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

9.31 Figure 9.30 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0cm. What is the refractive index of the liquid? FIGURE 9.30 253 Reprint 2025-26 Physics Notes Reprint 2025-26 Wave Optics Chapter Ten WAVE OPTICS 10.1 INTRODUCTION In 1637 Descartes gave the corpuscular model of light and derived Snell’s law. It explained the laws of reflection and refraction of light at an interface. The corpuscular model predicted that if the ray of light (on refraction) bends towards the normal then the speed of light would be greater in the second medium. This corpuscular model of light was further developed by Isaac Newton in his famous book entitled OPTICKS and because of the tremendous popularity of this book, the corpuscular model is very often attributed to Newton. In 1678, the Dutch physicist Christiaan Huygens put forward the wave theory of light – it is this wave model of light that we will discuss in this chapter. As we will see, the wave model could satisfactorily explain the phenomena of reflection and refraction; however, it predicted that on refraction if the wave bends towards the normal then the speed of light would be less in the second medium. This is in contradiction to the prediction made by using the corpuscular model of light. It was much later confirmed by experiments where it was shown that the speed of light in water is less than the speed in air confirming the prediction of the wave model; Foucault carried out this experiment in 1850. The wave theory was not readily accepted primarily because of Newton’s authority and also because light could travel through vacuum255 Reprint 2025-26 Physics and it was felt that a wave would always require a medium to propagate from one point to the other. However, when Thomas Young performed his famous interference experiment in 1801, it was firmly established that light is indeed a wave phenomenon. The wavelength of visible light was measured and found to be extremely small; for example, the wavelength of yellow light is about 0.6 mm. Because of the smallness of the wavelength of visible light (in comparison to the dimensions of typical mirrors and lenses), light can be assumed to approximately travel in straight lines. This is the field of geometrical optics, which we had discussed in the previous chapter. Indeed, the branch of optics in which one completely neglects the finiteness of the wavelength is called geometrical optics and a ray is defined as the path of energy propagation in the limit of wavelength tending to zero. After the interference experiment of Young in 1801, for the next 40 years or so, many experiments were carried out involving the interference and diffraction of lightwaves; these experiments could only be satisfactorily explained by assuming a wave model of light. Thus, around the middle of the nineteenth century, the wave theory seemed to be very well established. The only major difficulty was that since it was thought that a wave required a medium for its propagation, how could light waves propagate through vacuum. This was explained when Maxwell put forward his famous electromagnetic theory of light. Maxwell had developed a set of equations describing the laws of electricity and magnetism and using these equations he derived what is known as the wave equation from which he predicted the existence of electromagnetic waves*. From the wave equation, Maxwell could calculate the speed of electromagnetic waves in free space and he found that the theoretical value was very close to the measured value of speed of light. From this, he propounded that light must be an electromagnetic wave. Thus, according to Maxwell, light waves are associated with changing electric and magnetic fields; changing electric field produces a time and space varying magnetic field and a changing magnetic field produces a time and space varying electric field. The changing electric and magnetic fields result in the propagation of electromagnetic waves (or light waves) even in vacuum. In this chapter we will first discuss the original formulation of the Huygens principle and derive the laws of reflection and refraction. In Sections 10.4 and 10.5, we will discuss the phenomenon of interference which is based on the principle of superposition. In Section 10.6 we will discuss the phenomenon of diffraction which is based on Huygens- Fresnel principle. Finally in Section 10.7 we will discuss the phenomenon of polarisation which is based on the fact that the light waves are transverse electromagnetic waves. * Maxwell had predicted the existence of electromagnetic waves around 1855; it was much later (around 1890) that Heinrich Hertz produced radiowaves in the laboratory. J.C. Bose and G. Marconi made practical applications of the Hertzian 256 waves Reprint 2025-26 Wave Optics

9.27 (a) m = ( fO/fe) = 28 f O  f O  (b) m = 1 + = 33.6 f e  25  349 Reprint 2025-26 Physics 9.28 (a) fO + fe = 145 cm (b) Angle subtended by the tower = (100/3000) = (1/30) rad. Angle subtended by the image produced by the objective h h = = f O 140 Equating the two, h = 4.7 cm. (c) Magnification (magnitude) of the eye-piece = 6. Height of the final image (magnitude) = 28 cm. 9.29 The image formed by the larger (concave) mirror acts as virtual object for the smaller (convex) mirror. Parallel rays coming from the object at infinity will focus at a distance of 110 mm from the larger mirror. The distance of virtual object for the smaller mirror = (110 –20) = 90 mm. The focal length of smaller mirror is 70 mm. Using the mirror formula, image is formed at 315 mm from the smaller mirror. 9.30 The reflected rays get deflected by twice the angle of rotation of the mirror. Therefore, d/1.5 = tan 7°. Hence d = 18.4 cm. 9.31 n = 1.33 CHAPTER 10 10.1 (a) Reflected light: (wavelength, frequency, speed same as incident light) l = 589 nm, n = 5.09 ´ 1014 Hz, c = 3.00 ´ 108 m s–1 (b) Refracted light: (frequency same as the incident frequency) n = 5.09 ´ 1014Hz v = (c/n) = 2.26 × 108 m s–1, l = (v/n) = 444 nm 10.2 (a) Spherical (b) Plane (c) Plane (a small area on the surface of a large sphere is nearly planar). 10.3 (a) 2.0 × 108 m s–1 (b) No. The refractive index, and hence the speed of light in a medium, depends on wavelength. [When no particular wavelength or colour of light is specified, we may take the given refractive index to refer to yellow colour.] Now we know violet colour deviates more than red in a glass prism, i.e. nv > nr. Therefore, the violet component of white light travels slower than the red component. 1.2 10 – 2  0.28 10 – 3 10.4  m = 600 nm 4 14. 10.5 K/4 10.6 (a) 1.17 mm (b) 1.56 mm 10.7 0.15° 350 10.8 tan–1(1.5) ~ 56.3o Reprint 2025-26 Answers