9.6 REFRACTION THROUGH A PRISM Figure 9.21 shows the passage of light through a triangular prism ABC. The angles of incidence and refraction at the first face AB are i and r1, while the angle of incidence (from glass to air) at the second face AC is r2 and the angle of refraction or emergence e. The angle between the emergent ray RS and the direction of the incident ray PQ is called the angle of deviation, d. In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°. FIGURE 9.21 A ray of light passing through a triangular glass prism. ÐA + ÐQNR = 180° From the triangle QNR, r1 + r2 + ÐQNR = 180° Comparing these two equations, we get r1 + r2 = A (9.34) The total deviation d is the sum of deviations at the two faces, d = (i – r1 ) + (e – r2 ) that is, d = i + e – A (9.35) Thus, the angle of deviation depends on the angle of incidence. A plot between the angle of deviation and angle of incidence is shown in Fig. 9.22. You can see that, in general, any given value of d, except for i = e, corresponds to two values i and hence of e. This, in fact, is expected from the symmetry of i and e in Eq. (9.35), i.e., d remains the same if i 239 Reprint 2025-26 Physics and e are interchanged. Physically, this is related to the fact that the path of ray in Fig. 9.21 can be traced back, resulting in the same angle of deviation. At the minimum deviation Dm, the refracted ray inside the prism becomes parallel to its base. We have d = Dm, i = e which implies r1 = r2. Equation (9.34) gives A 2r = A or r = (9.36) 2 In the same way, Eq. (9.35) gives Dm = 2i – A, or i = (A + Dm)/2 (9.37) The refractive index of the prism is FIGURE 9.22 Plot of angle of deviation (d) n 2 sin[( A + D m )/2] versus angle of incidence (i) for a n 21 = = (9.38) triangular prism. n1 sin[ A /2] The angles A and Dm can be measured experimentally. Equation (9.38) thus provides a method of determining refractive index of the material of the prism. For a small angle prism, i.e., a thin prism, Dm is also very small, and we get sin[( A + Dm )/2] A + Dm ) /2 ≃ ( n 21 = sin[ A /2] A /2 Dm = (n21–1)A It implies that, thin prisms do not deviate light much. 9.7 OPTICAL INSTRUMENTS A number of optical devices and instruments have been designed utilising reflecting and refracting properties of mirrors, lenses and prisms. Periscope, kaleidoscope, binoculars, telescopes, microscopes are some examples of optical devices and instruments that are in common use. Our eye is, of course, one of the most important optical device the nature has endowed us with. We have already studied about the human eye in Class X. We now go on to describe the principles of working of the microscope and the telescope. 9.7.1 The microscope A simple magnifier or microscope is a converging lens of small focal length (Fig. 9.23). In order to use such a lens as a microscope, the lens is held near the object, one focal length away or less, and the eye is positioned close to the lens on the other side. The idea is to get an erect, magnified and virtual image of the object at a distance so that it can be viewed comfortably, i.e., at 25 cm or more. If the object is at a distance f, the 240 image is at infinity. However, if the object is at a distance slightly less Reprint 2025-26 Ray Optics and Optical Instruments FIGURE 9.23 A simple microscope; (a) the magnifying lens is located such that the image is at the near point, (b) the angle subtanded by the object, is the same as that at the near point, and (c) the object near the focal point of the lens; the image is far off but closer than infinity. than the focal length of the lens, the image is virtual and closer than infinity. Although the closest comfortable distance for viewing the image is when it is at the near point (distance D @ 25 cm), it causes some strain on the eye. Therefore, the image formed at infinity is often considered most suitable for viewing by the relaxed eye. We show both cases, the first in Fig. 9.23(a), and the second in Fig. 9.23(b) and (c). The linear magnification m, for the image formed at the near point D, by a simple microscope can be obtained by using the relation 241 Reprint 2025-26 Physics v  1 1   v  m = = v – 1 – u  v f =  f  Now according to our sign convention, v is negative, and is equal in magnitude to D. Thus, the magnification is  D  m = 1 + (9.39)  f  Since D is about 25 cm, to have a magnification of six, one needs a convex lens of focal length, f = 5 cm. Note that m = h¢/h where h is the size of the object and h¢ the size of the image. This is also the ratio of the angle subtended by the image to that subtended by the object, if placed at D for comfortable viewing. (Note that this is not the angle actually subtended by the object at the eye, which is h/u.) What a single-lens simple magnifier achieves is that it allows the object to be brought closer to the eye than D. We will now find the magnification when the image is at infinity. In this case we will have to obtained the angular magnification. Suppose the object has a height h. The maximum angle it can subtend, and be clearly visible (without a lens), is when it is at the near point, i.e., a distance D. The angle subtended is then given by  h  tan θo =  D » qo (9.40) We now find the angle subtended at the eye by the image when the object is at u. From the relations h ′ v = m = h u we have the angle subtended by the image h ′ h v h tan θi = = ⋅ = »q. The angle subtended by the object, when it −v −v u −u is at u = –f.  h  θi = (9.41)  f  as is clear from Fig. 9.23(c). The angular magnification is, therefore  θi  D m = (9.42)  θo = f This is one less than the magnification when the image is at the near point, Eq. (9.39), but the viewing is more comfortable and the difference in magnification is usually small. In subsequent discussions of optical instruments (microscope and telescope) we shall assume the image to be 242 at infinity. Reprint 2025-26 Ray Optics and Optical Instruments FIGURE 9.24 Ray diagram for the formation of image by a compound microscope. The A simple microscope has a limited maximum magnification (£ 9) for realistic focal lengths. For much larger magnifications, one uses two lenses, one compounding the effect of the other. This is known as a compound world’s microscope. A schematic diagram of a compound microscope is shown in Fig. 9.24. The lens nearest the object, called the objective, forms a largestreal, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, produces the final image, which is enlarged optical and virtual. The first inverted image is thus near (at or within) the focal plane of the eyepiece, at a distance appropriate for final image formation at infinity, or a little closer for image formation at the near point. Clearly,the final image is inverted with respect to the original object. http://astro.nineplanets.org/bigeyes.html telescopes We now obtain the magnification due to a compound microscope. The ray diagram of Fig. 9.24 shows that the (linear) magnification due to the objective, namely h¢/h, equals h ′ L m O = = (9.43) h f o where we have used the result  h   h ′  tanβ =  f o =  L  Here h¢ is the size of the first image, the object size being h and fo being the focal length of the objective. The first image is formed near the focal point of the eyepiece. The distance L, i.e., the distance between the second focal point of the objective and the first focal point of the eyepiece (focal length fe) is called the tube length of the compound microscope. 243 Reprint 2025-26 Physics As the first inverted image is near the focal point of the eyepiece, we use the result from the discussion above for the simple microscope to obtain the (angular) magnification me due to it [Eq. (9.39)], when the final image is formed at the near point, is  D  m e =  1 +  [9.44(a)]  f e  When the final image is formed at infinity, the angular magnification due to the eyepiece [Eq. (9.42)] is me = (D/fe ) [9.44(b)] Thus, the total magnification [(according to Eq. (9.33)], when the image is formed at infinity, is L   D  m = m om e =    (9.45)  f o   f e  Clearly, to achieve a large magnification of a small object (hence the name microscope), the objective and eyepiece should have small focal lengths. In practice, it is difficult to make the focal length much smaller than 1 cm. Also large lenses are required to make L large. For example, with an objective with fo = 1.0 cm, and an eyepiece with focal length fe = 2.0 cm, and a tube length of 20 cm, the magnification is L   D  m = m o m e =     f o   f e  20 25 250 1 2 Various other factors such as illumination of the object, contribute to the quality and visibility of the image. In modern microscopes, multi- component lenses are used for both the objective and the eyepiece to improve image quality by minimising various optical aberrations (defects) in lenses. 9.7.2 Telescope The telescope is used to provide angular magnification of distant objects (Fig. 9.25). It also has an objective and an eyepiece. But here, the objective has a large focal length and a much larger aperture than the eyepiece. Light from a distant object enters the objective and a real image is formed in the tube at its second focal point. The eyepiece magnifies this image producing a final inverted image. The magnifying power m is the ratio of the angle b subtended at the eye by the final image to the angle a which the object subtends at the lens or the eye. Hence h f o f o m . (9.46) 244 f e h f e Reprint 2025-26 Ray Optics and Optical Instruments In this case, the length of the telescope tube is fo + fe. Terrestrial telescopes have, in addition, a pair of inverting lenses to make the final image erect. Refracting telescopes can be used both for terrestrial and astronomical observations. For example, consider a telescope whose objective has a focal length of 100 cm and the eyepiece a focal length of 1 cm. The magnifying power of this telescope is m = 100/1 = 100. Let us consider a pair of stars of actual separation 1¢ (one minute of arc). The stars appear as though they are separated by an angle of 100 × 1¢ = 100¢ =1.67°. FIGURE 9.25 A refracting telescope. The main considerations with an astronomical telescope are its light gathering power and its resolution or resolving power. The former clearly depends on the area of the objective. With larger diameters, fainter objects can be observed. The resolving power, or the ability to observe two objects distinctly, which are in very nearly the same direction, also depends on the diameter of the objective. So, the desirable aim in optical telescopes is to make them with objective of large diameter. The largest lens objective in use has a diameter of 40 inch (~1.02 m). It is at the Yerkes Observatory in Wisconsin, USA. Such big lenses tend to be very heavy and therefore, difficult to make and support by their edges. Further, it is rather difficult and expensive to make such large sized lenses which form images that are free from any kind of chromatic aberration and distortions. For these reasons, modern telescopes use a concave mirror rather than a lens for the objective. Telescopes with mirror objectives are called reflecting telescopes. There is no chromatic aberration in a mirror. Mechanical support is much less of a problem since a mirror weighs much less than a lens of equivalent optical quality, and can be supported over its entire back surface, not just over its rim. One obvious problem with a reflecting telescope is that the objective mirror focusses light inside 245 Reprint 2025-26 Physics FIGURE 9.26 Schematic diagram of a reflecting telescope (Cassegrain). the telescope tube. One must have an eyepiece and the observer right there, obstructing some light (depending on the size of the observer cage). This is what is done in the very large 200 inch (~5.08 m) diameters, Mt. Palomar telescope, California. The viewer sits near the focal point of the mirror, in a small cage. Another solution to the problem is to deflect the light being focussed by another mirror. One such arrangement using a convex secondary mirror to focus the incident light, which now passes through a hole in the objective primary mirror, is shown in Fig. 9.26. This is known as a Cassegrain telescope, after its inventor. It has the advantages of a large focal length in a short telescope. The largest telescope in India is in Kavalur, Tamil Nadu. It is a 2.34 m diameter reflecting telescope (Cassegrain). It was ground, polished, set up, and is being used by the Indian Institute of Astrophysics, Bangalore. The largest reflecting telescopes in the world are the pair of Keck telescopes in Hawaii, USA, with a reflector of 10 metre in diameter. SUMMARY 1. Reflection is governed by the equation Ði = Ðr¢ and refraction by the Snell’s law, sini/sinr = n, where the incident ray, reflected ray, refracted ray and normal lie in the same plane. Angles of incidence, reflection and refraction are i, r ¢ and r, respectively. 2. The critical angle of incidence ic for a ray incident from a denser to rarer medium, is that angle for which the angle of refraction is 90°. For i > ic, total internal reflection occurs. Multiple internal reflections in diamond (ic @ 24.4°), totally reflecting prisms and mirage, are some examples of total internal reflection. Optical fibres consist of glass fibres coated with a thin layer of material of lower refractive index. Light incident at an angle at one end comes out at the other, after multiple internal reflections, even if the fibre is bent. Reprint 2025-26 Ray Optics and Optical Instruments 3. Cartesian sign convention: Distances measured in the same direction as the incident light are positive; those measured in the opposite direction are negative. All distances are measured from the pole/optic centre of the mirror/lens on the principal axis. The heights measured upwards above x-axis and normal to the principal axis of the mirror/ lens are taken as positive. The heights measured downwards are taken as negative. 4. Mirror equation: 1 1 1 + = v u f where u and v are object and image distances, respectively and f is the focal length of the mirror. f is (approximately) half the radius of curvature R. f is negative for concave mirror; f is positive for a convex mirror. 5. For a prism of the angle A, of refractive index n 2 placed in a medium of refractive index n1, n 2 sin ( A + D m ) / 2  n 21 = = n 1 sin ( A / 2 ) where Dm is the angle of minimum deviation. 6. For refraction through a spherical interface (from medium 1 to 2 of refractive index n1 and n 2, respectively) n 2 n 1 n 2 − n 1 − = v u R Thin lens formula 1 1 1 − = v u f Lens maker’s formula 1 ( n 2 − n1 )  1 1  = − f n1  R1 R 2  R1 and R2 are the radii of curvature of the lens surfaces. f is positive for a converging lens; f is negative for a diverging lens. The power of a lens P = 1/f. The SI unit for power of a lens is dioptre (D): 1 D = 1 m–1. If several thin lenses of focal length f1, f2, f3,.. are in contact, the effective focal length of their combination, is given by 1 1 1 1 = + + + … f f 1 f 2 f 3 The total power of a combination of several lenses is P = P1 + P2 + P3 + … 7. Dispersion is the splitting of light into its constituent colour. 247 Reprint 2025-26 Physics 8. Magnifying power m of a simple microscope is given by m = 1 + (D/f), where D = 25 cm is the least distance of distinct vision and f is the focal length of the convex lens. If the image is at infinity, m = D/f. For a compound microscope, the magnifying power is given by m = me × m0 where me = 1 + (D/fe), is the magnification due to the eyepiece and mo is the magnification produced by the objective. Approximately, L D m = × f o f e where fo and fe are the focal lengths of the objective and eyepiece, respectively, and L is the distance between their focal points. 9. Magnifying power m of a telescope is the ratio of the angle b subtended at the eye by the image to the angle a subtended at the eye by the object. β f o m = = α f e where f0 and fe are the focal lengths of the objective and eyepiece, respectively. POINTS TO PONDER 1. The laws of reflection and refraction are true for all surfaces and pairs of media at the point of the incidence. 2. The real image of an object placed between f and 2f from a convex lens can be seen on a screen placed at the image location. If the screen is removed, is the image still there? This question puzzles many, because it is difficult to reconcile ourselves with an image suspended in air without a screen. But the image does exist. Rays from a given point on the object are converging to an image point in space and diverging away. The screen simply diffuses these rays, some of which reach our eye and we see the image. This can be seen by the images formed in air during a laser show. 3. Image formation needs regular reflection/refraction. In principle, all rays from a given point should reach the same image point. This is why you do not see your image by an irregular reflecting object, say the page of a book. 4. Thick lenses give coloured images due to dispersion. The variety in colour of objects we see around us is due to the constituent colours of the light incident on them. A monochromatic light may produce an entirely different perception about the colours on an object as seen in white light. 5. For a simple microscope, the angular size of the object equals the angular size of the image. Yet it offers magnification because we can keep the small object much closer to the eye than 25 cm and hence have it subtend a large angle. The image is at 25 cm which we can see. Without the microscope, you would need to keep the small object at 25 cm which would subtend a very small angle. Reprint 2025-26 Ray Optics and Optical Instruments EXERCISES

9.4 Alkynes 1 2 3 4 5 Like alkenes, alkynes are also unsaturated II. H3C–C≡ C– CH2– CH3 Pent–2-yne hydrocarbons. They contain at least one triple 4 3 2 1 bond between two carbon atoms. The number III. H3C–CH–C≡ CH 3-Methyl but–1-yne |of hydrogen atoms is still less in alkynes as CH3compared to alkenes or alkanes. Their general Structures I and II are position isomers formula is CnH2n–2. and structures I and III or II and III are chain The first stable member of alkyne series isomers. is ethyne which is popularly known as acetylene. Acetylene is used for arc welding Problem 9.13 purposes in the form of oxyacetylene flame Write structures of different isomers obtained by mixing acetylene with oxygen corresponding to the 5 th member of gas. Alkynes are starting materials for a large alkyne series. Also write IUPAC names of number of organic compounds. Hence, it all the isomers. What type of isomerism is interesting to study this class of organic is exhibited by different pairs of isomers? compounds. Solution 9.4.1 Nomenclature and Isomerism th 5 member of alkyne has the molecular In common system, alkynes are named as formula C6H10. The possible isomers are: derivatives of acetylene. In IUPAC system, they Table 9.2 Common and IUPAC Names of Alkynes (CnH2n–2) Value of n Formula Structure Common name IUPAC name 2 C2H2 H-C≡CH Acetylene Ethyne 3 C3H4 CH3-C≡CH Methylacetylene Propyne 4 C4H6 CH3CH2-C≡CH Ethylacetylene But-1-yne 4 C4H6 CH3-C≡C-CH3 Dimethylacetylene But-2-yne Reprint 2025-26 Hydrocarbons 315 (a) HC ≡ C – CH2 – CH2 – CH2 – CH3 Hex-1-yne (b) CH3 – C ≡ C – CH2 – CH2 – CH3 Hex-2-yne (c) CH3 – CH2 – C ≡ C – CH2– CH3 Hex-3-yne 3-Methylpent-1-yne 4-Methylpent-1-yne 4-Methylpent-2-yne Fig. 9.6 Orbital picture of ethyne showing (a) sigma overlaps (b) pi overlaps. orbitals of the other carbon atom, which undergo lateral or sideways overlapping to 3,3-Dimethylbut-1-yne form two pi (π) bonds between two carbon atoms. Thus ethyne molecule consists of one Position and chain isomerism shown by C–C σ bond, two C–H σ bonds and two C–C different pairs. π bonds. The strength of C≡C bond (bond enthalpy 823 kJ mol -1) is more than those 9.4.2 Structure of Triple Bond of C=C bond (bond enthalpy 681 kJ mol –1) Ethyne is the simplest molecule of alkyne and C–C bond (bond enthalpy 348 kJ mol–1). series. Structure of ethyne is shown in The C≡C bond length is shorter (120 pm) Fig. 9.6. than those of C=C (133 pm) and C–C (154 pm). Electron cloud between two carbon Each carbon atom of ethyne has two sp atoms is cylindrically symmetrical about thehybridised orbitals. Carbon-carbon sigma (σ) internuclear axis. Thus, ethyne is a linear bond is obtained by the head-on overlapping molecule. of the two sp hybridised orbitals of the two carbon atoms. The remaining sp hybridised 9.4.3 Preparation orbital of each carbon atom undergoes 1. From calcium carbide: On industrial overlapping along the internuclear axis with scale, ethyne is prepared by treating the 1s orbital of each of the two hydrogen calcium carbide with water. Calcium atoms forming two C-H sigma bonds. carbide is prepared by heating quick lime H-C-C bond angle is of 180°. Each carbon with coke. Quick lime can be obtained byhas two unhybridised p orbitals which are heating limestone as shown in the followingperpendicular to each other as well as to the reactions:plane of the C-C sigma bond. The 2p orbitals of one carbon atom are parallel to the 2p CaCO3 ∆ CaO + O2 (9.55) Reprint 2025-26 316 chemistry CaO + 3C CaC2 + CO (9.56) the sp hybridised carbon2 atoms whereas they are attached to sp hybridised carbon Calcium 3 atoms in ethene and sp hybridised carbons carbide in ethane. Due to the maximum percentage of CaC2 + 2H2O Ca(OH)2 + C2H2 (9.57) s character (50%), the sp hybridised orbitals of carbon atoms in ethyne molecules have2. From vicinal dihalides : Vicinal dihalides highest electronegativity; hence, these attract on treatment with alcoholic potassium the shared electron pair of the C-H bond of hydroxide undergo dehydrohalogenation. ethyne to a greater extent than that of the One molecule of hydrogen halide is 2 sp hybridised orbitals of carbon in ethene eliminated to form alkenyl halide which 3 and the sp hybridised orbital of carbon in on treatment with sodamide gives alkyne. ethane. Thus in ethyne, hydrogen atoms can be liberated as protons more easily as compared to ethene and ethane. Hence, hydrogen atoms of ethyne attached to triply bonded carbon atom are acidic in nature. You may note that the hydrogen atoms attached to the triply bonded carbons are acidic but not all the hydrogen atoms of alkynes. HC ≡ CH + Na → HC ≡ C–Na++ 1/2 H2 9.4.4 Properties Monosodium Physical properties ethynide Physical properties of alkynes follow the same (9.59) trend of alkenes and alkanes. First three HC ≡ C– Na + Na → Na+ C–Na+ ≡ C–Na++ 1/2 H2members are gases, the next eight are liquids and the higher ones are solids. All alkynes Disodium ethynide are colourless. Ethyene has characteristic (9.60)odour. Other members are odourless. Alkynes are weakly polar in nature. They are lighter CH3 – C ≡ C – H + Na+ NH–2 than water and immiscible with water but ↓ soluble in organic solvents like ethers, carbon CH3 – C ≡ C– Na+ + NH3 tetrachloride and benzene. Their melting Sodium propynide (9.61) point, boiling point and density increase with These reactions are not shown by alkenesincrease in molar mass. and alkanes, hence used for distinction Chemical properties between alkynes, alkenes and alkanes. What Alkynes show acidic nature, addition reactions about the above reactions with but-1-yne and and polymerisation reactions as follows : but-2-yne ? Alkanes, alkenes and alkynes A. Acidic character of alkyne: Sodium follow the following trend in their acidic metal and sodamide (NaNH2) are strong behaviour : bases. They react with ethyne to form sodium i) CH ≡ CH > H2C – CH2 > CH3 –CH3acetylide with the liberation of dihydrogen gas. These reactions have not been observed ii) HC ≡ CH > CH3 –C≡ CH >> CH3 –C≡C–CH3in case of ethene and ethane thus indicating that ethyne is acidic in nature in comparison B. Addition reactions: Alkynes contain a to ethene and ethane. Why is it so ? Has triple bond, so they add up, two molecules of it something to do with their structures dihydrogen, halogen, hydrogen halides etc. and the hybridisation ? You have read that Formation of the addition product takes place hydrogen atoms in ethyne are attached to according to the following steps. Reprint 2025-26 Hydrocarbons 317 The addition product formed depends upon stability of vinylic cation. Addition in unsymmetrical alkynes takes place according to Markovnikov rule. Majority of the reactions of alkynes are the examples of electrophilic addition reactions. A few addition reactions (9.66)are given below: (i) Addition of dihydrogen (iv) Addition of water Pt/Pd/Ni H2 Like alkanes and alkenes, alkynes are alsoHC≡CH+H2 [H2C = CH2] CH3–CH3 immiscible and do not react with water. (9.62) However, one molecule of water adds to alkynes on warming with mercuric sulphate CH3–C≡CH + H2 Pt/Pd/Ni [CH3–CH=CH2] and dilute sulphuric acid at 333 K to form Propyne Propene carbonyl compounds. ↓H2 CH3–CH2–CH3 Propane (9.63) (ii) Addition of halogens (9.67) (9.64) Reddish orange colour of the solution of bromine in carbon tetrachloride is decolourised. This is used as a test for unsaturation. (iii) Addition of hydrogen halides (9.68) Two molecules of hydrogen halides (HCl, HBr, (v) Polymerisation HI) add to alkynes to form gem dihalides (in (a) Linear polymerisation: Under suitable which two halogens are attached to the same conditions, linear polymerisation of ethyne carbon atom) takes place to produce polyacetylene or H–C≡C–H+H–Br [CH2 = CH–Br]→ CHBr2 polyethyne which is a high molecular Bromoethene weight polyene containing repeating units of CH3 (CH = CH – CH = CH ) and can be represented 1,1-Dibromoethane as —(CH = CH – CH = CH)n— Under special (9.65) conditions, this polymer conducts electricity. Reprint 2025-26 318 chemistry Thin film of polyacetylene can be used as but in a majority of reactions of aromatic electrodes in batteries. These films are good compounds, the unsaturation of benzene ring conductors, lighter and cheaper than the is retained. However, there are examples of metal conductors. aromatic hydrocarbons which do not contain a (b) Cyclic polymerisation: Ethyne on benzene ring but instead contain other highly unsaturated ring. Aromatic compoundspassing through red hot iron tube at 873K containing benzene ring are known asundergoes cyclic polymerization. Three benzenoids and those not containing amolecules polymerise to form benzene, which benzene ring are known as non-benzenoids.is the starting molecule for the preparation of Some examples of arenes are givenderivatives of benzene, dyes, drugs and large below:number of other organic compounds. This is the best route for entering from aliphatic to aromatic compounds as discussed below: Benzene Toluene Naphthalene (9.69) Biphenyl Problem 9.14 How will you convert ethanoic acid into 9.5.1 Nomenclature and Isomerism benzene? The nomenclature and isomerism of aromatic Solution hydrocarbons has already been discussed in Unit 8. All six hydrogen atoms in benzene are equivalent; so it forms one and only one type of monosubstituted product. When two hydrogen atoms in benzene are replaced by two similar or different monovalent atoms or groups, three different position isomers are possible. The 1, 2 or 1, 6 is known as the ortho (o–), the 1, 3 or 1, 5 as meta (m–) and the 1, 4 as para (p–) disubstituted compounds. A few examples of derivatives of benzene are given below:

9.15 Apply mirror equation and the condition: (a) f < 0 (concave mirror); u < 0 (object on left) (b) f > 0; u < 0 (c) f > 0 (convex mirror) and u < 0 (d) f < 0 (concave mirror); f < u < 0 to deduce the desired result.