DC Pandey Optics and Modern Physics PDF

 Understanding Physics JEE Main & Advanced OPTICS AND MODERN PHYSICS  Understanding Physics JEE Main & Advanced OPTICS AND MODERN PHYSICS DC PANDEY [B.Tech, M.Tech, Pantnagar, ID 15722] ARIHANT PRAKASHAN (Series), MEERUT  Understanding Physics JEE Main...

 Understanding Physics JEE Main & Advanced OPTICS AND MODERN PHYSICS  Understanding Physics JEE Main & Advanced OPTICS AND MODERN PHYSICS DC PANDEY [B.Tech, M.Tech, Pantnagar, ID 15722] ARIHANT PRAKASHAN (Series), MEERUT  Understanding Physics JEE Main & Advanced ARIHANT PRAKASHAN (Series), MEERUT All Rights Reserved © SARITA PANDEY No part of this publication may be re-produced, stored in a retrieval system or by any means, electronic, mechanical, photocopying, recording, scanning, web or otherwise without the written permission of the publisher. Arihant has obtained all the information in this book from the sources believed to be reliable and true. However, Arihant or its editors or authors or illustrators don’t take any responsibility for the absolute accuracy of any information published and the damage or loss suffered thereupon. All disputes subject to Meerut (UP) jurisdiction only. ADMINISTRATIVE & PRODUCTION OFFICES Regd. Office 'Ramchhaya' 4577/15, Agarwal Road, Darya Ganj, New Delhi -110002 Tele: 011- 47630600, 43518550; Fax: 011- 23280316 Head Office Kalindi, TP Nagar, Meerut (UP) - 250002 Tel: 0121-2401479, 2512970, 4004199; Fax: 0121-2401648 SALES & SUPPORT OFFICES Agra, Ahmedabad, Bengaluru, Bareilly, Chennai, Delhi, Guwahati, Hyderabad, Jaipur, Jhansi, Kolkata, Lucknow, Meerut, Nagpur & Pune. ISBN 978-93-13190-59-2 Published by ARIHANT PUBLICATIONS (I) LTD. For further information about the books published by Arihant, log on to www.arihantbooks.com or e-mail at [email protected] /arihantpub /@arihantpub Arihant Publications /arihantpub Understanding Physics JEE Main & Advanced PREFACE The overwhelming response to the previous editions of this book gives me an immense feeling of satisfaction and I take this an opportunity to thank all the teachers and the whole student community who have found this book really beneficial. In the present scenario of ever-changing syllabus and the test pattern of JEE Main & Advanced, the NEW EDITION of this book is an effort to cater all the difficulties being faced by the students during their preparation of JEE Main & Advanced. The exercises in this book have been divided into two sections viz., JEE Main & Advanced. Almost all types and levels of questions are included in this book. My aim is to present the students a fully comprehensive textbook which will help and guide them for all types of examinations. An attempt has been made to remove all the printing errors that had crept in the previous editions. I am extremely thankful to (Dr.) Mrs. Sarita Pandey, Mr. Anoop Dhyani and Nisar Ahmad for their endless efforts during the project. Comments and criticism from readers will be highly appreciated and incorporated in the subsequent editions. DC Pandey  Understanding Physics JEE Main & Advanced CONTENTS 29. ELECTROMAGNETIC WAVES 1-19 29.1 Introduction 29.3 Electromagnetic Waves 29.2 Displacement Current 29.4 Electromagnetic Spectrum 30. REFLECTION OF LIGHT 21-70 30.1 Introduction 30.3 Reflection of Light 30.2 General Concepts used in 30.4 Reflection from a Spherical Surface Geometrical Optics 31. REFRACTION OF LIGHT 71-190 31.1 Refraction of Light and Refractive 31.5 Refraction from Spherical Surface Index of a Medium 31.6 Lens Theory 31.2 Law of Refraction (Snell’s Law) 31.7 Total Internal Reflection (TIR) 31.3 Single Refraction from Plane 31.8 Refraction Through Prism Surface 31.9 Deviation 31.4 Shift due to a Glass Slab 31.10 Optical Instruments 32. INTERFERENCE AND DIFFRACTION OF LIGHT 191-251 32.1 Principle of Superposition 32.4 Young's Double Slit 32.2 Resultant Amplitude and Intensity Experiment (YDSE) due to Coherent Sources 32.5 Introduction to Diffraction 32.3 Interference 32.6 Diffraction from a Narrow Slit 33. MODERN PHYSICS-I 253-323 33.1 Dual Nature of Electromagnetic 33.3 Momentum and Radiation Pressure Waves 33.4 de-Broglie Wavelength of Matter 33.2 Electromagnetic Spectrum Wave  Understanding Physics JEE Main & Advanced 33.5 Early Atomic Structure 33.8 X-Rays 33.6 The Bohr Hydrogen Atom 33.9 Emission of Electrons 33.7 Hydrogen Like Atoms 33.10 Photoelectric Effect 34. MODERN PHYSICS-II 325-374 34.1 Nuclear Stability and 34.5 Binding Energy and Nuclear Stability Radioactivity 34.6 Nuclear Fission (Divide and Conquer) 34.2 The Radioactive Decay Law 34.7 Nuclear Fusion 34.3 Successive Disintegration 34.4 Equivalence of Mass and Energy 35. SEMICONDUCTORS 375-414 35.1 Introduction 35.5 Junction Diode as a Rectifier 35.2 Energy Bands in Solids 35.6 Applications of p-n Junction Diodes 35.3 Intrinsic and Extrinsic 35.7 Junction Transistors Semiconductors 35.8 Transistor as an Amplifier 35.4 p-n Junction Diode 35.9 Digital Electronics and Logic Gates 36. COMMUNICATION SYSTEM 415-431 36.1 Introduction 36.6 Modulation 36.2 Different Terms Used in 36.7 Amplitude Modulation Communication System 36.8 Production of Amplitude 36.3 Bandwidth of Signals Modulated Wave 36.4 Bandwidth of Transmission Medium 36.9 Detection of Amplitude 36.5 Propagation of Electromagnetic Modulated Wave Waves or Communication Channels Hints & Solutions 433-518 JEE Main & Advanced Previous Years' Questions (2018-13) 1-28 Understanding Physics JEE Main & Advanced SYLLABUS JEE Main ELECTROMAGNETIC WAVES Electromagnetic waves and their characteristics. Transverse nature of electromagnetic waves. Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, g- rays). Applications of e.m. waves. OPTICS Reflection and refraction of light at plane and spherical surfaces, Mirror formula, Total internal reflection and its applications, Deviation and Dispersion of light by a prism, Lens formula, Magnification, Power of a lens, Combination of thin lenses in contact, Microscope and astronomical telescope (reflecting and refracting) and their magnifying powers. WAVE OPTICS Wave front and Huygens’ principle, Laws of reflection and refraction using Huygen’s principle. Interference, Young’s double slit experiment and expression for fringe width. Diffraction due to a single slit, width of central maximum. Resolving power of microscopes and astronomical telescopes, Polarisation, Plane polarized light; Brewster’s law, uses of plane polarized light and polaroids. DUAL NATURE OF MATTER AND RADIATION Dual nature of radiation. Photoelectric effect, Hertz and Lenard’s observations; Einstein’s photoelectric equation; Particle nature of light. Matter waves-wave nature of particle, de-Broglie relation. Davisson-Germer experiment. Understanding Physics JEE Main & Advanced ATOMS AND NUCLEI a-particle scattering experiment; Rutherford’s model of atom; Bohr model, Energy levels, Hydrogen spectrum. Composition and size of nucleus, Atomic masses, Isotopes, Isobars; Isotones. Radioactivity - a, b and g particles/rays and their properties; radioactive decay law. Mass-energy relation, Mass defect; Binding energy per nucleon and its variation with mass number, Nuclear fission and fusion. ELECTRONIC DEVICES Semiconductors; semiconductor diode: I-V characteristics in forward and reverse bias; Diode as a rectifier; I-V characteristics of LED, photodiode, solar cell and Zener diode; Zener diode as a voltage regulator. Junction Transistor, transistor action, Characteristics of a transistor; transistor as an amplifier (common emitter configuration) and oscillator. Logic gates (OR, AND, NOT, NAND and NOR). Transistor as a switch. COMMUNICATION SYSTEMS Propagation of electromagnetic waves in the atmosphere; Sky and space wave propagation, Need for modulation, Amplitude and frequency modulation, Bandwidth of signals, Bandwidth of transmission medium, Basic elements of a communication system (block diagram only). Understanding Physics JEE Main & Advanced JEE Advanced GENERAL Focal length of a concave mirror and a convex lens using U V method. OPTICS Rectilinear propagation of light, Reflection and refraction at plane and spherical surfaces, Total internal reflection, Deviation and dispersion of light by a prism, Thin lenses, Combinations of mirrors and thin lenses, Magnification. WAVE NATURE OF LIGHT Huygens principle, Interference limited to Young’s double-slit experiment. MODERN PHYSICS Atomic nucleus, a, b and g radiations, Law of radioactive decay, Decay constant, Half-life and mean life, Binding energy and its calculation, Fission and fusion processes, Energy calculation in these processes. Photoelectric effect, Bohr’s theory of hydrogen-like atoms, Characteristic and continuous X-rays, Moseley’s law, de-Broglie wavelength of matter waves. Understanding Physics JEE Main & Advanced This book is dedicated to my honourable grandfather (Late) Sh. Pitamber Pandey a Kumaoni poet and a resident of Village Dhaura (Almora), Uttarakhand 29.1 Introduction 29.2 Displacement Current 29.3 Electromagnetic Waves 29.4 Electromagnetic Spectrum  2 — Optics and Modern Physics 29.1 Introduction Earlier we have learned that a time varying magnetic field produces an electric field. Is the converse also true? Does a time varying electric field can produce a magnetic field? James Clerk Maxwell argued that not only an electric current but also a time varying electric field generates magnetic field. Maxwell formulated a set of equations (known as Maxwell’s equations) involving electric and magnetic fields. Maxwell’s equations and Lorentz force formula make all the basic laws of electromagnetism. The most important outcome of Maxwell’s equations is the existence of electromagnetic waves. The changing electric and magnetic fields form the basis of electromagnetic waves. A combination of time varying electric and magnetic fields (referred as electromagnetic wave) propagate in space very close to the speed of light (= 3 ´ 10 8 m/s) obtained from optical measurements. We shall take a brief discussion of electromagnetic waves mainly developed by Maxwell around 1864. 29.2 Displacement Current An electric current produces magnetic field. Value of magnetic field (due to an electric current) at some point can be obtained by Biot-Savart law or Ampere’s circuital law. We have stated Ampere’s law as ò B× d l = m 0 i...(i) where left hand side of this equation is the line integral of magnetic field over a closed path and i is the electric current crossing the surface bounded by that closed path. Ampere’s law in this form is not valid if the electric field at the surface varies with time. For an example if we place a magnetic needle in between the plates of a capacitor during its charging or discharging then it deflects. Although, there is no current between the plates, so magnetic field should be zero. Hence, the needle should not show any deflection. But deflection of needle shows that there is a magnetic field in the region between plates of capacitor during charging or discharging. So, there must be some other source (other than current) of magnetic field. This other source is nothing but the changing electric field. Because at the time of charging or discharging of capacitor electric field between the plates changes. The relation between the changing electric field and the magnetic field resulting from it is given by df E ò B × d l = m 0e 0 dt...(ii) Here, f E is the flux of the electric field through the area bounded by the closed path along which line integral of B is calculated. Combining Eqs. (i) and (ii), we can make a general expression of Ampere’s circuital law and that is df E æ df E ö ò B × d l = m 0 i + m 0e 0 dt =m0 ç i +e0 è dt ø ÷ or ò B × d l = m 0 (i + id )...(iii)  Chapter 29 Electromagnetic Waves — 3 df E Here, id = e 0...(iv) dt is called the displacement current and which is produced by the change in electric field. The current due to the flow of charge is often called conduction current and is denoted by ic. Thus, Eq. (iii) can also be written as ò B × d l = m 0 (ic + id )...(v) Example In the figure, a capacitor is charged by a battery through a resistance R. Charging of capacitor will be exponential. A time varying charging current i flows in the circuit (due to flow of charge) till charging continues. A time varying electric field is also produced between the plates. This causes a displacement current id between the plates. There is no current between the plates due to flow of charge, as a medium between the plates is insulator. Consider two closed paths a and b as shown in figure. Ampere’s circuital law in these two paths is Along path a ò B × dl = m 0i or ò B × d l = m 0 ic Along path b ò B × d l = m 0 id E - + i id a b i = id R Fig. 29.1 df E Here, id = e 0 is in the direction shown in figure. In sample example 29.1, we have shown that dt ic = id Extra Points to Remember Faraday’s law of electromagnetic induction says that changing magnetic field gives rise to an electric field and its line integral (= emf ) is given by the equation df ò E × d l = - dtB A changing electric field gives rise to a magnetic field is the symmetrical counterpart of Faraday’s law. This is a consequence of the displacement current and given by df ò B × d l = m 0 e0 dtE = m 0 id Thus, time dependent electric and magnetic fields give rise to each other. 4 — Optics and Modern Physics Maxwell’s Equations ˜ ò E × d s = qin e0 (Gauss’s law for electricity) ˜ ò B × ds = 0 (Gauss’s law for magnetism) dfB ˜ ò E × dl = - dt (Faraday’s law) dfE ˜ ò B × d l = m 0 ( i c + i d ) = m 0 i c + m 0 e0 dt (Ampere-Maxwell’s law) V Example 29.1 During charging of a capacitor show that the displacement current between the plates is equal to the conduction current in the connecting wire. q Solution Let A is the area of plates, q is the charge on capacitor at + - some instant and d the separation between the plates. ic dq Conduction current, i c =...(i) dt Electric field between the plates, s q A q E= = = e0 e0 Ae 0 Fig. 29.2 The flux of the electric field through the given area is æ q ö q fE = EA = çç ÷A= ÷ è A e0 ø e0 dfE 1 æ dq ö \ = ç ÷ dt e 0 è dt ø Displacement current, dfE id = e 0 dt é 1 dq ù = e0 ê × ú ë e 0 dt û dq or id =...(ii) dt From Eqs. (i) and (ii), we can see that ic = id Hence Proved. INTRODUCTORY EXERCISE 29.1 1. A parallel-plate capacitor with plate area A and separation between the plates d , is charged by a constant current i. Consider a plane surface of area A/2 parallel to the plates and drawn symmetrically between the plates. Find the displacement current through this area.  Chapter 29 Electromagnetic Waves — 5 29.3 Electromagnetic Waves Stationary charges produce only electric field. Charges in uniform motion (or steady currents) produce both electric and magnetic fields. Accelerated charges radiate electromagnetic waves. It is an important result of Maxwell’s theory. Thus, an accelerated charge produces all three electric field, magnetic field and electromagnetic waves. Consider an oscillating charged particle. Let f is the frequency of its oscillations. This oscillating charged particle produces an oscillating electric field (of same frequency f ). Now, this oscillating electric field becomes a source of oscillating magnetic field (Ampere-Maxwell’s law). This oscillating magnetic field again becomes a source of oscillating electric field (Faraday’s law) and so on. The oscillating electric and magnetic fields regenerate each other and electromagnetic wave propagates through the space. The frequency of the electromagnetic wave is equal to the frequency of oscillation of the charge. Frequency of visible light is of the order of 1014 Hz, while the maximum frequency that we can get with modern electronic circuits is of the order of 1011 Hz. Therefore, it is difficult to experimentally demonstrate the production of visible light. Hertz’s experiment (in 1887) demonstrated the production of electromagnetic waves of low frequency (in radio wave region). Jagdish Chandra Bose succeeded in producing the electromagnetic waves of much higher frequency in the laboratory. Extra Points to Remember ˜ When electromagnetic waves propagate in space then electric and magnetic fields oscillate in mutually perpendicular directions. Further, they are perpendicular to the direction of propagation of electromagnetic wave also. ˜ Consider a plane electromagnetic wave propagating along the z-direction. The electric field E x is along the x-axis and varies sinusoidally. The magnetic field By is along the y-axis and again varies sinusoidally. We can write E x and By as E x = E0 sin (wt - kz) and By = B0 sin (wt - kz) Wavelength Electric field Magnetic field Direction of wave Fig. 29.3 Thus, electromagnetic wave travels in the direction of E ´ B. ˜ From Maxwell’s equations and the knowledge of waves we can write the following expressions, k = 2p l and w = 2 pf Speed of light (in vacuum) w E 1 c = = fl= 0 = k B0 e0m 0 where, f is the frequency of electromagnetic wave and l its wavelength. 6 — Optics and Modern Physics ˜ Unlike a mechanical wave (like sound wave) an electromagnetic wave does not require any material medium for the oscillations of electric and magnetic fields. They can travel in vacuum also. Oscillations of electric and magnetic fields are self sustaining in free space or vacuum. ˜ In a material medium (like glass, water etc.), electric and magnetic fields are different from the external fields. They are described in terms of permittivity e and magnetic permeabilitym. In Maxwell’s equations, e0 and m 0 are thus replaced by e and m and the velocity of light becomes, 1 v= em Thus, the velocity of light depends on electric and magnetic properties of the medium. ˜ Like other waves, electromagnetic waves also carry energy and momentum.In previous chapters, we have 1 B2 studied that, energy density in electric field = e0 E 2 and energy density in magnetic field = × 2 2 m0 An electromagnetic wave contains both electric and magnetic field. Therefore, energy density is associated with both the fields. ˜ Consider a plane perpendicular to the direction of propagation of the electromagnetic wave. If the total energy transferred to a surface in time t is E, then total momentum delivered to this surface for complete absorption is E Dp= (complete absorption) c If the wave is totally reflected, the momentum delivered is 2E Dp= (completely reflected) c ˜ The energy transferred per unit area per unit time perpendicular to the direction of propagation of electromagnetic wave is called the intensity of wave. It is given by 1 I = e0 E 2c 2 Here, E is the rms value of electric field or Erms. V Example 29.2 A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time,. $j) V/m. What is B at this point? E = ( 63 E0 E Solution c= or = B0 B E \ B= c Substituting the values in SI units, 6.3 B= 3 ´ 108 = 2.1´ 10-8 T From the relation c = E´ B We can see that B is along positive z- direction. Because, E is along $j direction and c along $i direction.  Chapter 29 Electromagnetic Waves — 7 \ B = ( 2.1´ 10-8 T ) k$ Ans. V Example 29.3 The magnetic field in a plane electromagnetic wave is given by B y = 2 ´ 10 -7 T sin ( 05. ´ 103 x + 15. ´ 1011 t) T (a) What is the wavelength and frequency of the wave? (b) Write an expression for the electric field. Solution (a) Wavelength From the given equation, we can see that k = 0.5 ´ 103 m -1 2p But, k= l 2p 2p \ l= = k 0.5 ´ 103. ´ 10-2 m = 125 Ans. Frequency Angular frequency,. ´ 1011 rad/s w = 15 But, w = 2pf w 15. ´ 1011 \ f= = \ 2p 2p = 2.39 ´ 1010 Hz Ans. E0 (b) c = B0 \ E 0 = cB 0 = ( 3.0 ´ 108 ) ( 2 ´ 10-7 ) = 60 V/m From the relation, c = E´ B We can see that E is along z-direction. \ E z = ( 60 V/ m ) sin ( 0.5 ´ 103 x + 15. ´ 1011 t ) V/m Ans. V Example 29.4 Light with an energy flux of 18 W/cm 2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 20 cm2 , find the average force exerted on the surface during a 30 minute time span. Solution Total energy incident on the given surface in the given time interval is E = ( 18 ´ 104 W / m 2 ) ( 20 ´ 10-4 m 2 ) ( 30 ´ 60 s ) = 6.48 ´ 105 J 8 — Optics and Modern Physics Therefore, the total momentum transferred to the given surface for complete absorption is E 6.48 ´ 105 Dp = = c 3.0 ´ 108 = 2.16 ´ 10-3 kg - m/s Dp 2.16 ´ 10-3 \ Fav = = Dt 30 ´ 60 = 1.2 ´ 10-6 N Ans. V Example 29.5 In the above example what is the average force if surface is perfectly reflecting? Solution (a) If the surface is perfectly reflecting, then 2E Dp = c Therefore, average force is doubled or Fav = 2.4 ´ 10-6 N Ans. V Example 29.6 Calculate the electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3 m. Assume that the efficiency of the bulb is 25. % and it is a point source. Solution Intensity at a distance r from a point source of power P is given by P I= 4pr 2 So, intensity at a distance of 3 m from the bulb with 2.5% efficiency will be 100 2. 5 I= 2 ´ 4p ( 3) 100 = 0.022 W/ m 2 Half of the intensity is provided by electric field and half by magnetic field. I \ I E = = 0.011 W/ m 2 2 1 But, I E is given by e 0 E 2 c 2 1 2 IE \ IE = e 0 E 2 c or E = 2 e0 c Substituting the values, we have 2 ´ 0.011 E= ( 8.85 ´ 10-12 ) ( 3 ´ 108 ) = 2. 9 V m Ans.  Chapter 29 Electromagnetic Waves — 9 Note that this is actually the rms value of electric field. From the equation, c = E/ B E 2.9 B= = c 3.0 ´ 108 = 9.6 ´ 10-9 T Ans. This is again the rms value of magnetic field. INTRODUCTORY EXERCISE 29.2 1 1. Show that the unit of is m/s. e0 m 0 2. A capacitor is connected to an alternating current source. Is there a magnetic field between the plates? 3. The sunlight reaching the earth has maximum electric field of 810 Vm -1. What is the maximum magnetic field in this light? 4. The electric field in an electromagnetic wave is given by E = (50 NC-1) sin w (t - x /c ). Find the energy contained in a cylinder of cross-section 10 cm 2 and length 50 cm along the x-axis. 29.4 Electromagnetic Spectrum The basic source of electromagnetic wave is an accelerated charge. This produces the changing electric and magnetic fields which constitute an electromagnetic wave. An electromagnetic wave may have its wavelength varying from zero to infinity. Not all of them are known till date. Today, we are familiar with electromagnetic waves having wavelengths as small as 30 fm (1 fm = 10 -15 m ) to as large as 30 km. The boundaries separating different regions of spectrum are not sharply defined, with visible light ranging from 4000 Å to 7000 Å. An approximate range of wavelengths associated with each colour are violet ( 4000 Å - 4500 Å), blue ( 4500 Å - 5200 Å), green (5200 Å - 5600 Å), yellow (5600 Å - 6000 Å), orange (6000 Å - 6250 Å) and red (6250 Å - 7000 Å). The classification of electromagnetic waves according to frequency or wavelength is called electromagnetic spectrum. Table below gives range of wavelengths and frequencies for different waves. Table 29.1 S.No. Type Wavelength range Frequency range 1. Radio waves > 01. m < 3 ´ 109 Hz 2. Micro waves 0.1 m to 1 mm 3 ´ 109 Hz to 3 ´ 1011 Hz 3. Infrared 1 mm to 7000 Å 3 ´ 1011 Hz to 4.3 ´ 1014 Hz 10 — Optics and Modern Physics S.No. Type Wavelength range Frequency range 4. Visible light 7000 Å to 4000 Å 4.3 ´ 1014 Hz to 7.5 ´ 1014 Hz 5. Ultraviolet 4000 Å to 10 Å 7.5 ´ 1014 Hz to 3 ´ 1017 Hz 6. X-rays 10 Å to 0.01 Å 3 ´ 1017 Hz to 3 ´ 1020 Hz 7. Gamma rays < 0.01 Å > 3 ´ 1020 Hz Note In the above table, wavelength is decreasing from top to bottom. But, frequency is increasing. Now, let us discuss them in brief in the order of increasing wavelength. 1. Gamma Rays These high frequency radiations are usually produced in nuclear reactions and also emitted by radioactive nuclei. They are used in medicines to destroy cancer cells. 2. X-Rays X-rays were discovered in 1895 by W.Roentgen. These are produced by the rapid deceleration of electrons that bombard a heavy metal target. These are also produced by electronic transitions between the energy levels in an atom. X-rays are used as a diagnostic tool in medicine and as a treatment for certain forms of cancer. 3. Ultraviolet Rays Ultraviolet radiation is produced by special lamps and very hot bodies. The sun is an important source of ultraviolet light. It plays an important role in the production of vitamin-D. But prolonged doses of UV radiation can induce skin cancers in human beings. Ozone layer in atmosphere at an altitude of about 40-50 km plays a protective role in this regarding. Depletion of this layer by chlorofluorocarbon (CFC) gas (such as Freon) is a matter of international concern now a days. 4. Visible Light It is most familiar form of electromagnetic waves.Our eye is sensitive to visible light. Visible light emitted or reflected from objects around us provides information about world. Process of photosynthesis in plants needs visible light. Visible light is produced by the transition of electrons in an atom from one energy level to other. 5. Infrared Radiation Infrared rays also sometimes referred as heat waves are produced by hot bodies. They are perceived by us as heat. In most of the materials, water molecules are present. These molecules readily absorb infrared rays. After absorption, their thermal motion increases, i.e. they heat up and heat their surroundings. Infrared rays are used for early detection of tumors. Infrared detectors are also used to observe growth of crops and for military purposes. 6. Microwaves Microwaves may be generated by the oscillations of electrons in a device called klystron. Microwave ovens are used in kitchens. In microwave ovens frequency of the microwaves is selected to match the resonant frequency of water molecules so that energy from the waves is transferred to the kinetic energy of the molecules. This raises the temperature of any food containing water. 7. Radio Waves Radio waves are generated when charges are accelerating through conducting wires. They are generated in L - C oscillators and are used in radio and television communication systems.  Chapter 29 Electromagnetic Waves — 11 Extra Points to Remember ˜ Our eyes are sensitive to visible light (lbetween 4000 Å to7000 Å). Similarly, different animals are sensitive to different ranges of wavelengths. For example, snakes can detect infrared waves. ˜ The basic difference between different types of electromagnetic waves lies in their frequencies and wavelengths. All of them travel with same speed in vacuum. Further they differ in their mode of interaction with matter. For example infrared waves vibrate the electrons, atoms and molecules of a substance. These vibrations increase the internal energy and temperature of the substance. This is why infrared waves are also called heat waves. ˜ Electromagnetic waves interact with matter via their electric and magnetic fields, which set in oscillation with the charges present in all matter. This interaction depends on the wavelength of the electromagnetic wave and the nature of atoms or molecules in the medium. ˜ Microwave Oven In electromagnetic spectrum frequency and energy of microwaves is smaller than the visible light. Water content is required for cooking food in microwave oven. Almost all food items contain some water content. Microwaves interact with water molecules and atoms via their electric and magnetic fields. Temperature of water molecules increases by this. These water molecules share this energy with neighboring food molecules, heating up the food. Porcelain vessels are used for cooking food in microwave oven. Because its large molecules vibrate and rotate with much smaller frequencies and do not get heated up. We cannot use metal vessels. Metal vessels interact with microwaves. These vessels may melt due to heating. Solved Examples V Example 1 Long distance radio broadcasts use short wave bands. Explain why? Solution Short radio waves are reflected by ionosphere. V Example 2 It is necessary to use satellites for long distance TV transmission. Explain why? Solution TV waves (part of radio waves) range from 54 MHz to 890 MHz. Unlike short wave bands (used in radio broadcasts) which are reflected by ionosphere, TV waves are not properly reflected by ionosphere. This is why, satellites are used for long distance TV transmission. V Example 3 The ozone layer on the top of the stratosphere is crucial for human survival. Explain why? Solution Ozone layer protects ourselves from ultraviolet radiations. Over exposure to UV radiation can cause skin cancer in human beings. Ozone layer absorbs UV radiations. But unfortunately over use of Chlorofluoro Carbon Gases (CFCs) is depleting this ozone layer and it is a matter of international concern now a days. V Example 4 Optical and radio telescopes are built on ground but X-ray astronomy is possible only from satellites orbiting the earth. Explain why? Solution Visible and radio waves can penetrate the atmosphere, while X-rays are absorbed by the atmosphere. This is why X-ray telescopes are installed in satellites orbiting the earth. V Example 5 If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now? Solution Due to presence of atmosphere green house effect takes place. Heat radiated by earth is trapped due to green house effect. In the absence of atmosphere, temperature of the earth would be lower because the green house effect of the atmosphere would be absent. V Example 6 Some scientists have predicted the global nuclear war on the earth would be followed by a severe nuclear winter with a devastating effect on life on earth. What might be the basis of this prediction? Solution After nuclear war, clouds would perhaps cover the atmosphere of earth preventing solar light from reaching many parts of earth. This would cause a winter. V Example 7 Why is the orientation of the portable radio with respect to broadcasting station important? Solution Electromagnetic waves are plane polarised, so the receiving antenna should be parallel to electric and magnetic part of wave.  Chapter 29 Electromagnetic Waves — 13 V Example 8 A plane electromagnetic wave propagating in the x-direction has a wavelength of 5.0 mm. The electric field is in the y-direction and its maximum magnitude is 30 Vm -1. Write suitable equations for the electric and magnetic fields as a function of x and t. Solution Given, l = 5 mm = 5 ´ 10-3 m 2p 2p k= = m -1 l 5 ´ 10-3 = 1257 m-1 w From the equation, c= k w = c k = (3 ´ 108 ) (1257) = 3.77 ´ 1011 rad/s E 0 = 30 V/m E c= 0 B0 E0 30 \ B0 = = c 3 ´ 108 = 10-7 T Now, E y = E 0 sin (wt - kx) = (30 V /m) sin [(3.77 ´ 1011 s -1 ) t - (1257 m-1 ) x] and Bz = (10-7 T) sin [(3.77 ´ 1011 s -1 ) t - (1257 m-1 ) x] Ans. V Example 9 A light beam travelling in the x-direction is described by the electric field E y = ( 300 Vm -1 ) sin w ( t - x c). An electron is constrained to move along the. ´ 10 7 ms -1. Find the maximum electric force and y-direction with a speed of 20 the maximum magnetic force on the electron. Solution Maximum Electric Force Maximum electric field, E 0 = 300 V /m \ Maximum electric force F = qE 0 = (1.6 ´ 10-19 ) (300) = 4.8 ´ 10-17 N Ans. Maximum Magnetic Force E0 From the equation, c= B0 E0 Maximum magnetic field, B0 = c 300 or B0 = = 10-6 T 3.0 ´ 108 \ Maximum magnetic force = B0qv sin 90o = B0qv Substituting the values, we have Maximum magnetic force = (10-6 )(1.6 ´ 10-19 )(2. 0 ´ 107 ) = 3.2 ´ 10-18 N Ans. 14 — Optics and Modern Physics V Example 10 A parallel plate capacitor having plate area A and plate separation d is joined to a battery of emf V and internal resistance R at t = 0. Consider a plane surface of area A 2, parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time. [The charge on the capacitor at time t is given by q = CV ( 1 - e - t t ), where t = CR] Solution Given, q = CV (1 - e- t t ) q CV \ Surface charge density, s= = (1 - e- t t ) A A Electric field between the plates of capacitor, s CV E= = (1 - e- t t ) e0 e0 A Electric flux from the given area, EA CV fE = = (1 - e- t t ) 2 2e0 dfE Displacement current, i d = e0 dt d é CV ù CV - t or i d = e0 ê (1 - e- t t )ú = e t dt ë 2e0 û 2t Substituting, t = CR V - t CR We have, id = e 2R e A Again substituting, C= 0 d td - V e 0AR id = e Ans. 2R V Example 11 About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation (a) at a distance of 1 m from the bulb? (b) at a distance of 10 m? Assume that the radiation is emitted isotropically and neglect reflection. Solution Effective power (energy radiated per second) = 5% of 100 W P=5W S r  Chapter 29 Electromagnetic Waves — 15 This energy will distribute on a sphere. At a distance r from the point source, area on which light is incident is S = 4 pr 2 \ Intensity at distance r from the point source, P 5 I= = = Energy incident per unit area per unit time S 4pr 2 (a) At r = 1m, 5 I= 4p (1)2 = 0.4 W/m2 Ans. (b) At r = 10 m, 5 I= 4p (10)2 = 0.004 W/m2 Ans. V Example 12 Suppose that the electric field of an electromagnetic wave in vacuum is E = {( 3.0 N /C ) cos [( 1.8 rad / m) y + ( 5.4 ´ 10 6 rad / s) t]} $i. (a) What is the direction of propagation of wave? (b) What is the wavelength l? (c) What is the frequency f? (d) What is the amplitude of the magnetic field of the wave? (e) Write an expression for the magnetic field of the wave. Solution (a) From the knowledge of wave we can see that electromagnetic wave is travelling along negative y -direction, as wt and ky both are positive. (b) k = 1.8 rad /m 2p k= l 2p 2p \ l= = = 3.5 m Ans. k 1.8 (c) w = 5.4 ´ 106 rad /s w = 2pf w 5.4 ´ 106 \ f = = 2p 2p = 8.6 ´ 105 Hz Ans. (d) E 0 = 3.0 N/C E From the relation, c= 0 B0 E0 3.0 We have, B0 = = c 3.0 ´ 108 = 10-8 T Ans. $ (e) E is along i direction, wave is travelling along negative y-direction. Therefore, oscillations of B are along z-direction or B = (10-8 T) cos [(1.8 rad /m) y + (5.4 ´ 106 rad /s ) t ] k $ Ans. 16 — Optics and Modern Physics V Example 13 A parallel plate capacitor made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V AC supply with a (angular) frequency of 300 rad / s. (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. Solution (a) Capacitive reactance, 1 XC = wC 1 = 300 ´ 100 ´ 10-12 108 = W 3 There is only capacitance in the circuit. V rms \ i rms = XC 230 = (108 / 3) = 6.9 ´ 10-6 A Ans. (b) Yes, the derivation in example 29.1 is true even if current is alternating. (c) Here, i d is displacement current and i the conduction current. Magnetic field at a distance r from the axis, m i B = 0 d2 r 2p R m 0 i rms \ Brms = r (i d = i = i rms ) 2 p R2 Substituting the values, we have (2 ´ 10-7 ) (6.9 ´ 10-6 ) -2 Brms = (3 ´ 10 ) (6 ´ 10-2)2 = 1.15 ´ 10-11 T \ B0 = 2 Brms = ( 2 ) (1.15 ´ 10-11 ) T = 1.63 ´ 10-11 T Ans. Exercises Single Correct Option 1. One requires 11eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in (a) visible region (b) infrared region (c) ultraviolet region (d) microwave region 2. If E and B represent electric and magnetic field vectors of the electromagnetic wave, the direction of propagation of electromagnetic wave is along (a) E (b) B (c) B ´ E (d) E ´ B 3. The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is (a) c : 1 (b) c2 : 1 (c) 1 : 1 (d) c : 1 4. Light with an energy flux of 20 W/cm 2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 30 cm 2. The total momentum delivered (for complete absorption) during 30 minutes is (a) 36 ´ 10-5 kg -m/s (b) 36 ´ 10-4 kg -m/s (c) 1.08 ´ 104 kg -m/s (d) 1.08 ´ 107 kg -m/s More than One Correct Options 5. A plane electromagnetic wave propagating along x-direction can have the following pairs of E and B (a) E x , By (b) E y , Bz (c) Bx , E y (d) E z , By 6. The source of electromagnetic waves can be a charge (a) moving with a constant velocity (b) moving in a circular orbit (c) at rest (d) falling in an electric field 7. An electromagnetic wave of intensity I falls on a surface kept in vacuum and exerts radiation pressure p on it. Which of the following are true? (a) Radiation pressure is I /c if the wave is totally absorbed (b) Radiation pressure is I /c if the wave is totally reflected (c) Radiation pressure is 2 I /c if the wave is totally reflected (d) Radiation pressure is in the range I /c < p < 2 I /c for real surfaces 8. A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. The electromagnetic waves produced (a) will have frequency of 109 Hz (b) will have frequency of 2 ´ 109 Hz (c) will have a wavelength of 0.3 m (d) fall in the region of radio waves 18 — Optics and Modern Physics Subjective Questions 9. Can an electromagnetic wave be deflected by an electric field? By a magnetic field? 10. What physical quantity is the same for X-rays of wavelength 10-10m, red light of wavelength 6800 Å and radio waves of wavelength 500 m? 11. 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? 12. A radio can tune into any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength of band? 13. 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? 14. Figure shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A. (a) Calculate the capacitance and the rate of change of potential difference between the plates. (b) Obtain the displacement current across the plates. (c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain. 15. Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N C and that its frequency is 50.0 MHz. (a) Determine B0 , w, k and l, (b) Find expressions for E and B. 16. A variable frequency AC source is connected to a capacitor. How will the displacement current change with decrease in frequency? 17. A laser beam has intensity 2.5 ´ 1014 Wm -2. Find the amplitudes of electric and magnetic fields in the beam. 18. In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2. 0 ´ 1010 Hz and amplitude 48 Vm -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 field E equals the average energy density of the field B. [c = 3 ´ 108 ms -1 ]. 19. The charge on a parallel plate capacitor varies as q = q0 cos 2pft. The plates are very large and close together (area = A, separation = d ). Neglecting the edge effects, find the displacement current through the capacitor.  Answers Introductory Exercise 29.1 1. i/2 Introductory Exercise 29.2 2. Yes. mT 3. 27 4. 5.55 ´ 10 -12 J Exercises 1. (c) 2. (d) 3. (c) 4. (b) 5. (b,d) 6. (b,d ) 7. (a,c,d) 8. (a,c,d) 9. No, No 10. The speed in vacuum is the same for all 11. E and B lie in x-y plane and are mutually perpendicular, 10 m 12. Wavelength band from 40 m to 25 m 13. 153 N/C 14. (a) 8.0 pF, 1.87 ´ 1010 Vs -1 (b) 0.15 A (c) Yes, provided by current we mean the sum of conduction and displacement currents. 15. (a) 400 nT, 3.14 ´ 10 8 rad/s , 105. rad/m, 6.00 m (b) E = (120 N/C) sin [(1.05 rad/m)] x - (3.14 ´ 10 8 rad/s)t] B = ( 400 nT) sin [(1.05 rad/m)] x - (3.14 ´ 10 8 rad/s) t] 16. Displacement current will decrease 17. 4.3 ´ 10 8 N/C , 1.44 T. ´ 10 -2 m 18. (a) 15. ´ 10 -7 T (b) 16 19. -2pq 0 f sin 2p ft 30.1 Introduction 30.2 General concepts used in Geometrical Optics 30.3 Reflection of light 30.4 Reflection from a spherical surface  22 — Optics and Modern Physics 30.1 Introduction The branch of physics called optics deals with the behaviour of light and other electromagnetic waves. Light is the principal means by which we gain knowledge of the world. Consequently, the nature of light has been the source of one of the longest debates in the history of science. Electromagnetic radiation with wavelengths in the range of about 4000 Å to 7000 Å, to which eye is sensitive is called light. In the present and next two chapters we investigate the behaviour of a beam of light when it encounters simple optical devices like mirrors, lenses and apertures. Under many circumstances, the wavelength of light is negligible compared with the dimensions of the device as in the case of ordinary mirrors and lenses. A light beam can then be treated as a ray whose propagation is governed by simple geometric rules. The part of optics that deals with such phenomena is known as geometric optics. However, if the wavelength is not negligible compared with the dimensions of the device (for example a very narrow slit), the ray approximation becomes invalid and we have to examine the behaviour of light in terms of its wave properties. This study is known as physical optics. 30.2 General Concepts used in Geometrical Optics Some general concepts which are used in whole geometrical optics are given below. 1. Normal incidence means angle of incidence (with normal) is 0°. If angle of incidence is 90°, it is called grazing incidence. 90° Normal incidence Grazing incidence Ð i = 0° Ð i = 90° Fig. 30.1 2. An image is formed either by reflection or refraction. Minimum two (reflected or refracted) rays are required for image formation. More the number of rays, more will be the intensity of image. 3. A light is reflected only from a silvered surface. Without any reflecting surface on the path of ray of light it keeps on moving ahead. p q (i) (ii) Fig. 30.2 In figure (i) For image formation, if reflected ray-p is required to the left of concave lens, then we will take it as dotted line.  Chapter 30 Reflection of Light — 23 In figure (ii) For image formation, if reflected ray-q is required to the right of concave mirror, then we will take it as dotted line. 4. Real object, virtual object, real image, virtual image The point where the rays meet (or appear to meet) before refraction or reflection is called object and the point where the rays meet (or appear to meet) after refraction or reflection is called image. Further, object (or image) is real if dark lines meet and virtual if dotted lines meet. In figure (a), object is real, while image is virtual. In figure (b), object is virtual while its image is real. O I I O (a) (b) O I2 I1 (c) Fig. 30.3 In figure (c), the object O is real. Its image formed by the lens (i.e. I 1 ) is real. But, it acts as a virtual object for mirror which forms its real image I 2. 5. The virtual images cannot be taken on screen. But, they can be seen by our eye. Because our eye lens forms their real image on our retina. Thus, if we put a screen at I in the above figure (a) no image will be formed on it. At the same time if we put the screen at I in figure (b), image will be formed. 6. Normally, the object is kept on the left hand side of the optical instrument (mirror, lens etc.), i.e. the ray of light travels from left to right. Sometimes, it may happen that the light is travelling in opposite direction. See the figure. O P O C I I I M O (a) O ® Object I ® Image (b) (c) Fig. 30.4 24 — Optics and Modern Physics In figures (a) and (b), light is travelling from left to right and in figure (c) it is travelling from right to left. 7. Sign convention The distances measured along the incident light are taken as positive while the distances against incident light are taken as negative. For example, in figures (a) and (b) the incident light travels from left to right. So, the distances measured in this direction are positive. While in figure (c) the incident light travels from right to left. So, in this case right to left direction will be positive. Distances are measured from pole of the mirror [point P in figure (b)], optical centre of the lens [point C in figure (a)] and the centre of the refracting surface [point M in figure (c)]. It may happen in some problem that sign convention does not remain same for the whole problem. For example, in the Fig. 30.5 shown, the ray of light starting A from O first undergoes refraction at A, then reflection at B B and then finally refraction at C. For refraction and reflection M P at A and B the incident light is travelling from left to right, so O distances measured along this direction are positive. For C final refraction at C the incident light travels from right to Fig. 30.5 left, so now the sign convention will change or right to left is positive. 8. Object distance (from P , C or M along the optic axis) in Fig. 30.4 is shown by u and image distance by v. 9. In front of mirror, object (or image) is always real, lines are always dark and u (or v ) are always negative. Behind the mirror object (or image) is always virtual, lines are always dotted and u (or v ) are always positive. This is because light always falls from front side of the mirror. Real Virtual +ve –ve +ve Fig. 30.6 10. In most of the cases, objects are real (whether refraction or reflection) and u for them is negative. +ve +ve O O u = –ve u = –ve Fig. 30.7  Chapter 30 Reflection of Light — 25 11. In case of mirror light always falls from front of the mirror. But in a lens (or some other refracting surface like slab) light can fall from both sides. or Fig. 30.8 A 12. Total steps and reduced steps 1 2 In Fig. 30.9 total steps are five. Four of them are refraction. Only 3 third is reflection. But we have made a lens formula for steps 1 D B and 2 or 4 and 5. So, the reduced steps are three, 5 4 lens ® mirror ® lens. C 13. Image real or virtual In Fig.30.9, if we wish to find the nature Fig. 30.9 of I 2 (after 2nd refraction), then it is real if it is formed to the right of ABC because ray of light has moved to this side and it is virtual to the left of ABC. Similarly, I 5 is real to the left of ADC and virtual to the right of ADC. 14. Final image coincides with the object In most of these cases there will be one mirror, plane or spherical (convex or concave) and light will be falling normal ( Ði = 0° ) to this mirror. In case of spherical mirror it is normal if ray of light passes through centre of curvature. In case of normal incidence, ray of light retraces its path and final image coincides with the object. C C Fig. 30.10 In all above figures, Ð i = 0° (normal incidence). Ray of light retraces its path. 15. Image at infinity means rays after refraction or reflection have become parallel to the optic axis. If a screen is placed directly in between these parallel rays no image will be formed on the screen. But if a lens (or a mirror) is placed on the path of these parallel rays, then image is formed at focus. Sometimes, our eye plays the role of this lens and the image is formed on our retina. Screen Parallel rays or F Retina Parallel rays Eye lens Fig. 30.11  26 — Optics and Modern Physics 16. Visual angle q Angle subtended by an object on our eye is called the visual angle. The apparent size depends on the visual angle. As the object moves away from the eye, actual size remains the same but visual angle decreases. Therefore, apparent size decreases. h h q1 q2 q2 < q1 Fig. 30.12 30.3 Reflection of Light When waves of any type strike the interface between two different materials, new waves are generated which move away from the interface. Experimentally, it is found that the rays corresponding to the incident and reflected waves make equal angles with the normal to the interface and that the reflected ray lies in the plane of incidence formed by the incident ray and the normal. Thus, the two laws of reflection can be summarised as under: (i) Ð i = Ð r (ii) Incident ray, reflected ray and normal lie on the same plane. Normal Incident Reflected ray ray i r Fig. 30.13 Two Important Points in Reflection Laws 1. The first law Ð i = Ð r can be applied for any type of surface. The main point is, normal at point of incidence. In spherical surface (convex mirror or concave mirror) normal at any point passes through centre of curvature. r i i r C C Concave mirror Convex mirror Fig. 30.14 2. Incident ray, reflected ray and normal are sometimes represented in the form of three vectors. Then, these three vectors should be coplanar.  Chapter 30 Reflection of Light — 27 Reflection from a Plane Surface (or Plane Mirror) Almost everybody is familiar with the image formed by a plane mirror. If the object is real, the image formed by a plane mirror is virtual, erect, of same size and at the same distance from the mirror. The ray diagram of the image of a point object and of an extended object is as shown below. B B¢ O I A A¢ Fig. 30.15 Important Points in Reflection from Plane Mirror 1. Relation between object distance ( u) and the image distance ( v ) in case of plane mirror is v=-u Here, v and u are measured from the plane surface. Two conclusions can be drawn from this equation. (i) Negative sign implies that object and image are on opposite sides of the mirror. So, if object is real then image is virtual and vice-versa. (ii) | v | = | u | And this implies that perpendicular distance of the object from the mirror is equal to the perpendicular distance of image from the mirror. I M O I M O Correct Wrong OM = MI OM = MI Fig. 30.16 2. Ray Diagram Let us draw the ray diagram of a point object and an extended object. r b i O d a b¢ d I a¢ (i) (ii) O ® Real ab ® Real I ® Virtual a¢b¢ ® Virtual Fig. 30.17 28 — Optics and Modern Physics Just as we have drawn the ray diagram of point object O in figure (i), we can also draw the ray diagrams of points a and b in figure (ii). 3. Field of view of an object for a given mirror Suppose a point object O is placed in front of a small mirror as shown in Fig. 30.18 (a), then a question arises in mind whether this mirror will make the image of this object or not. Or suppose an elephant is standing in front of a small mirror, will the mirror form the image of the elephant or not. The answer is yes, it will form. A mirror whatever may be the size of it forms the images of all objects lying in front of it. But every object has its own field of view for the given mirror. The field of view is the region between the extreme reflected rays and depends on the location of the object in front of the mirror. If our eye lies in the field of view then only we can see the image of the object otherwise not. The field of view of an object placed at different locations in front of a plane mirror are shown in Fig. 30.18 (b) and (c). The region between extreme reflected rays (reflected from the end points of the mirror) is called the field of view. To see the image of object eye should lie in this region, as all reflected rays lie in this region. O I O O I (a) (b) (c) Fig. 30.18 4. Suppose a mirror is rotated by an angle q (say anti-clockwise), keeping the incident ray fixed then the reflected ray rotates by 2q along the same direction, i.e. anti-clockwise. Y N N¢ R¢ I R i – 2q i–q I q i–q i i X q (a) (b) Fig. 30.19 In figure (a), I is the incident ray, N the normal and R the reflected ray. In figure (b), I remains as it is N and R shift to N ¢ and R ¢. From the two figures, we can see that earlier the reflected ray makes an angle i with y-axis while after rotating the mirror it makes an angle i – 2q. Thus, we may conclude that the reflected ray has been rotated by an angle 2q.  Chapter 30 Reflection of Light — 29 H 5. The minimum length of a plane mirror to see one’s full height is , where H is the height of 2 person. But, the mirror should be placed in a fixed position which is shown in Fig. 30.20. A x F B x C ( x + y) y D G y E Person Fig. 30.20 A ray starting from head (A) after reflecting from upper end of the mirror (F) reaches the eye at C. Similarly, the ray starting from the foot (E) after reflecting from the lower end (G) also reaches the eye at C. In two similar triangles ABF and BFC, AB = BC = x (say), Similarly in triangles CDG and DGE, CD = DE = y (say) Now, we can see that height of the person is 2 ( x + y) and that of mirror is (x + y), i.e. height of the mirror is half the height of the person. Note The mirror can be placed anywhere between the centre lines BF (of AC) and DG (of CE ). As the mirror is moved away on this line, image also moves away from the person. So, apparent size keeps on decreasing. 6. A person is standing exactly at midway between a wall and a mirror and he wants to see the full height of the wall (behind him) in a plane mirror (in front of him). The minimum length of mirror H in this case should be , where H is the height of wall. The ray diagram in this case is shown in 3 Fig. 30.21 A H 2x F x B I x (x + y) C (x + y) y K G y E 2y J D Wall Person Mirror d d Fig. 30.21 In triangles HBI and IBC, HI = IC = x (say). Now, in triangles HBI and ABF, AF FB AF 2d = or = or AF = 2x HI BI x d Similarly, we can prove that DG = 2 y if, CK = KJ = y Now, we can see that height of the wall is 3 ( x + y) while that of the mirror is ( x + y). 30 — Optics and Modern Physics 7. Object and image velocity There are four important points related to object and image velocity. (i) Image speed is equal to the object speed. (ii) Image velocity and object velocity make same angles from the plane mirror on two opposite sides of the mirror. (iii) Components of velocities which are along the mirror are equal. (iv) Components of velocities which are perpendicular to the mirror are equal and opposite. The following four figures demonstrate the above four points. v v v O I v q q v v O v v I O I O I q q Fig. 30.22 Three Types of Problems in Reflection from Plane Mirror Type 1. Based on law of reflection Ð i = Ð r Concept These problems are purely based on geometry. Proper normal at point of incidence is very important. V Example 30.1 Two plane mirrors M 1 and M 2 are inclined at angle q as shown. A ray of light 1, which is parallel to M 1 strikes M 2 and after two reflections, the ray 2 becomes parallel to M 2. Find the angle q. M2 2 1 q M1 Fig. 30.23 Solution Different angles are as shown in Fig. 30.24. In triangle ABC, q B a q a aa a =90°– q q q q A C Fig. 30.24 q + q + q = 180° \ q = 60° Ans.  Chapter 30 Reflection of Light — 31 V Example 30.2 Prove that for any value of angle i, rays 1 and 2 are parallel. 1 i 2 90° Fig. 30.25 Solution PQ and MN are mutually parallel. Rays 1 and 2 are making equal angles ( = Ði ) from PQ and MN. So, they are mutually parallel. 1 i Q 2 P i 90° – i 90° – i i M N Fig. 30.26 Important Result Two plane mirrors kept at 90° deviate each ray of light by 180° from its original path. Type 2. Based on field of view Concept The region between extreme reflected rays is called field of view. To see the image of object our eye should lie in the field of view. V Example 30.3 A point source of light S, placed at a distance L in front of the centre of a mirror of width d, S hangs vertically on a wall. A man walks in front of the d mirror along a line parallel to the mirror at a distance L 2L from it as shown. The greatest distance over which he can see the image of the light source in the mirror is 2L (a) d/2 (b) d Fig. 30.27 (c) 2 d (d) 3 d (JEE 2000) G Solution (d) The ray diagram will be as shown in Fig. 30.28. C HI = AB = d A f d H DS = CD = f D 2 S Since, AH = 2AD E I d B \ GH = 2 CD = 2 = d 2 F Similarly, IJ = d J \ GJ = GH + HI + IJ = d + d + d = 3d Fig. 30.28 32 — Optics and Modern Physics V Example 30.4 A pole of height 4 m is kept in front of a vertical plane mirror of length 2 m. The lower end of the mirror is at a height of 6 m from the ground. The horizontal distance between the mirror and the pole is 2 m. Upto what minimum and maximum heights a man can see the image of top of the pole at a horizontal distance of 4 m (from the mirror) standing on the same horizontal line which is passing through the pole and the horizontal point below the mirror? Solution PQ = Pole, MN = Image of pole E HG BD = GN BN ( HG ) ( BN ) ( 2) ( 6) \ BD = = D GN 2 I =6m 2m Minimum height required = AD = BD + AB = 10 m C H Q 2m IG BE N Further, = B G GN BN 4m ( IG ) ( BN ) ( 4 ) ( 6) \ BE = = = 12 m A P F M GN 2 2m 2m 2m \ Maximum height required = AE = BE + AB = 16 m Fig. 30.29 Type 3. Based on object and image velocity Concept We have already discussed four points on this topic. V Example 30.5 A plane mirror is lying in x-y plane. Object velocity is v 0 = ( 2$i - 3$j + 4 k $ ) m/ s. Find the image velocity. Solution Components of object velocity parallel to plane mirror (or lying in x - y plane) remain unchanged. But, component perpendicular to plane mirror changes its direction but magnitude remains the same. Hence, the image velocity is v = ( 2$i - 3$j - 4k$ ) m/s I Ans. V Example 30.6 An object is falling vertically downwards with velocity 10 m/s. In terms of $i and $j , find the image velocity. O 10 m/s ^j ^ 30° i Fig. 30.30  Chapter 30 Reflection of Light — 33 Solution Image speed will also be 10 m/s. Further, image vI =10 m/s velocity will make same angle ( = 60° ) from the mirror on opposite ^j 30° side of it. 60° 30° In terms of i$ and $j , ^i 60° 30° vI = 10 cos 30° $i + 10 sin 30° $j v0 =10 m/s or vI = ( 5 3 $i + 5 $j ) m/s Ans. Fig. 30.31 V Example 30.7 A point object is moving with a speed v M1 before an arrangement of two mirrors as shown in v figure. Find the magnitude of velocity of image in q M2 mirror M 1 with respect to image in mirror M 2. Fig. 30.32 Solution Using the same concept used in above problem, we have vI1 = v to find magnitude of relative velocity between vI 1 and vI 2. The angle between these two vectors is M1 2 q. q Hence, vr = vI 1 - vI 2 q 2q v0 = vI 2 = v \ | vr | = | vI 1 - vI 2 | M2 q This is nothing but magnitude of subtraction of two Fig. 30.33 velocity vectors of equal magnitudes v each and angle between them equal to 2 q. Hence, | vr | = v 2 + v 2 - 2( v ) ( v ) cos 2 q Solving these two equations, we get | v2 | = 2 v sin q Ans. INTRODUCTORY EXERCISE 30.1 1. A man approaches a vertical plane mirror at speed of 2 m/s. At what rate does he approach his image? 2. An object M is placed at a distance of 3 m from a mirror with its lower end at 2 m from ground as shown in Fig. 30.34. There is a person at a 2m distance of 4 m from object. Find minimum and maximum height of person to see the image of object. 2m Person 4 m M 3 m 3. In terms of q find the value of i , so that ray of light retraces its path after Fig. 30.34 third reflection. i q Fig. 30.35  34 — Optics and Modern Physics 30.4 Reflection from a Spherical Surface We shall mainly consider the spherical mirrors, i.e. those which are part of a spherical surface. Terms and Definitions There are two types of spherical mirrors, concave and convex.