25 smaller than D. It makes β greater then α and v  25cm.Correspondingu  6 cm the same object appears bigger. v Angular magnification or magnifying ∴ m= = 6. Thus, image size is 6 times u power, in this case, is given by that of the object, and appears also 6 times larger. Example 9.12: A magni