REFLECTION AND REFRACTION OF LIGHT

Transcription

1 Relection and Reraction o Light MODULE REFLECTION AND REFRACTION OF LIGHT Light makes us to see things and is responsible or our visual contact with our immediate environment. It enables us to admire and adore various beautiul maniestations o mother nature in lowers, plants, birds, animals, and other orms o lie. Can you imagine how much shall we be deprived i we were visually impaired? Could we appreciate the brilliance o a diamond or the majesty o a rainbow? Have you ever thought how light makes us see? How does it travel rom the sun and stars to the earth and what is it made o? Such questions have engaged human intelligence since the very beginning. You will learn about some phenomena which provide answers to such questions. Look at light entering a room through a small opening in a wall. You will note the motion o dust particles, which essentially provide simple evidence that light travels in a straight line. An arrow headed straight line represents the direction o propagation o light and is called a ray; a collection o rays is called a beam. The ray treatment o light constitutes geometrical optics. In lesson 22, you will learn that light behaves as a wave. But a wave o short wavelength can be well opproximated by the ray treatment. When a ray o light alls on a mirror, its direction changes. This process is called relection. But when a ray o light alls at the boundary o two dissimilar suraces, it bends. This process is known as reraction. You will learn about relection rom mirrors and reraction rom lenses in this lesson. You will also learn about total internal relection. These phenomena ind a number o useul applications in daily lie rom automobiles and health care to communication. OBJECTIVES Ater studying this lesson, you should be able to: explain relection at curved suraces and establish the relationship between the ocal length and radius o curvature o spherical mirrors; 87

2 MODULE - 6 Relection and Reraction o Light state sign convention or spherical suraces; derive the relation between the object distance, the image distance and the ocal length o a mirror as well as a spherical reractive surace; state the laws o reraction; explain total internal relection and its applications in everyday lie; derive Newton's ormula or measuring the ocal length o a lens; describe displacement method to ind the ocal length o a lens; and derive an expression or the ocal length o a combination o lenses in contact. 20. REFLECTION OF LIGHT FROM SPHERICAL SURFACES In your earlier classes, you have learnt the laws o relection at a plane surace. Let us recall these laws : Law The incident ray, the relected ray and the normal to the relecting surace at the point o incidence always lie in the same plane. Law 2 The angle o incidence is equal to the angle o relection : i r Fig. 20. : Relection o light rom a plane surace These are illustrated in Fig Though initially stated or plane suraces, these laws are also true or spherical mirrors. This is because a spherical mirror can be regarded as made up o a large number o extremely small plane mirrors. A well-polished spoon is a amiliar example o a spherical mirror. Have you seen the image o your ace in it? Fig. 20.2(a) and 20.2 (b) show two main types o spherical mirrors. Fig : Spherical mirrors : a) a convex mirror, and b) a concave mirror 88

3 Relection and Reraction o Light Note that the relecting surace o a convex mirror curves outwards while that o a concave mirror curves inwards. We now deine a ew important terms used or spherical mirrors. The centre o the sphere, o which the mirror is a part, is called the centre o curvature o the mirror and the radius o this sphere deines its radius o curvature. The middle point O o the relecting surace o the mirror is called its pole. The straight line passing through C and O is said to be the principal axis o the mirror. The circular outline (or periphery) o the mirror is called its aperture and the angle ( MCM ) which the aperture subtends at C is called the angular aperture o the mirror. Aperture is a measure o the size o the mirror. A beam o light incident on a spherical mirror parallel to the principal axis converges to or appears to diverge rom a common point ater relection. This point is known as principal ocus o the mirror. The distance between the pole and the principal ocus gives the ocal length o the mirror. A plane passing through the ocus perpendicular to the principal axis is called the ocal plane. We will consider only small aperture mirrors and rays close to the principal axis, called paraxial rays. (The rays away rom the principal axis are called marginal or parapheral rays.) MODULE - 6 INTEXT QUESTIONS 20.. Answer the ollowing questions : (a) Which mirror has the largest radius o curvature : plane, concave or convex? (b) Will the ocal length o a spherical mirror change when immersed in water? (c) What is the nature o the image ormed by a plane or a convex mirror? (d) Why does a spherical mirror have only one ocal point? 2. Draw diagrams or concave mirrors o radii 5cm, 7cm and 0cm with common centre o curvature. Calculate the ocal length or each mirror. Draw a ray parallel to the common principal axis and draw relected rays or each mirror. 3. The radius o curvature o a spherical mirror is 30cm. What will be its ocal length i (i) the inside surace is silvered? (ii) outside surace is silvered? 4. Why are dish antennas curved? 20.. Ray Diagrams or Image Formation Let us again reer to Fig. 20.2(a) and 20.2(b). You will note that the ray o light through centre o curvature retraces its path. 89

4 MODULE - 6 Relection and Reraction o Light the ray o light parallel to the principal axis, on relection, passes through the ocus; and the ray o light through F is relected parallel to the principal axis. To locate an image, any two o these three rays can be chosen. The images are o two types : real and virtual. Real image o an object is ormed when relected rays actually intersect. These images are inverted and can be projected on a screen. They are ormed on the same side as the object in ront o the mirror (Fig. 20.3(a)). Virtual image o an object is ormed by relected rays that appear to diverge rom the mirror. Such images are always erect and virual; these cannot be projected on a screen. They are ormed behind the mirror (Fig. 20.3(b)). Fig : Image ormed by a) concave mirror, and b) convex mirror Sign Convention We ollow the sign convention based on the cartesian coordinate system. While using this convention, the ollowing points should be kept in mind:. All distances are measured rom the pole (O) o the mirror. The object is always placed on the let so that the incident ray is always taken as travelling rom let to right. 2. All the distances on the let o O are taken as negative and those on the right o O as positive. Fig : Sign convention 3. The distances measured above and normal to the principal axis are taken as positive and the downward distances as negative. The radius o curvature and the ocal length o a concave mirror are negative and those or a convex mirror are positive. 90

5 Relection and Reraction o Light 20.2 DERIVATION OF MIRROR FORMULA We now look or a relation between the object distance (u), the image distance (v) and the local length o a spherical mirror. We make use o simple geometry to arrive at a relation, which surprisingly is applicable in all situations. Reer to Fig. 20.5,which shows an object AB placed in ront o a concave mirror. The mirror produces an image A B. AX and AY are two rays rom the point A on the object AB, M is the concave mirror while XA and YA are the relected rays. Using sign conventions, we can write object distance, OB u, ocal length, OF, image distance, OB v, and radius o curvature OC 2 Consider ΔABF and ΔFOY. These are similar triangles. We can, thereore, write AB OY FB OF Similarly, rom similar triangles ΔXOF and ΔB A F, we have (20.) XO A B OF (20.2) FB But AB XO, as AX is parallel to the principal axis. Also A B OY. Since let hand sides o Eqns. (20.) and (20.2) are equal, we equate their right hand sides. Hence, we have FB OF OF FB Putting the values in terms o u, v and in Eqn. (20.3), we can write u ( ) Fig : Image ormation by a concave mirror: mirror ormula v ( ) In optics it is customary to denote object distance by v. You should not conuse it with velocity. (20.3) MODULE - 6 u+ v + 9

6 MODULE - 6 On cross multiplication, we get uv u v Relection and Reraction o Light or uv u + v Dividing throughout by uv, we get the desired relation between the ocal length and the object and image distances : v + (20.4) u We next introduce another important term magniication. This indicates the ratio o the size o image to that o the object : m size o the image size o the object h h 2 But A B AB v u h v 2 m h u Since a real image is inverted, we can write A B m AB v u To solve numerical problems, remember the ollowing steps :. (20.5) (20.5b). For any spherical mirror, use the mirror ormula: v + u 2. Substitute the numerical values o the given quantities with proper signs. 3. Do not give any sign to the quantity to be determined; it will automatically be obtained with the relevant sign. 4. Remember that the linear magniication is negative or a real image and positive or a virtual image. 5. It is always better to draw a igure beore starting the (numerical) work. INTEXT QUESTIONS A person standing near a mirror inds his head look smaller and his hips larger. How is this mirror made? 2. Why are the shaving mirrors concave while the rear view mirrors convex? Draw ray diagrams to support your answer. 92

7 Relection and Reraction o Light 3. As the position o an object in ront o a concave mirror o ocal length 25cm is varied, the position o the image also changes. Plot the image distance as a unction o the object distance letting the latter change rom x to + x. When is the image real? Where is it virtual? Draw a rough sketch in each case. 4. Give two situations in which a concave mirror can orm a magniied image o an object placed in ront o it. Illustrate your answer by a ray diagram. 5. An object 2.6cm high is 24cm rom a concave mirror whose radius o curvature is 6cm. Determine (i) the location o its image, and (ii) size and nature o the image. 6. A concave mirror orms a real image our times as tall as the object placed 5cm rom it. Find the position o the image and the radius o curvature o the mirror. 7. A convex mirror o radius o curvature 20cm orms an image which is hal the size o the object. Locate the position o the object and its image. 8. A monkey gazes in a polished spherical ball o 0cm radius. I his eye is 20cm rom the surace, where will the image o his eye orm? MODULE REFRACTION OF LIGHT When light passes obliquely rom a rarer medium (air) to a denser medium (water, glass), there is a change in its direction o propagation. This bending o light at the boundary o two dissimilar media is called reraction. When a ray o light is reracted at an interace, it obeys the ollowing two laws : Law I : The incident ray, the reracted ray and the normal to the surace at the point o incidence always lie in the same plane. Law II : The ratio o the sine o the angle o incidence to the sine o the angle o reraction is constant or a given pair o media. It is independent o the angle o incidence when light propagates rom a rarer to a denser medium. Moreover, or a light o given colour, the ratio depends only on the pair o media. This law was pronounced by the Dutch scientist Willebrord van Roijen Snell and in his honour is oten reerred to as Snell s law. According to Snell s law sin i sin r μ 2 where μ 2 is a constant, called the reractive index o second medium with respect to the irst medium, and determines how much bending would take place at the interace separating the two media. It may also be expressed as the ratio o the 93

8 MODULE - 6 Relection and Reraction o Light velocity o light in the irst medium to the velocity o light in the second medium c μ 2 c2 Table 20. : Reractive indices o some common materials Medium μ Vacuum/air Water.33 Ordinary glass.50 Crown glass.52 Reractive indices o a ew typical substances are given in Table 20..Note that these values are with respect to air or vacuum. The medium having larger reractive index is optically denser medium while the one having smaller reractive index is called rarer medium. So water is denser than air but rarer than glass. Similarly, crown glass is denser than ordinary glass but rarer than lint glass. I we consider reraction rom air to a medium like glass, which is optically denser than air[fig (a)], then r is less than i. On the other hand, i the ray passes rom water to air, r is greater than i [Fig. 20.6(b)]. That is, the reracted ray bends towards the normal on the air glass interace and bends away rom the normal on water air interace. Dense lint glass.65 Diamond 2.42 (a) (b) Fig : a) Reraction on air glass interace, and b) reraction on water air interace Willebrord Van Roijen Snell ( ) Willebrord Snell was born in 580 in Lieden. He began to study mathematics at a very young age. He entered the University o Leiden and initially studied law. But, soon he turned his attention towards mathematics and started teaching at the university by the time he was 20. In 63, Snell succeded his ather as Proessor o Mathematics. He did some important work in mathematics, including the method o calculating the approximate value o π by polygon. His method o using 96 94

9 Relection and Reraction o Light sided polygon gives the correct value o π up to seven places while the classical method only gave this value upto two correct places. Snell also published some books including his work on comets. However, his biggest contribution to science was his discovery o the laws o reraction. However, he did not publish his work on reraction. It became known only in 703, seventy seven years ater his death, when Huygens published his results in Dioptrics. MODULE Reversibility o light Reer to Fig. 20.6(b) again. It illustrates the principle o reversibility. It appears as i the ray o light is retracing its path. It is not always necessary that the light travels rom air to a denser medium. In act, there can be any combination o transparent media. Suppose, light is incident at a water-glass interace. Then, by applying Snell s law, we have sin i sin i w g μ wg (20.6) Now, let us consider separate air-glass and air-water interaces. By Snell s law, we can write sin sin i i a g μ ag and sin sin i i a w μ aw On combining these results, we get This can be rewritten as μ ag sin i g μ aw sin i w (20.7) sin i sin i w g μ μ ag aw (20.8) On comparing Eqns. (20.6) and (20.8), we get μ wg μ μ ag aw (20.9) 95

10 MODULE - 6 Relection and Reraction o Light This result shows than when light travels rom water to glass, the reractive index o glass with respect to water can be expressed in terms o the reractive indics o glass and water with respect to air. Example 20. : A ray o light is incident at an angle o 30 o at a water-glass interace. Calculate the angle o reraction. Given μ ag.5, μ aw.3. Solution : From Eqn. (20.8), we have sin iw μ sin i μ g ag aw sin 30 sin i g or sin i g or i g 25 o 4 Example 20.2 : Calculate the speed o light in water i its reractive index with respect to air is 4/3. Speed o light in vacuum ms. Solution : We know that μ c v or v c μ 8 (3 0 ms ) 4/ ms Example 20.3 : The reractive indices o glass and water are.52 and.33 respectively. Calculate the reractive index o glass with respect to water. Solution : Using Eqn. (20.9), we can write μ wg μ μ ag aw

11 Relection and Reraction o Light INTEXT QUESTIONS What would be the lateral displacement when a light beam is incident normally on a glass slab? 2. Trace the path o light i it is incident on a semicircular glass slab towards its centre when i < i c and i > i c. 3. How and why does the Earth s atmosphere alter the apparant shape o the Sun and Moon when they are near the horizon? 4. Why do stars twinkle? 5. Why does a vessel illed with water appear to be shallower (less deep) than when without water? Draw a neat ray diagram or it. 6. Calculate the angle o reraction o light incident on water surace at an angle o 52º. Take μ 4/3. MODULE TOTAL INTERNAL REFLECTION ACTIVITY 20. Take a stick, cover it with cycle grease and then dip it in water or take a narrow glass bottle, like that used or keeping Homeopathic medicines, and dip it in water. You will observe that the stick or the bottle shine almost like silver. Do you know the reason? This strange eect is due to a special case o reraction. We know that when a ray o light travels rom an optically denser to an optically rarer medium, say rom glass to air or rom water to air, the reracted ray bends away rom the normal. This means that the angle o reraction is greater than the angle o incidence. What happens to the reracted ray when the angle o incidence is increased? The bending o reracted ray also increases. However, the maximum value o the angle o reraction can be 90 o. The angle o incidence in the denser medium or which the angle o reraction in rarer medium, air in this case, equals 90º is called the critical angle, i C. The reracted ray then moves along the boundary separating the two media. I the angle o incidence is greater than the critical angle, the incident ray will be relected back in the same medium, as shown in Fig. 20.7(c). Such a relection is called Total Internal Relection and the incident ray is said to be totally internally relected. For total internal relection to take place, the ollowing two conditions must be satisied. Light must travel rom an optically denser to an optically rarer medium. The angle o incidence in the denser medium must be greater than the critical angle or the two media. 97

12 MODULE - 6 Relection and Reraction o Light The glass tube in water in Activity 20. appeared silvery as total internal relection took place rom its surace. An expression or the critical angle in terms o the reractive index may be obtained readily, using Snell s law. For reraction at the glass-air interace, we can write sin i sin r μga Fig : Reraction o light as it travels rom glass to air or a) i < i c, b) i i c and c) i > i c Table 20.2 :Critical angles or a ew substances Substance μ Critical angle Water o Crown glass o Diamond o Dense lint glass o Putting r 90º or i i c, we have sin i sin 90 c o μga or sin i c μga Hence μag sin i μ ga c The critical angles or a ew substances are given in Table 20.2 Example 20.4 : Reractive index o glass is.52. Calculate the critical angle or glass air interace. Solution : We know that μ /sin i c sin i c /μ.52 i c 42º Much o the shine in transparent substances is due to total internal relection. Can you now explain why diamonds sparkle so much? This is because the critical angle is quite small and most o the light entering the crystal undergoes multiple internal relections beore it inally emerges out o it. In ordinary relection, the relected beam is always weaker than the incident beam, even i the relecting surace is highly polished. This is due to the act that some 98

13 Relection and Reraction o Light light is always transmitted or absorbed. But in the case o total internal relection, cent percent (00%) light is relected at a transparent boundary Applications o Reraction and Total Internal Relection There are many maniestations o these phenomena in real lie situations. We will consider only a ew o them. (a) Mirage : Mirage is an optical illusion which is observed in deserts or on tarred roads in hot summer days. This, you might have observed, creates an illusion o water, which actually is not there. Due to excessive heat, the road gets very hot and the air in contact with it also gets heated up. The densities and the reractive indices o the layers immediately above the road are lower than those o the cooler higher layers. Since there is no abrupt change in medium (see Fig. 20.9), a ray o light rom a distant object, such as a tree, bends more and more as it passes through these layers. And when it alls on a layer at an angle greater than the critical angle or the two consecutive layers, total internal relection occurs. This produces an inverted image o the tree giving an illusion o relection rom a pool o water. MODULE - 6 Fig : Formation o mirage Totally Relecting Prisms : A prism with right angled isosceles trianglur base or a totally relecting prism with angles o 90 o, 45 o and 45 o is a very useul device or relecting light. Reer to Fig. 20.9(a). The symmetry o the prism allows light to be incident on O at an angle o 45º, which is greater than the critical angle or glass i.e. 42 o. As a result, light suers total internal relection and is deviated by 90º. Fig : Totally relecting prisms 99

14 MODULE - 6 Relection and Reraction o Light Choosing another ace or the incident rays, it will be seen (Fig. 20.9(b)) that the ray gets deviated through 80 0 by two successive total internal relections taking place at O and O. Optical Fibres Fig : Multiple relection in an optical ibre An optical ibre is a hair-thin structure o glass or quartz. It has an inner core which is covered by a thin layer (called cladding) o a material o slightly lower reractive index. For example, the reractive index o the core is about.7 and that o the cladding is.5. This arrangement ensures total internal relection. You can easily understand it, i you recall the conditions or total internal relection. When light is incident on one end o the ibre at a small angle, it undergoes multiple total internal relections along the ibre (Fig. 20.0). The light inally emerges with undiminished intensity at the other end. Even i the ibre is bent, this process is not aected. Today optical ibres are used in a big way. A lexible light pipe using optical ibres may be used in the examination o inaccessible parts o the body e.g. laproscopic examination o stomach, urinary bladder etc. Other medical applications o optical ibres are in neurosurgery and study o bronchi. Besides medical applications, optical ibres have brought revolutionary changes in the way we communicate now. Each ibre can carry as many as 0,000 telephone messages without much loss o intensity, to ar o places. That is why millions o people across continents can interact simultaneously on a ibre optic network. 200 INTEXT QUESTIONS Why can t total internal relection take place i the ray is travelling rom a rarer to a denser medium? 2. Critical angle or glass is 42. Would this value change i a piece o glass is immersed in water? Give reason or your answer.

15 Relection and Reraction o Light 3. Show, with the help a ray diagram how, a ray o light may be deviated through 90 o using a i) plane mirror, and ii) totally relecting prism. Why is the intensity o light greater in the second case? 4. A liquid in a container is 25cm deep. What is its apparant depth when viewed rom above, i the reractive index o the liquid is.25? What is the critical angle or the liquid? 20.5 REFRACTION AT A SPHERICAL SURFACE We can study ormation o images o objects placed around spherical suraces such as glass marbles (Kanchas), water drops, glass bottle, etc. For measuring distances rom spherical reracing suraces, we use the same sign convention as applicable or spherical mirrors. Reer to Fig SPS is a convex reracting surace separating two media, air and glass. C is its centre o curvature. P is a point on SPS almost symmetrically placed. You may call it the pole. CP is then the principal axis. O is a point object. OA is an incident ray and AB is the reracted ray. Another ray OP alls on the surace normally and goes undeviated ater reraction. PC and AB appear to come rom I. Hence I is the virtual image o O. MODULE - 6 Fig. 20. : Reraction at a spherical surace Let OAN i, the angle o incidence and CAB r, the angle o reraction. Using the proper sign convention, we can write PO u ; PI v ; PC + R Let α, β, and γ be the angles subtended by OA, IA and CA, respectively with the principal axis and h the height o the normal dropped rom A on the principal axis. In ΔOCA and ΔICA, we have i α + γ (i is exterior angle) (20.0) and r β + γ (r is exterior angle) (20.) 20

16 MODULE - 6 From Snell s law, we recall that Relection and Reraction o Light sin i sin r μ where μ is the reractive index o the glass surace with respect to air. For a surace o small aperture, P will be close to A and so i and r will be very small (sin i ~ i, sin r ~ r). The above equation, thereore, gives i μr (20.2) Substituting the values o i and r in Eqn. (20.2) rom Eqns. (20.0) and (20.) respectively, we get α + γ μ (β + γ) or α μβ γ (μ ) (20.3) As α, β and γ are very small, we can take tan α ~ α, and tan β ~ β, and tan γ ~ γ. Now reerring to ΔOAM in Fig. 20., we can write α tan α AM MO AP PO h (i M is very near to P) u β tan β AM MI AM PI h v and γ tan γ AM MC AM PC h R Substituting or α, β and γ in Eqn. (20.3), we get hu μh v (μ ) h R μ or v u μ (20.4) R This important relationship correlates the object and image distances to the reractive index μ and the radius o curvature o the reracting surace Relection through lenses A lens is a thin piece o transparent material (usually glass) having two suraces, one or both o which are curved (mostly spherical). You have read in your earlier classes that lenses are mainly o two types, namely, convex lens and concave lens. Each o them is sub-divided into three types as shown in Fig Thus, you can have plano-convex and plano-concave lenses too. 202

17 Relection and Reraction o Light Basic Nomenclature Thin lens : I the thickness o a lens is negligible in comparison to the radii o curvature o its curved suraces, the lens is reerred to as a thin lens. We will deal with thin lenses only. Principal axis is the line joining the centres o curvature o two suraces o the lens. Optical centre is the point at the centere o the lens situated on the principal axis. The rays passing through the optical centre do not deviate. Fig : Types o lenses Principal ocus is the point at which rays parallel and close to the principal axis converge to or appear to diverge. It is denoted by F (Fig. 20.3) Rays o light can pass through a lens in either direction. So every lens has two principal ocii, one on its either side. Focal length is the distance between the optical centre and the principal ocus. In Fig. 20.3, OF is ocal length (). As per the sign convention, OF is positive or a convex lens and negative or a concave lens. Focal plane is the plane passing through the ocus o a lens perpendicular to its principal axis. MODULE - 6 Fig : Foci o a) convex, and b) concave lenses Lens Maker s Formula and Magniication You can now guess that the ocal length must be related to the radius o curvature and the reractive index o the material o the lens. Suppose that a thin convex lens L is held on an optical bench (Fig. 20.4). Let the reractive index o the material o the lens with respect to air be μ and the radii o curvatures o its two suraces be R and R 2, respectively. Let a point object be situated on the principal 203

18 MODULE - 6 Relection and Reraction o Light axis at P. C and C 2 are the centres o curvature o the curved suraces and 2, respectively. µ µ 2 I µ N A B N 2 P Q 2 C 2 O R I I' Since the lens used is actually thin, points A and B may be considered very close to point a and hence C A is taken equal to C Q and C 2 B as C 2 Q. u R 2 L Fig : Point image o a point object or by a thin double convex lens A ray rom P strikes surace at A. C N is normal to surace at the point A. The ray PA travels rom the rarer medium (air) to the denser medium (glass), and bends towards the normal to proceed in the direction AB. The ray AB would meet the principal axis C 2 C at the point I in the absence o the surace 2. Similarly, another ray rom P passing through the optical centre O passes through the Point I. I is thus the virtual image o the object P. Then object distance OP u and image distance OI v (say). Using Eqn. (20.4) we can write v v u' μ v u μ R (20.5) Due to the presence o surace 2, the ray AB strikes it at B. C 2 N 2 is the normal to it at point B. As the ray AB is travelling rom a denser medium (glass) to a rarer medium (air), it bends away rom the normal C 2 N 2 and proceeds in the direction BI and meets another ray rom P at I. Thus I is image o the object P ormed by the lens. It means that image distance OI v. Considering point object O, its virtual image is I (due to surace ) and the inal image is I. I is the virtual object or surace 2 and I is the inal image. Then or the virtual object I and the inal image I, we have, object distance OI u v and image distance OI v. On applying Eqn. (20.2) and cosidering that the ray AB is passing rom glass to air, we have (/ μ ) + v v (/ μ ) R 2 or, μv v μ μr 2 204

19 Relection and Reraction o Light Multiping both sides by μ, we get μ μ v v R 2 Adding Eqns. (20.5) and (20.6), we have (20.6) MODULE - 6 v u (μ ) (20.7) R R2 It u, that is the object is at ininity, the incoming rays are parallel and ater reraction will converge at the ocus (v ). Then Eqn. (20.7) reduces to (μ ) R R2 (20.8) This is called lens maker s ormula. From Eqns. (20.7) and (20.8), we can conclude that The ocal length o a lens depends on the radii o curvature o spherical suraces. Focal length o a lens o larger radii o curvature will be more. Focal length o a lens is smaller i the reractive index o its material is high. In case a lens is dipped in water or any other transparent medium, the value o μ changes and you can actually work out that ocal length will increase. However, i the density o the medium is more than that o the material o the lens, say carbon disulphide, the rays may even diverge Newton s Formula Fig shows the image o object AB ormed at A B by a convex lens F and F 2 are the irst and second principal ocii respectively. Let us measure the distances o the object and image rom the irst ocus and second ocus respectively. Let x be the distance o object rom the irst ocus and x 2 be the distance o image rom the second ocus and and 2 the irst and second ocal lengths, respectively as shown in Fig B u L Y A F x O L 2 F 2 v x 2 F A 2 Y B Fig

20 MODULE - 6 Now, in similar s, ABF and OL F y y x Relection and Reraction o Light Also rom similar Δs, OLF 2 and A B F 2 y x y Comparing these two equations we get x x or 2 (say), then x x or 2 x x This relation is called Newton s ormula and can be conveniently used to measure the ocal length Displacement Method to ind the Position o Images (Conjugates points) In the igure , A B is the image o the object AB as ormed by a lens L. OA u and OA v. The principle o reversibility o light rays tells us that i we move the lens towards the right such that AO v, then again the image will be ormed at the same place. B L x 2 A F u v O O v F A A u 2 B Fig Thus AA D u + v...(i) and the separation between the two positions o the lens: OO x (v u)...(ii) 206

21 Relection and Reraction o Light Adding (i) and (ii) we get x+ D v 2 and, subtracting (ii) rom (i) we get D x u 2 Substituting these values in the lens ormula, we get. MODULE - 6 v ( u ) x+ D D x D+ x D x 2( D x+ D+ x) D2 x2 4D D2 x2 or D2 x2 4D Thus, keeping the positions o the object and screen ixed we can obtain equally clear, bright and sharp images o the object on the screen corresponding to the two positions o the lens. This again is a very convenient way o inding o a lens FORMATION OF IMAGES BY LENSES The ollowing properties o the rays are used in the ormation o images by lenses: A ray o light through the optical centre o the lens passes undeviated. A parallel ray, ater reraction, passes through the principal ocus. A ray o light through F or F is rendered parallel to the principal axis ater reraction. Any two o these rays can be chosen or drawing ray diagrams. The lens ormula v u suggests the dependence o the image distance (v) on the object distance (u) and the ocal length () o the lens. 207

22 MODULE - 6 Relection and Reraction o Light The magniication o a lens is deined as the ratio o the height o the image ormed by the lens to the height o the object and is denoted by m : m I O v u where I is height o the image and O the height o the object. Example 20.5 : The radii o curvature o a double convex lens are 5cm and 30cm, respectively. Calculate its ocal length. Also, calculate the ocal length when it is immersed in a liquid o reractive index.65. Take μ o glass.5. Solution : From Eqn. (20.8) we recall that (μ ) R R2 Here R + 5cm, and R 2 30cm. On substituting the given data, we get (.5 ) cm When the lens is immersed in a liquid, μ will be replaced by μ lg : μ lg μ μ ag al Thereore l (μ lg ) R R cm As is negative, the lens indeed behaves like a concave lens POWER OF A LENS A practical application o lenses is in the correction o the deects o vision. You may be using spectacles or seen other learners, parents and persons using 208

23 Relection and Reraction o Light spectacles. However, when asked about the power o their lens, they simply quote a positive or negative number. What does this number signiy? This number is the power o a lens in dioptre. The power o a lens is deined as the reciprocal o its ocal length in metre: MODULE - 6 P The S unit o power o a lens is m. Dioptre is only a commercial unit generally used by opticians. The power o a convex lens is positive and that o a concave lens is negative. Note that greater power implies smaller ocal length. Using lens maker s ormula, we can relate power o a lens to its radii o curvature: (μ ) R R2 or P (μ ) R R2 Example 20.6 : Calculate the radius o curvature o a biconvex lens with both suraces o equal radii, to be made rom glass (μ.54), in order to get a power o dioptre. Solution : P (μ ) R R2 P dioptre μ.54 R R and R 2 R Substuting the given values in lens makers ormula, we get (0.54) R R m 39 cm 20.8 COMBINATION OF LENSES Reer to Fig Two thin convex lenses A and B having ocal lengths and 2, respectively have been kept in contact with each other. O is a point object placed on the common principal axis o the lenses. 209

24 MODULE - 6 Relection and Reraction o Light Fig : Two thin convex lenses in contact Note that lens A orms the image o object O at I. This image serves as the virtual object or lens B and the inal image is thus ormed at I. I v be the object distance and v the image distance or the lens A, then using the lens ormula, we can write v u (20.9) I v is the inal image distance or the lens B, we have v v 2 (20.20) Note that in writing the above expression, we have taken v as the object distance or the thin lens B. Adding Eqns. (20.9) and (20.20), we get v u + 2 (20.2) I the combination o lenses is replaced by a single lens o ocal length F such that it orms the image o O at the same position I, then this lens is said to be equivalent to both the lenses. It is also called the equivalent lens or the combination. For the equivalent lens, we can write v u F (20.22) where F +. (20.23) 2 20

25 Relection and Reraction o Light I P is power o the equivalent lens and P and P 2 are respectively the powers o individual lenses, then P P + P 2 (20.24) Note that Eqns.(20.23) and (20.24) derived by assuming two thin convex lenses in contact also hold good or any combination o two thin lenses in contact (the two lenses may both be convex, or concave or one may be concave and the other convex). Example 20.7 : Two thin convex lenses o ocal lengths 20cm and 40cm are in contact with each other. Calculate the ocal length and the power o the equivalent lens. Solution : The ormula or the ocal length o the combination F gives F MODULE or F cm 0.33cm Power o the equivalent lens is P F dioptre. INTEXT QUESTIONS On what actors does the ocal length o a lens depend? 2. A lens, whose radii o curvature are dierent, is used to orm the image o an object placed on its axis. I the ace o the lens acing the object is reversed, will the position o the image change? 3. The reractive index o the material o an equi-double convex lens is.5. Prove that the ocal length is equal to the radius o curvature. 4. What type o a lens is ormed by an air bubble inside water? 5. A lens when immersed in a transparent liquid becomes invisible. Under what condition does this happen? 6. Calculate the ocal length and the power o a lens i the radii o curvature o its two suraces are +20cm and 25cm (μ.5). 2

26 MODULE - 6 Relection and Reraction o Light 7. Is it possible or two lenses in contact to produce zero power? 8. A convex lens o ocal length 40cm is kept in contact with a concave lens o ocal length 20cm. Calculate the ocal length and the power o the combination. Deects in image ormation Lenses and mirrors are widely used in our daily lie. It has been observed that they do not produce a point image o a point object. This can be seen by holding a lens against the Sun and observing its image on a paper. You will note that it is not exactly circular. Mirrors too do not produce a perect image. The deects in the image ormation are known as aberrations. The aberrations depend on (i) the quality o lens or mirror and (ii) the type o light used. Two major aberrations observed in lenses and mirrors, are (a) spherical aberration and (b) chromatic aberration. These aberration produce serious deects in the images ormed by the cameras, telescopes and miscroscopes etc. Spherical Aberration This is a monochromatic deect in image ormation which arises due to the sphericity and aperture o the reracting or relecting suraces. The paraxial rays and the marginal rays orm images at dierent points I p and I m respectively (Fig. 20.6) Fig :Spherical aberration in a) spherical mirror, and b) lens. I p is image ormed by the paraxial rays and I m that ormed by the marginal rays. The spherical aberration in both mirrors and lenses can be reduced by allowing only the paraxial rays to be incident on the surace. It is done by using stops. Alternatively, the paraxial rays may be cut-o by covering the central portion, thus allowing only the marginal or parapheral rays to orm the image. However, the use o stops reduces the brightness o the image. A much appreciated method is the use o elliptical or parabolic mirrors. 22

27 Relection and Reraction o Light The other methods to minimize spherical aberration in lenses are : use o plano convex lenses or using a suitable combination o a convex and a concave lens. Chromatic Aberration in Lenses A convex lens may be taken as equivalent to two small-angled prisms placed base to base and the concave lens as equivalent to such prisms placed vertex to vertex. Thus, a polychromatic beam incident on a lens will get dispersed. The parallel beam will be ocused at dierent coloured ocii. This deect o the image ormed by spherical lenses is called chromatic aberration. It occurs due to the dispersion o a polychromatic incident beam (Fig Obviously the red colour is ocused arther rom the lens while the blue colour is ocused nearer the lens (in a concave lens the ocusing o the red and blue colours takes place in the same manner but on the opposite side o it). MODULE - 6 F v F r F r F v Fig. 20.7: Chromatic aberration To remove this deect we combine a convergent lens o suitable material and ocal length when combined with a divergent lens o suitable ocal length and material. Such a lens combination is called an achromatic doublet. The ocal length o the concave lens can be ound rom the necessary condition or achromatism given by ω 2 + ω 0 2 WHAT YOU HAVE LEARNT Real image is ormed when relected rays actually intersect ater relection. It can be projected on a screen. The ocal length is hal o the radius o curvature. 2 R 23

28 MODULE - 6 Relection and Reraction o Light The object and image distances are related to magniication as m u v Reraction o light results in change in the speed o light when it travels rom one medium to another. This causes the rays o light to bend towards or away rom the normal. The reractive index μ determines the extent o bending o light at the interace o two media. Snell s law is mathematically expressed as sini μ2 sin r where i is the angle o incidence in media and r is the angle o reraction in media 2. Total internal relection is a special case o reraction wherein light travelling rom a denser to a rarer media is incident at an angle greater than the critical angle: μ sin i Any transparent media bounded by two spherical suraces or one spherical and one plane surace orms a lens. Images by lenses depend on the local length and the distance o the object rom it. Convex lenses are converging while concave lenses are diverging. C (μ ) R R2 m u v and v u are simple relationships between the ocal length (), the reractive index, the radii o curvatures (R, R 2 ), the object distance (u) and the image distance (v). Newton's ormula can be used to measure the ocal length o a lens. Displacement method is a very convenient way o inding ocal length o a lens. 24

29 Relection and Reraction o Light Power o a lens indicates how diverging or converging it is: MODULE - 6 P Power is expressed in dioptre. (or m in SI units) The ocal length F o an equivalent lens when two their lenses o ocal lengths and 2 one kept in contact is given by F + 2 TERMINAL EXERCISES. List the uses o concave and convex mirrors. 2. What is the nature and position o image ormed when the object is at (i) ininity (ii) 2 (iii) in case o concave mirror and convex mirror. 3. List the actors on which lateral displacement o an incident ray depends as it suers reraction through a parallel-sided glass slab? Why is the lateral displacement larger i angle o incidence is greater. Show this with the help o a ray diagram. 4. State conditions or total internal relection o light to take place. 5. How is +.5 dioptre dierent rom.5 dioptre? Deine dioptre. 6. Why does the intensity o light become less due to reraction? 7. A lamp is 4m rom a wall. Calculate the ocal length o a concave mirror which orms a ive times magniied image o the lamp on the wall. How ar rom the wall must the mirror be placed? 8. A dentist s concave mirror has a radius o curvature o 30cm. How ar must it be placed rom a cavity in order to give a virtual image magniied ive times? 9. A needle placed 45cm rom a lens orms an image on a screen placed 90cm on the other side o the lens. Identiy the type o the lens and determine its ocal length. What is the size o the image, i the size o the needle is 5.0cm? 0. An object o size 3.0cm is placed 4cm in ront o a concave lens o ocal length 2 cm. Describe the nature o the image by the lens. What happens i the object is moved arther rom the lens?. An object is placed at a distance o 00cm rom a double convex lens which orms a real image at a distance o 20cm. The radii o curvature o the suraces 25

30 MODULE - 6 Relection and Reraction o Light o a lens are 25cm and 2.5 cm respectively. Calculate the reractive index o the material o the lens. 2. A ray o light is travelling rom diamond to glass. Calculate the value o the critical angle or the ray, i the reractive index o glass is.5 and that o diamond A small object is placed at a distance o 5cm rom two coaxial thin convex lenses in contact. I the ocal length o each lens is 20cm. Calculate the ocal length and the power o the combination and the distance between the object and its image. 4. While inding the ocal length o a convex lens, an object was kept at a distance o 65.0 cm rom the screen. Tow positions o the lens or which clear image o the object was ormed on the screen were obtained. The distance between these two positions was ound to be 5 cm. Calculate the ocal length o the given lens. ANSEWERS TO INTEXT QUESTIONS 20.. (a) plane mirror (its radius o curvature is ininitely large). (b) No. The ocal length o a spherical mirror is hal o its radius o curvature ( ~ R/2) and has nothing to do with the medium in which it is immersed. (c) Virtual (d) This is because the rays parallel to the principal axis converge at the ocal point F; and the rays starting rom F, ater relection rom the mirror, become parallel to the principal axis. Thus, F serves both as the irst and the second ocal point. 2. Focal lengths : 2.5cm, 3.5cm, 5cm. F 3 F 2 F P 3 P 2 P 5cm 7cm 0cm 5cm 35cm 2.5cm 26

31 Relection and Reraction o Light 3. 5cm; +5cm. 4. The dish antennas are curved so that the incident parallel rays can be ocussed on the receiver The upper part o the mirror must be convex and its lower part concave. 2. Objects placed close to a concave mirror give an enlarged image. Convex mirrors give a diminished erect image and have a larger ield o view. MODULE or u >, we get real image; u 2 is a special case when an object kept as the centre o curvature o the mirror orms a real image at this point itsel (v 2 ). For u <, we get virtual image. 4. When (i) u <, and (ii) < u < 2. 27

32 MODULE - 6 Relection and Reraction o Light 5. (i) 2cm in ront o mirror, real and inverted, (ii) 0.8cm 6. v 60cm, R 24cm 7. u 0cm, v +5cm 8. v 4cm No lateral displacement. 2. r > i when i < i Total internal relection where i > i c 3. The density o air and hence its reractive index decrease as we go higher in altitude. As a result, the light rays rom the Sun, when it is below the horizon, pass rom the rarer to the denser medium and bend towards the normal, till they are received by the eye o the observer. This causes the shape to appear elongated. 4. Due to the change in density o the dierent layers o air in the atmosphere, μ changes continuously. Thereore, the reractive index o air varies at dierent levels o atmosphere. This along with air currents causes twinkling o stars. 5. Due to reraction point P appears at P Total internal relection cannot take place i the ray travels rom a rarer to a denser medium as the angle o reraction will always be less than the angle o incidence. 2. Yes the critical angle will change as μ ag sin i c μ μ ωg μ ag aw 28

33 Relection and Reraction o Light MODULE The intensity in the second case is more due to total internal relection cm, i c sin No. Changing the position o R and R 2 in the lens maker s ormula does not aect the value o. So the image will be ormed in the same position. 3. Substitute R R; R 2 R and μ.5 in the lens maker s ormula. You will get R. 4. Concave lens. But it is shaped like a convex lens. 5. This happens when the reractive index o the material o the lens is the same as that o the liquid cm and P 4.5 dioptre 7. Yes, by placing a convex and a concave lens o equal ocal length in contact cm, 2.5 dioptre Answers to Problems in Terminal Exercise , 5m. 8. 2cm 9. 30cm, size o image 0cm, converging lens 0. The image is erect, virtual and diminished in size, and located at 8.4cm rom the lens on the same side as the object. As the object is moved away rom the lens, the virtual image moves towards the ocus o the lens but never beyond and progressively diminshes in size.. μ º 3. 0cm, 0 dioptre, 45 cm cm 29

34 MODULE - 6 Dispersion and Scattering o Light 2 DISPERSION AND SCATTERING OF LIGHT In the previous lesson you have learnt about relection, reraction and total internal relection o light. You have also learnt about image ormation by mirrors and lenses and their uses in daily lie. When a narrow beam o ordinary light is reracted by a prism, we see colour bands. This phenomenon has to be other than relection or reraction. The splitting o white light into its constituent colours or wavelengths by a medium is called dispersion. In this lesson, you will study about this phenomenon. A beautiul maniestation o this phenomenon in nature is in the orm o rainbow. You will also learn in this lesson about the phenomenon o scattering o light, which gives sky its blue colour and the sun red colour at sunrise and sunset. Elementary idea o Raman eect will also be discussed in this lesson. OBJECTIVES Ater studying this lesson, you should be able to : explain dispersion o light; derive relation between the angle o deviation (δ), angle o prism (A) and reractive index o the material o the prism (μ); relate the reractive index with wavelength and explain dispersion through a prism; explain ormation o primary and secondary rainbows; explain scattering o light and list its applications.and; and explain Raman eect. 2. DISPERSION OF LIGHT Natural phenomena like rings around planets (halos) and ormation o rainbow etc. cannot be explained by the rectilinear propagation o light. To understand 220

35 Dispersion and Scattering o Light such events, light is considered as having wave nature. (You will learn about it in the next lesson.) As you know, light waves are transverse electromagnetic waves which propagate with speed ms in vacuum. O the wide range o electromagnetic spectrum, the visible light orms only a small part. Sunlight consists o seven dierent wavelengths corresponding to seven colours. Thus, colours may be identiied with their wavelengths. You have already learnt that the speed and wavelength o waves change when they travel rom one medium to another. The speed o light waves and their corresponding wavelengths also change with the change in the medium. The speed o a wave having a certain wavelength becomes less than its speed in ree space when it enters an optically denser medium. The reractive index μ has been deined as the ratio o the speed o light in vacuum to the speed o light in the medium. It means that the reractive index o a given medium will be dierent or waves having wavelengths m and m because these waves travel with dierent speeds in the same medium. This variation o the reractive index o a material with wavelength is known as dispersion. This phenomenon is dierent rom reraction. In ree space and even in air, the speeds o all waves o the visible light are the same. So, they are not separated. (Such a medium is called a non-dispersive medium.) But in an optically denser medium, the component wavelengths (colours) travel with dierent speeds and thereore get separated. Such a medium is called dispersive medium. Does this suggest that light will exhibit dispersion whenever it passes through an optically denser medium. Let us learn about it now. MODULE Dispersion through a Prism The separation o colours by a medium is not a suicient condition to observe dispersion o light. These colours must be widely separated and should not mix up again ater emerging rom the dispersing medium. A glass slab (Fig. 2.) is not suitable or observing dispersion as the rays o the emergent beam are very close and parallel to the incident beam Fig. 2. : Passage o light through a glass slab Newton used a prism to demonstrate dispersion o light. Reer to Fig White light rom a slit alls on the ace AB o the prism and light emerging rom ace AC is seen to split into dierent colours. Coloured patches can be seen on a screen. The ace AC increases the separation between the rays reracted at the ace AB. The incident white light PQ thus splits up into its component seven colours : Violet, indigo, blue, green, yellow, orange and red (VIBGYOR). The wavelengths travelling with dierent speeds are reracted through dierent angles and are thus 22