8.2 A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to 213 a 230 V ac supply with a (angular) frequency of 300 rad s–1. Reprint 2025-26 Physics (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates. FIGURE 8.6 8.3 What physical quantity is the same for X-rays of wavelength 10–10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500m? 8.4 A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength? 8.5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? 8.6 A charged particle oscillates about its mean equilibrium position with a frequency of 10 9 Hz. What is the frequency of the electromagnetic waves produced by the oscillator? 8.7 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave? 8.8 Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is n = 50.0 MHz. (a) Determine, B0,w, k, and l. (b) Find expressions for E and B. 8.9 The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hn (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation? 8.10 In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m–1. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field. [c = 3 × 108 m s–1.] Reprint 2025-26

10.3 (a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in vacuum is 3.0 × 108 m s–1) (b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?

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