10.6 A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. (a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? 273 Reprint 2025-26 Physics Chapter Eleven DUAL NATURE OF RADIATION AND MATTER 11.1 INTRODUCTION The Maxwell’s equations of electromagnetism and Hertz experiments on the generation and detection of electromagnetic waves in 1887 strongly established the wave nature of light. Towards the same period at the end of 19th century, experimental investigations on conduction of electricity (electric discharge) through gases at low pressure in a discharge tube led to many historic discoveries. The discovery of X-rays by Roentgen in 1895, and of electron by J. J. Thomson in 1897, were important milestones in the understanding of atomic structure. It was found that at sufficiently low pressure of about 0.001 mm of mercury column, a discharge took place between the two electrodes on applying the electric field to the gas in the discharge tube. A fluorescent glow appeared on the glass opposite to cathode. The colour of glow of the glass depended on the type of glass, it being yellowish-green for soda glass. The cause of this fluorescence was attributed to the radiation which appeared to be coming from the cathode. These cathode rays were discovered, in 1870, by William Crookes who later, in 1879, suggested that these rays consisted of streams of fast moving negatively charged particles. The British physicist J. J. Thomson (1856-1940) confirmed this hypothesis. By applying mutually perpendicular electric and magnetic fields across the discharge 274 tube, J. J. Thomson was the first to determine experimentally the speed Reprint 2025-26 Dual Nature of Radiation and Matter and the specific charge [charge to mass ratio (e/m)] of the cathode ray particles. They were found to travel with speeds ranging from about 0.1 to 0.2 times the speed of light (3 ×108 m/s). The presently accepted value of e/m is 1.76 × 1011 C/kg. Further, the value of e/m was found to be independent of the nature of the material/metal used as the cathode (emitter), or the gas introduced in the discharge tube. This observation suggested the universality of the cathode ray particles. Around the same time, in 1887, it was found that certain metals, when irradiated by ultraviolet light, emitted negatively charged particles having small speeds. Also, certain metals when heated to a high temperature were found to emit negatively charged particles. The value of e/m of these particles was found to be the same as that for cathode ray particles. These observations thus established that all these particles, although produced under different conditions, were identical in nature. J. J. Thomson, in 1897, named these particles as electrons, and suggested that they were fundamental, universal constituents of matter. For his epoch-making discovery of electron, through his theoretical and experimental investigations on conduction of electricity by gasses, he was awarded the Nobel Prize in Physics in 1906. In 1913, the American physicist R. A. Millikan (1868-1953) performed the pioneering oil-drop experiment for the precise measurement of the charge on an electron. He found that the charge on an oil-droplet was always an integral multiple of an elementary charge, 1.602 × 10–19 C. Millikan’s experiment established that electric charge is quantised. From the values of charge (e) and specific charge (e/m), the mass (m) of the electron could be determined.

10.7 POLARISATION Consider holding a long string that is held horizontally, the other end of which is assumed to be fixed. If we move the end of the string up and down in a periodic manner, we will generate a wave propagating in the +x direction (Fig. 10.17). Such a wave could be described by the following equation FIGURE 10.17 (a) The curves represent the displacement of a string at t = 0 and at t = Dt, respectively when a sinusoidal wave is propagating in the +x-direction. (b) The curve represents the time variation of the displacement at x = 0 when a sinusoidal wave is propagating in the +x-direction. At x = Dx, the time variation of the displacement will be slightly displaced to the right. y (x,t) = a sin (kx – wt) (10.15) where a and w(= 2pn) represent the amplitude and the angular frequency of the wave, respectively; further, 2 π λ = (10.16) k represents the wavelength associated with the wave. We had discussed propagation of such waves in Chapter 14 of Class XI textbook. Since the displacement (which is along the y direction) is at right angles to the direction of propagation of the wave, we have what is known as a transverse wave. Also, since the displacement is in the y direction, it is often referred to as a y-polarised wave. Since each point on the string moves on a straight line, the wave is also referred to as a linearly polarised 269 Reprint 2025-26 Physics wave. Further, the string always remains confined to the x-y plane and therefore it is also referred to as a plane polarised wave. In a similar manner we can consider the vibration of the string in the x-z plane generating a z-polarised wave whose displacement will be given by z (x,t) = a sin (kx – wt) (10.17) It should be mentioned that the linearly polarised waves [described by Eqs. (10.15) and (10.17)] are all transverse waves; i.e., the displacement of each point of the string is always at right angles to the direction of propagation of the wave. Finally, if the plane of vibration of the string is changed randomly in very short intervals of time, then we have what is known as an unpolarised wave. Thus, for an unpolarised wave the displacement will be randomly changing with time though it will always be perpendicular to the direction of propagation. Light waves are transverse in nature; i.e., the electric field associated with a propagating light wave is always at right angles to the direction of propagation of the wave. This can be easily demonstrated using a simple polaroid. You must have seen thin plastic like sheets, which are called polaroids. A polaroid consists of long chain molecules aligned in a particular direction. The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get absorbed. Thus, if an unpolarised light wave is incident on such a polaroid then the light wave will get linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules; this direction is known as the pass-axis of the polaroid. Thus, if the light from an ordinary source (like a sodium lamp) passes through a polaroid sheet P1, it is observed that its intensity is reduced by half. Rotating P1 has no effect on the transmitted beam and transmitted intensity remains constant. Now, let an identical piece of polaroid P2 be placed before P1. As expected, the light from the lamp is reduced in intensity on passing through P2 alone. But now rotating P1 has a dramatic effect on the light coming from P2. In one position, the intensity transmitted by P2 followed by P1 is nearly zero. When turned by 90° from this position, P1 transmits nearly the full intensity emerging from P2 (Fig. 10.18). The experiment at figure 10.18 can be easily understood by assuming that light passing through the polaroid P2 gets polarised along the pass- axis of P2. If the pass-axis of P2 makes an angle q with the pass-axis of P1, then when the polarised beam passes through the polaroid P2, the component E cos q (along the pass-axis of P2) will pass through P2. Thus, as we rotate the polaroid P1 (or P2), the intensity will vary as: I = I0 cos2q (10.18) where I0 is the intensity of the polarized light after passing through P1. This is known as Malus’ law. The above discussion shows that the Reprint 2025-26 Wave Optics FIGURE 10.18 (a) Passage of light through two polaroids P2 and P1. The transmitted fraction falls from 1 to 0 as the angle between them varies from 0° to 90°. Notice that the light seen through a single polaroid P1 does not vary with angle. (b) Behaviour of the electric vector when light passes through two polaroids. The transmitted polarisation is the component parallel to the polaroid axis. The double arrows show the oscillations of the electric vector. intensity coming out of a single polaroid is half of the incident intensity. By putting a second polaroid, the intensity can be further controlled from 50% to zero of the incident intensity by adjusting the angle between the pass-axes of two polaroids. Polaroids can be used to control the intensity, in sunglasses, windowpanes, etc. Polaroids are also used in photographic cameras and 3D movie cameras. Example 10.2 Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids? Solution Let I0 be the intensity of polarised light after passing through the first polariser P1. Then the intensity of light after passing through second polariser P2 will be I = I 0cos 2θ, where q is the angle between pass axes of P1 and P2. Since P1 and P3 are crossed the angle between the pass axes of P2 and P3 will be (p/2–q). Hence the intensity of light emerging from P3 will be  π  – θ I = I 0 cos 2θ cos 2  2  EXAMPLE = I0 cos2q sin2q =(I0/4) sin22q Therefore, the transmitted intensity will be maximum when q = p/4. 10.2 271 Reprint 2025-26 Physics SUMMARY 1. Huygens’ principle tells us that each point on a wavefront is a source of secondary waves, which add up to give the wavefront at a later time. 2. Huygens’ construction tells us that the new wavefront is the forward envelope of the secondary waves. When the speed of light is independent of direction, the secondary waves are spherical. The rays are then perpendicular to both the wavefronts and the time of travel is the same measured along any ray. This principle leads to the well known laws of reflection and refraction. 3. The principle of superposition of waves applies whenever two or more sources of light illuminate the same point. When we consider the intensity of light due to these sources at the given point, there is an interference term in addition to the sum of the individual intensities. But this term is important only if it has a non-zero average, which occurs only if the sources have the same frequency and a stable phase difference. 4. Young’s double slit of separation d gives equally spaced interference fringes. 5. A single slit of width a gives a diffraction pattern with a central λ 2λ maximum. The intensity falls to zero at angles of ± , ± , etc., a a with successively weaker secondary maxima in between. 6. Natural light, e.g., from the sun is unpolarised. This means the electric vector takes all possible directions in the transverse plane, rapidly and randomly, during a measurement. A polaroid transmits only one component (parallel to a special axis). The resulting light is called linearly polarised or plane polarised. When this kind of light is viewed through a second polaroid whose axis turns through 2p, two maxima and minima of intensity are seen. POINTS TO PONDER 1. Waves from a point source spread out in all directions, while light was seen to travel along narrow rays. It required the insight and experiment of Huygens, Young and Fresnel to understand how a wave theory could explain all aspects of the behaviour of light. 2. The crucial new feature of waves is interference of amplitudes from different sources which can be both constructive and destructive, as shown in Young’s experiment. 3. Diffraction phenomena define the limits of ray optics. The limit of the ability of microscopes and telescopes to distinguish very close objects is set by the wavelength of light. 4. Most interference and diffraction effects exist even for longitudinal waves like sound in air. But polarisation phenomena are special to transverse waves like light waves. Reprint 2025-26 Wave Optics EXERCISES

12.6 DE BROGLIE’S EXPLANATION OF BOHR’S SECOND POSTULATE OF QUANTISATION Of all the postulates, Bohr made in his model of the atom, perhaps the most puzzling is his second postulate. It states that the angular momentum of the electron orbiting around the nucleus is quantised (that is, Ln = nh/2p; n = 1, 2, 3 …). Why should the angular momentum have only those values that are integral multiples of h/2p? The French physicist Louis de Broglie explained this puzzle in 1923, ten years after Bohr proposed his model. We studied, in Chapter 11, about the de Broglie’s hypothesis that material particles, such as electrons, also have a wave nature. C. J. Davisson and L. H. Germer later experimentally verified the wave nature of electrons in 1927. Louis de Broglie argued that the electron in its circular orbit, as proposed by Bohr, must be seen as a particle wave. In analogy to waves travelling on a string, particle waves too can lead to standing waves under resonant conditions. From FIGURE 12.8 A standing wave Chapter 14 of Class XI Physics textbook, we know that when is shown on a circular orbit a string is plucked, a vast number of wavelengths are excited. where four de Broglie wavelengths fit into theHowever only those wavelengths survive which have nodes circumference of the orbit. at the ends and form the standing wave in the string. It means that in a string, standing waves are formed when the total distance travelled by a wave down the string and back is one wavelength, two wavelengths, or any integral number of wavelengths. Waves with other wavelengths interfere with themselves upon reflection and their amplitudes quickly drop to zero. For an electron moving in nth circular orbit of radius rn, the total distance is the circumference of the orbit, 3012prn. Thus Reprint 2025-26 Physics 2p rn = nl, n = 1, 2, 3... (12.12) Figure 12.8 illustrates a standing particle wave on a circular orbit for n = 4, i.e., 2prn = 4l, where l is the de Broglie wavelength of the electron moving in nth orbit. From Chapter 11, we have l = h/p, where p is the magnitude of the electron’s momentum. If the speed of the electron is much less than the speed of light, the momentum is mvn. Thus, l = h/ mvn. From Eq. (12.12), we have 2p rn = n h/mvn or m vn rn = nh/2p This is the quantum condition proposed by Bohr for the angular momentum of the electron [Eq. (12.15)]. In Section 12.5, we saw that this equation is the basis of explaining the discrete orbits and energy levels in hydrogen atom. Thus de Broglie hypothesis provided an explanation for Bohr’s second postulate for the quantisation of angular momentum of the orbiting electron. The quantised electron orbits and energy states are due to the wave nature of the electron and only resonant standing waves can persist. Bohr’s model, involving classical trajectory picture (planet-like electron orbiting the nucleus), correctly predicts the gross features of the hydrogenic atoms*, in particular, the frequencies of the radiation emitted or selectively absorbed. This model however has many limitations. Some are: (i) The Bohr model is applicable to hydrogenic atoms. It cannot be extended even to mere two electron atoms such as helium. The analysis of atoms with more than one electron was attempted on the lines of Bohr’s model for hydrogenic atoms but did not meet with any success. Difficulty lies in the fact that each electron interacts not only with the positively charged nucleus but also with all other electrons. The formulation of Bohr model involves electrical force between positively charged nucleus and electron. It does not include the electrical forces between electrons which necessarily appear in multi-electron atoms. (ii) While the Bohr’s model correctly predicts the frequencies of the light emitted by hydrogenic atoms, the model is unable to explain the relative intensities of the frequencies in the spectrum. In emission spectrum of hydrogen, some of the visible frequencies have weak intensity, others strong. Why? Experimental observations depict that some transitions are more favoured than others. Bohr’s model is unable to account for the intensity variations. Bohr’s model presents an elegant picture of an atom and cannot be generalised to complex atoms. For complex atoms we have to use a new and radical theory based on Quantum Mechanics, which provides a more complete picture of the atomic structure. * Hydrogenic atoms are the atoms consisting of a nucleus with positive charge +Ze and a single electron, where Z is the proton number. Examples are hydrogen atom, singly ionised helium, doubly ionised lithium, and so forth. In these 302 atoms more complex electron-electron interactions are nonexistent. Reprint 2025-26 Atoms SUMMARY 1. Atom, as a whole, is electrically neutral and therefore contains equal amount of positive and negative charges. 2. In Thomson’s model, an atom is a spherical cloud of positive charges with electrons embedded in it. 3. In Rutherford’s model, most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus (typically one by ten thousand the size of an atom), and the electrons revolve around it. 4. Rutherford nuclear model has two main difficulties in explaining the structure of atom: (a) It predicts that atoms are unstable because the accelerated electrons revolving around the nucleus must spiral into the nucleus. This contradicts the stability of matter. (b) It cannot explain the characteristic line spectra of atoms of different elements. 5. Atoms of most of the elements are stable and emit characteristic spectrum. The spectrum consists of a set of isolated parallel lines termed as line spectrum. It provides useful information about the atomic structure. 6. To explain the line spectra emitted by atoms, as well as the stability of atoms, Niel’s Bohr proposed a model for hydrogenic (single elctron) atoms. He introduced three postulates and laid the foundations of quantum mechanics: (a) In a hydrogen atom, an electron revolves in certain stable orbits (called stationary orbits) without the emission of radiant energy. (b) The stationary orbits are those for which the angular momentum is some integral multiple of h/2p. (Bohr’s quantisation condition.) That is L = nh/2p, where n is an integer called the principal quantum number. (c) The third postulate states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency (n) of the emitted photon is then given by hn = Ei – Ef An atom absorbs radiation of the same frequency the atom emits, in which case the electron is transferred to an orbit with a higher value of n. Ei + hn = Ef 7. As a result of the quantisation condition of angular momentum, the electron orbits the nucleus at only specific radii. For a hydrogen atom it is given by  n 2   h  2 4 πε0 rn = 2  m   2 π  e The total energy is also quantised: me 4 E n = − 2 2 2 8n ε0 h = –13.6 eV/n2 The n = 1 state is called ground state. In hydrogen atom the ground state energy is –13.6 eV. Higher values of n correspond to excited states (n > 1). Atoms are excited to these higher states by collisions with other atoms or electrons or by absorption of a photon of right frequency. 303 Reprint 2025-26 Physics 8. de Broglie’s hypothesis that electrons have a wavelength λ = h/mv gave an explanation for Bohr’s quantised orbits by bringing in the wave- particle duality. The orbits correspond to circular standing waves in which the circumference of the orbit equals a whole number of wavelengths. 9. Bohr’s model is applicable only to hydrogenic (single electron) atoms. It cannot be extended to even two electron atoms such as helium. This model is also unable to explain for the relative intensities of the frequencies emitted even by hydrogenic atoms. POINTSPOINTSPOINTSPOINTSPOINTS TOTOTOTOTO PONDERPONDERPONDERPONDERPONDER 1. Both the Thomson’s as well as the Rutherford’s models constitute an unstable system. Thomson’s model is unstable electrostatically, while Rutherford’s model is unstable because of electromagnetic radiation of orbiting electrons. 2. What made Bohr quantise angular momentum (second postulate) and not some other quantity? Note, h has dimensions of angular momentum, and for circular orbits, angular momentum is a very relevant quantity. The second postulate is then so natural! 3. The orbital picture in Bohr’s model of the hydrogen atom was inconsistent with the u quantum mechanics in which Bohr’s orbits are regions where the electron may be found with large probability. 4. Unlike the situation in the solar system, where planet-planet gravitational forces are very small as compared to the gravitational force of the sun on each planet (because the mass of the sun is so much greater than the mass of any of the planets), the electron-electron electric force interaction is comparable in magnitude to the electron- nucleus electrical force, because the charges and distances are of the same order of magnitude. This is the reason why the Bohr’s model with its planet-like electron is not applicable to many electron atoms. 5. Bohr laid the foundation of the quantum theory by postulating specific orbits in which electrons do not radiate. Bohr’s model include only one quantum number n. The new theory called quantum mechanics supports Bohr’s postulate. However in quantum mechanics (more generally accepted), a given energy level may not correspond to just one quantum state. For example, a state is characterised by four quantum numbers (n, l, m, and s), but for a pure Coulomb potential (as in hydrogen atom) the energy depends only on n. 6. In Bohr model, contrary to ordinary classical expectation, the frequency of revolution of an electron in its orbit is not connected to the frequency of spectral line. The later is the difference between two orbital energies divided by h. For transitions between large quantum numbers (n to n – 1, n very large), however, the two coincide as expected. 7. Bohr’s semiclassical model based on some aspects of classical physics and some aspects of modern physics also does not provide a true picture of the simplest hydrogenic atoms. The true picture is quantum mechanical affair which differs from Bohr model in a number of fundamental ways. But then if the Bohr model is not strictly correct, why do we bother about it? The reasons which make Bohr’s model still useful are: Reprint 2025-26 Atoms (i) The model is based on just three postulates but accounts for almost all the general features of the hydrogen spectrum. (ii) The model incorporates many of the concepts we have learnt in classical physics. (iii) The model demonstrates how a theoretical physicist occasionally must quite literally ignore certain problems of approach in hopes of being able to make some predictions. If the predictions of the theory or model agree with experiment, a theoretician then must somehow hope to explain away or rationalise the problems that were ignored along the way. EXERCISES