# NCERT Solutions Class 12 Physics Chapter 8

## NCERT Solutions Class 12 Physics Chapter 8- Electromagnetic waves

The NCERT Solutions Class 12 Physics Chapter 8 provides information about Electromagnetic waves and is available on the Extramarks platform. Students preparing to appear for the Class 12 board exams can use these notes to study and check the details of all topics and sub-topics included in this chapter. Our notes offer solutions to all in-text and exercise questions from the NCERT books. The step-by-step answers help students to revise all concepts thoroughly. The physics chapter 8 class 12 notes are based on the latest CBSE curriculum NCERT books for 2022-2023. Students can benefit from the Extamarks NCERT Solutions Class 12 Physics Chapter 8 for their Class 12 exam preparation.

Chapter 8 Class 12 Physics talks about various concepts related to Electromagnetic waves. Students must have some basic knowledge of Class 11 physics for a quick recap. They can refer to NCERT Solutions Class 11 available on Extramarks. The chapter includes the basic introduction to Electromagnetic Waves, Displacement Current, Sources of electromagnetic waves, Electromagnetic Spectrum, and different types of electromagnetic waves. Students will learn about the nature of the concepts that will help them prepare for Chapter 8 Physics Class 12 thoroughly.

NCERT Solutions Class 12 Physics Chapter 8 helps students attain high scores. With help from these solutions, students can comprehend all vital concepts and theories from this chapter. Students can also refer to NCERT Solutions Class 8, NCERT Solutions Class 9, and NCERT Solutions Class 10 to get a quick recap of the subject.

### Key Topics Covered In NCERT Solutions Class 12 Physics Chapter 8

In Extramarks NCERT Solutions Class 12 Physics Chapter 8, students can be assured that all topics are included and defined in detail. By referring to these solutions, students can prepare for board examinations and entrance tests such as NEET and JEE (Main and Advanced) exams. Subject elites at Extramarks have covered the following key topics in the NCERT Solutions Class 12 Physics Chapter 8:

 Exercise Topic 8.1 Introduction 8.2 Displacement Current 8.3 Electromagnetic Waves 8.4 Electromagnetic Spectrum

8.1 Introduction

In the introductory section, students will gain complete information regarding all vital concepts of Electromagnetic waves. The definition of the EM waves and their behavior in the three axes has been included in a simplified manner. The various properties of electromagnetic waves and their behavior in different conditions are explained in detail in the NCERT Solutions Class 12 Physics Chapter 8. Students will also learn about different laws and equations of electromagnetic waves and some practical uses in our everyday lives.

Some main features of this chapter are

• Maxwell’s equations
• Ampere’s Circuital Law
• Oscillation of Electric and Magnetic Fields

8.2 Displacement Current

This section aids students to understand the concept of displacement current. It is explained with the help of this formula, id= 0 dEdt

It also includes Maxwell’s basic equations based on electricity and magnetism. With the help of NCERT Solutions class 12 Physics Chapter 8, students can get a clear idea of Gauss’ law of Magnetism and Electricity, Faradays’ law of electromagnetic induction, and Amperes’ modified form of Circuital law. These equations help students to get complete information on all electromagnetic interactions. In addition, Maxwell’s equations depict the relationship between electricity, magnetism, and ray optics. In this section of the NCERT Solutions Class 12 Physics Chapter 8, the similarities between conduction current and displacement current are also analysed.

8.3 Electromagnetic Waves

This section is bifurcated into two sub-parts, Sources of electromagnetic waves and Nature of electromagnetic (EM) waves. Students also learn about energy and some properties of EM waves. A detailed explanation of the oscillation of electric and magnetic fields and relation between oand odenoted by speed (c) is included in the NCERT Solutions Class 12 Physics Chapter 8.

Under the nature of the wave topic, important concepts are explained, such as electric field, wavenumber, and magnetic field of an EM wave. Using Maxwell’s equations, it is proved that the total energy carried per unit volume by an Electromagnetic wave = Et = E2o. The Velocity and Momentum of the EM wave are also explained with the help of their formulae.

There are various numerical questions asked based on the formulas mentioned in this section. Extramarks NCERT Solutions Class 12 Physics Chapter 8 covers all the essential formulae for reference.

8.4 Electromagnetic Spectrum

In this section of Chapter 8, students learn about the sequential distribution of electromagnetic waves based on frequency or wavelength. The NCERT solutions Class 12 Physics Chapter 8 also includes detailed information about the different types of EM waves. They are Radio Waves, Infrared waves, Microwaves, Visible rays, Ultraviolet (UV) rays, X-rays, and Gamma rays.

Students learn about the characteristics, wavelength, and applications of these waves with the help of the NCERT Solutions Class 12 Physics Chapter 8. The variations in the wavelength and amplitude of these waves depending upon the effect of an electric field and a magnetic field are also summarised. Students are expected to remember the EM spectrum in terms of decreasing or increasing order of wavelength.

### NCERT Solutions Class 12 Physics Chapter 8

NCERT Solutions Class 12 Physics Chapter 8 Electromagnetic waves are available on the Extramarks web portal. The notes include well-explained and step-by-step solutions for all questions of NCERT books. The physics chapter 8 class 12 notes cover all essential topics by following a specific format. This enables them to revise the chapter quickly. Students can also view NCERT Solutions of other chapters included in the NCERT Physics Syllabus of Class 12. With the help of these notes, they can try to solve the questions on their own.

Refer below for full details about NCERT Solutions Class 12 Physics Chapter 8:

Students can also check out and explore other NCERT Solutions from the latest edition of CBSE(NCERT) books on our Extramarks website:

• NCERT Solutions Class 1
• NCERT Solutions Class 2
• NCERT Solutions Class 3
• NCERT Solutions Class 4
• NCERT Solutions Class 5
• NCERT Solutions Class 6
• NCERT Solutions Class 7
• NCERT Solutions Class 8
• NCERT Solutions Class 9
• NCERT Solutions Class 10
• NCERT Solutions Class 11

NCERT Exemplar Class 12 Physics

The NCERT Exemplar Problems with Solutions is available on the Extramarks website. In addition, all CBSE students can use the NCERT Solutions Class 12 Physics Chapter 8 to get a clear understanding of the topic. It includes MCQs, HOTS, Worksheets, objective and subjective questions. These NCERT Solutions are an important resource that will help students in the CBSE exams and competitive entrance examination preparation.

Extramarks realizes the importance of Exemplar books for students of Class 12. Hence, we provide all-inclusive NCERT Exemplar Class 12 Physics Solutions. It will help students to get important tips and shortcut methods useful for solving physics problems accurately in no time. At Extramarks, students can also get different study materials like CBSE previous year question papers, CBSE sample papers, etc., to boost the exam preparation.

### Key Features of NCERT Solutions Class 12 Physics Chapter 8

In the NCERT Solutions Class 12 Physics Chapter 8, every question is solved with the utmost care by the experts at Extramarks. These solutions are with reference to the CBSE syllabus keeping in mind the marks allotted for each concept. It is an important reference material as it helps students gain confidence and thus establish a strong command of the subject.

Students can prepare for the annual examinations with the help of the NCERT Solutions Class 12 Physics Chapter 8. The notes are created in a simple and concise manner to make even the most difficult concept easy to understand. Here are some more features:

• The NCERT Solutions Class 12 Physics Chapter 8 helps students to increase their preparation level
• The NCERT Solutions prepares students to attempt Physics questions of all difficulty levels.
• Students may refer to the notes to clear their doubts, especially with difficult topics.

Q.1 Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm and separated by 5 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 valid at each plate of the capacitor Explain? Ans.

$\begin{array}{l}\text{a) As capacitance C of the capacitor is, C =}\frac{{\text{ε}}_{\text{0}}\text{A}}{\text{d}}\text{thus, we get}\\ \text{C =}\frac{{\text{(8.854×10}}^{\text{-12}}{\text{Fm}}^{-1}{\text{×πr}}^{\text{2}}\text{)}}{\text{d}}\\ \text{C=}\frac{{\text{8.854×10}}^{\text{-12}}{\text{Fm}}^{-1}\text{× 3.14 ×}{\left({\text{12×10}}^{\text{-2}}\text{}\mathrm{m}\right)}^{\text{2}}}{{\text{5×10}}^{\text{-2}}\text{}\mathrm{m}}\\ \text{C = 8 pF}\\ \text{As, V=}\frac{\text{Q}}{\text{C}}\text{ thus, }\frac{\text{dV}}{\text{dt}}\text{=}\frac{\text{dQ}}{\text{Cdt}}\\ \frac{\text{dV}}{\text{dt}}\text{=}\frac{\text{I}}{\text{C}}\text{=}\frac{\text{0.15 A}}{{\text{8×10}}^{\text{-12}}\text{}\mathrm{F}}{\text{ = 1.875×10}}^{\text{10}}{\text{Vs}}^{\text{-1}}\text{\hspace{0.17em}}\\ \left(\text{b}\right)\text{Displacement current}\\ {\text{I}}_{\text{d}}{\text{= ε}}_{\text{0}}\frac{\text{dΦ}}{\text{dt}}{\text{= ε}}_{\text{0}}\frac{\text{d}\left(\text{EA}\right)}{\text{dt}}{\text{= ε}}_{\text{0}}\text{A}\frac{\text{dE}}{\text{dt}}\text{}\\ {\text{I}}_{\text{d}}{\text{= ε}}_{\text{0}}\text{A}\frac{\text{dV}}{\text{d}\left(\text{dt}\right)}\text{= C}\frac{\text{dV}}{\text{dt}}\text{}\\ {\text{Thus, I}}_{\text{d}}\text{=}\left({\text{8×10}}^{\text{-12}}\mathrm{F}\right)\left({\text{1.875×10}}^{\text{10}}\text{}{\mathrm{Vs}}^{-1}\right)\text{=0.15 A}\\ \text{Displacement current is equal to conduction current.}\\ \left(\text{c}\right)\text{Yes, Kirchhoff’s first rule is valid at each plate of the capacitor as}\\ \text{conduction current entering one plate is equal to the displacement current leaving that plate.}\end{array}$

Q.2 A parallel plate capacitor (shown in figure) made of circular plates each of radius R = 6 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V a.c supply with an angular frequency of 300 rads-1.
(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 cm from the axis between the plates. Ans.

$\begin{array}{l}\left(\text{a}\right){\text{Here, V\hspace{0.17em}}}_{\text{rms}}{\text{= 230 V , C =100 pF = 10}}^{\text{-10}}{\text{F , ω = 300 rad s}}^{\text{-1}}\\ {\text{As, I\hspace{0.17em}}}_{\text{rms}}\text{=}\frac{{\text{V\hspace{0.17em}}}_{\text{rms}}}{{\text{X\hspace{0.17em}}}_{\text{c}}}\text{=}\frac{{\text{V\hspace{0.17em}}}_{\text{rms}}}{\text{C\hspace{0.17em}ω}}\\ {\text{Thus, I\hspace{0.17em}}}_{\text{rms}}{\text{= 230 V × 10}}^{\text{-10}}{\text{F × 300 rad s}}^{\text{-1}}\text{= 6.9 µA}\end{array}$ $\begin{array}{l}\left(\text{b}\right)\text{Yes, the conduction current is equal to the displacement current because}\\ {\text{I}}_{\text{d}}{\text{= ε}}_{\text{0}}\frac{\text{dΦ}}{\text{dt}}{\text{= ε}}_{\text{0}}\frac{\text{d}\left(\text{EA}\right)}{\text{dt}}\text{}\\ {\text{Thus, I}}_{\text{d}}{\text{= ε}}_{\text{0}}\text{A}\frac{\text{dV}}{\text{d}\left(\text{dt}\right)\text{}}\text{= C}\frac{\text{dV}}{\text{dt}}\\ \text{As, }\frac{\text{dV}}{\text{dt}}\text{=}\frac{{\text{I}}_{\text{C}}}{\text{C}}\text{ }\\ {\text{Thus, I}}_{\text{d}}\text{=}\frac{{\text{CI}}_{\text{C}}}{\text{C}}{\text{= I}}_{\text{C}}\left(\text{conduction current}\right)\end{array}$

$\begin{array}{l}\\ \left(\text{c}\right)\text{As per the figure given below, using Ampere’s circuital law we get,}\end{array}$ $\begin{array}{l}{\text{B = ε}}_{\text{0}}{\text{\hspace{0.17em}µ}}_{\text{0\hspace{0.17em}}}\text{r}\frac{\text{dE}}{\text{2 dt}}\text{}\\ \text{As, E =}\frac{\text{q}}{{\text{A\hspace{0.17em}ε}}_{\text{0}}}\\ \text{Thus,}\frac{\text{dE}}{\text{dt}}\text{=}\frac{\left(\frac{\text{dq}}{\text{dt}}\right)}{{\text{A\hspace{0.17em}ε}}_{\text{0}}}\\ \text{}\frac{\text{dE}}{\text{dt}}\text{=}\frac{{\text{I\hspace{0.17em}}}_{\text{rms}}}{\mathrm{\pi }{\text{\hspace{0.17em}R}}^{\text{2}}{\text{\hspace{0.17em}ε}}_{\text{0}}}\text{}\\ \text{Hence, B becomes; B =}\left(\frac{{\text{ε}}_{\text{0\hspace{0.17em}}}{\text{µ\hspace{0.17em}}}_{\text{0}}\text{r}}{\text{2}}\right)\text{x}\left(\text{}\frac{{\text{I}}_{\text{\hspace{0.17em}rms}}}{\mathrm{\pi }{\text{\hspace{0.17em}R\hspace{0.17em}}}^{\text{2}}{\text{\hspace{0.17em}ε\hspace{0.17em}}}_{\text{0}}}\text{}\right)\text{=}\frac{{\text{µ}}_{\text{0\hspace{0.17em}}}{\text{I\hspace{0.17em}}}_{\text{rms}}\text{\hspace{0.17em}r}}{\text{2}\mathrm{\pi }{\text{\hspace{0.17em}R}}^{\text{2}}}\\ {\text{Therefore, amplitude of magnetic field will be; B}}_{\text{0}}\text{=}\frac{{\text{µ}}_{\text{0}}{\text{\hspace{0.17em}I}}_{\text{0}}\text{\hspace{0.17em}r}}{\text{2}\mathrm{\pi }{\text{\hspace{0.17em}R}}^{\text{2}}}\\ {\text{or, B}}_{\text{0}}\text{=}\frac{{\text{µ}}_{\text{0}}\sqrt{\text{2}}{\text{\hspace{0.17em}I\hspace{0.17em}}}_{\text{rms}}}{\text{2\hspace{0.17em}}\mathrm{\pi }{\text{\hspace{0.17em}R}}^{\text{2}}}\text{}\\ {\text{B}}_{\text{0}}\text{=}\frac{\left(\text{4\hspace{0.17em}}\mathrm{\pi }{\text{×10}}^{\text{-7}}{\text{NA}}^{\text{-2}}\right)\text{}\left({\text{3×10}}^{\text{-2}}\text{m}\right)\text{}\left(\text{1.414}\right)\text{}\left({\text{6.9 ×10}}^{\text{-6}}\text{A}\right)}{\text{2\hspace{0.17em}}\mathrm{\pi }\text{×}{\left({\text{6 ×10}}^{\text{-2}}\text{m}\right)}^{\text{2}}\text{}}\text{}\\ \text{}⇒{\text{B}}_{\text{0}}{\text{= 1.63 ×10}}^{\text{-11}}\text{Tesla}\end{array}$

Q.3 What physical quantity is the same for X-rays of wavelength 10-10 m, red light of wavelength 6800 Angstrom and radio waves of wavelength 500 m?

Ans.

$\begin{array}{l}\text{The speed of light (i}\text{.e}\text{.) c = 3}×{\text{10}}^{\text{8}}{\text{ms}}^{\text{-1}}\text{remains the same in a vacuum for given X rays, red light and}\\ \text{radio waves}\text{.}\end{array}$

Q.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?

Ans.

The electric field vector E and magnetic field vector B are perpendicular to each other and they lie in x-y plane. $\begin{array}{l}\text{If the frequency of wave},\text{ }\mathrm{\nu }=30\text{\hspace{0.17em}}\mathrm{MHz}\\ \mathrm{\lambda }=\frac{\text{c}}{\mathrm{\nu }}=\frac{\text{3}×\text{1}{0}^{\text{8}}\text{}{\mathrm{ms}}^{-1}}{\text{3}0×\text{1}{0}^{\text{6}}\text{Hz}}=\text{1}0\text{m}\end{array}$

Q.5 A radio can tune into any station in the 7.5 MHz to 12 MHz. What is the corresponding wavelength band?

Ans.

$\begin{array}{l}\text{As, λ =}\frac{\text{c}}{\mathrm{\nu }}\text{=}\frac{{\text{3 ×10}}^{\text{8}}\text{}{\mathrm{ms}}^{-1}}{{\text{7.5 × 10}}^{\text{6}}\text{}\mathrm{Hz}}\text{= 40 m}\\ {\text{Similarly λ}}^{\text{’}}\text{=}\frac{\text{c}}{{\mathrm{\nu }}^{\text{’}}\text{}}\text{=}\frac{{\text{3 ×10}}^{\text{8}}\text{}{\mathrm{ms}}^{-1}}{{\text{12 ×10}}^{\text{6}}\text{}\mathrm{Hz}}\text{= 25 m}\\ \text{Thus, the corresponding wavelength band is 40 m to 25 m.}\end{array}$

Q.6 A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator?

Ans.

Frequency of the electromagnetic waves produced by the oscillator will be same as the frequency of oscillating charged particle (i.e.) 109 Hz.

Q.7 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of electric field part of the wave?

Ans.

$\begin{array}{l}\text{As},\text{c}=\frac{{\text{E}}_{0}}{{\text{B}}_{0}}\text{thus}\\ {}_{}{\text{E}}_{0}={\text{cB}}_{0}\\ =\text{3}×\text{1}{0}^{\text{8}}{\text{ms}}^{\text{-1}}×\text{51}0×\text{1}{0}^{-\text{9}}\text{T}\\ ={\text{153 NC}}^{\text{-1}}\end{array}$

Q.8 Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is ν = 50.0 MHz. (a) Determine, B0, ω, k, and λ. (b) Find expressions for E and B.

Ans.

$\begin{array}{l}\text{(a)}\left(\text{i}\right)\text{As, c =}\frac{{\text{E}}_{\text{0}}}{{\text{B}}_{\text{0}}}\text{\hspace{0.17em}}\\ \therefore {\text{B}}_{\text{0}}\text{=}\frac{{\text{E}}_{\text{0}}}{\text{c}}\text{}\\ {\text{B}}_{\text{0}}\text{=}\frac{{\text{120 NC}}^{\text{-1}}\text{}}{{\text{3×10}}^{\text{8}}{\text{ms}}^{\text{-1}}}\text{= 400 nT}\\ \left(\text{ii}\right)\text{Angular\hspace{0.17em}frequency\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}by\hspace{0.17em}ω = 2}\mathrm{\pi \nu }\\ \mathrm{\omega }{\text{= 3.14×50×10}}^{\text{6}}{\text{rad/s = 3.14×10}}^{\text{8}}\text{\hspace{0.17em}rad/s}\\ \left(\text{iii}\right)\text{Propagation\hspace{0.17em}constant,\hspace{0.17em}k =}\frac{\text{2}\mathrm{\pi }}{\text{λ}}\\ \text{k =}\frac{\text{2×3.14}}{\text{6 m}}\text{= 1.05 rad/m}\\ \left(\text{iv}\right)\text{λ =}\frac{\text{c}}{\mathrm{\nu }}\\ \text{λ =}\frac{{\text{3×10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{{\text{50×10}}^{\text{6}}{\text{s}}^{\text{-1}}}\text{= 6 m}\end{array}$

(b) Suppose a plane electromagnetic wave is propagating in X direction. Then, the electric field vector will be in Y-direction and magnetic field will be in Z-direction. This is because all three vectors are mutually perpendicular to each other. $\begin{array}{l}{\text{Electric\hspace{0.17em}field\hspace{0.17em}vector,​\hspace{0.17em}E = E}}_{\text{0}}\text{sin}\left(\text{kx – ωt}\right)\stackrel{^}{\mathrm{j}}\\ {\text{E = (120\hspace{0.17em} NC}}^{\text{-1}}\text{)\hspace{0.17em}sin\hspace{0.17em}}\left({\text{(1.05 \hspace{0.17em}radm}}^{\text{-1}}\text{)}\mathrm{x}{\text{– (3.14×10}}^{\text{8}}{\text{\hspace{0.17em}rads}}^{\text{-1}}\text{)\hspace{0.17em}t}\right)\stackrel{^}{\mathrm{j}}\text{and}\\ {\text{Magnetic\hspace{0.17em}field\hspace{0.17em}vector,​\hspace{0.17em}B = B}}_{\text{0}}\text{sin}\left(\text{kx – ωt}\right)\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ {\text{B = (400×10}}^{\text{-7}}\text{\hspace{0.17em}T)\hspace{0.17em}sin\hspace{0.17em}}\left({\text{(1.05\hspace{0.17em} radm}}^{\text{-1}}{\text{)x – (3.14×10}}^{\text{8}}{\text{\hspace{0.17em}rads}}^{\text{-1}}\text{)\hspace{0.17em}t}\right)\stackrel{^}{\mathrm{k}}\text{\hspace{0.17em} with usual units.}\end{array}$

Q.9 The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula

$\text{E}=\text{h}\nu$

(for energy of a quantum of radiation : photon) and obtain the photon energy in units of eV for different parts of electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?

Ans.

$\begin{array}{l}\text{As per the relation for photon energy E = h}\mathrm{\nu }\text{ joules}\\ \text{E =}\frac{\text{h}\mathrm{\nu }}{{\text{1.6 × 10}}^{\text{-19}}}\text{eV}\\ \text{As,}\mathrm{\nu }\text{=}\frac{\text{c}}{\text{λ}}\text{ then}\\ \text{we get, E =}\frac{\text{h\hspace{0.17em}c}}{\text{λ\hspace{0.17em}e-}}\\ \text{For, wavelength = λ, we obtain}\\ {\text{E}}_{\text{λ}}\text{=}\frac{\left({\text{6.63×10}}^{\text{-34}}\text{Js}\right){\text{( 3×10}}^{\text{8}}{\text{ms}}^{-1}\text{)}}{\left({\text{λ ×1.6×10}}^{\text{-19}}\text{}\mathrm{C}\right)}\text{}\\ \text{Similarly wavelengths of different parts of electromagnetic spectrum can be obtained as}\\ {\text{E}}_{\text{λ}}\text{=}\left(\frac{{\text{1.24 × 10}}^{\text{-6}}}{\text{λ}}\right)\text{eV\hspace{0.17em}}\\ \left(\text{a}\right)\text{For radio waves let λ = 1 km, then}\\ {\text{E}}_{\text{λ}}\text{=}\left(\frac{{\text{1.24 × 10}}^{\text{-6}}}{\text{1000 m}}\right)\text{eV =}\left({\text{1.24×10}}^{\text{-9}}\right)\text{eV }\\ \left(\text{b}\right)\text{For microwaves let λ = 1 cm, then}\\ {\text{E}}_{\text{λ}}\text{=}\left(\frac{{\text{1.24 × 10}}^{\text{-6}}}{{\text{10}}^{\text{-2}}}\right)\text{eV =}\left({\text{1.24×10}}^{\text{-4}}\right)\text{eV\hspace{0.17em}}\\ \left(\text{c}\right)\text{For visible light let λ = 1 µm, then}\\ {\text{E}}_{\text{λ}}\text{=}\left(\frac{{\text{1.24 × 10}}^{\text{-6}}}{{\text{10}}^{\text{-6}}}\right)\text{eV = 1.24 eV\hspace{0.17em}.}\\ \left(\text{d}\right){\text{For X rays let λ = 1nm = 10}}^{\text{-9}}\text{\hspace{0.17em}m, then}\\ {\text{E}}_{\text{λ}}\text{=}\left(\frac{{\text{1.24 × 10}}^{\text{-6}}}{{\text{10}}^{\text{-9}}}\right)\text{eV =}\left({\text{1.24 ×10}}^{\text{3}}\right)\text{eV\hspace{0.17em}}\\ \left(\text{e}\right){\text{For gamma rays let λ = 10}}^{\text{-12}}\text{\hspace{0.17em}m, then}\\ {\text{E}}_{\text{λ}}\text{=}\left(\frac{{\text{1.24 × 10}}^{\text{-6}}}{{\text{10}}^{\text{-12}}}\right)\text{eV =}\left({\text{1.24 ×10}}^{\text{6}}\right)\text{eV\hspace{0.17em}}\end{array}$

Q.10 In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2 × 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 E field equals the average energy density of the B field. (c = 3 x 108 ms-1).

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{As,wavelength of the​\hspace{0.17em}wave, λ =}\frac{\text{c}}{\mathrm{\nu }}\text{}\\ \text{λ =}\frac{\text{c}}{\mathrm{\nu }}\text{}\\ \text{=}\frac{{\text{3 × 10}}^{\text{8}}{\text{ms}}^{-1}\text{}}{{\text{2 ×10}}^{\text{10}}{\text{s}}^{-1}\text{}}\\ {\text{= 1.5 ×10}}^{\text{-2}}\text{\hspace{0.17em}m\hspace{0.17em}}\\ \left(\text{b}\right)\text{Using c =}\frac{{\text{E}}_{\text{0}}}{{\text{B}}_{\text{0}}}\text{,}\\ {\text{we get amplitude of the oscillating magnetic field, B}}_{\text{0}}\text{=}\frac{{\text{E}}_{\text{0}}}{\text{c}}\\ {\text{Thus B}}_{\text{0}}\text{=}\frac{{\text{48 Vm}}^{-1}}{\left({\text{3×10}}^{\text{8}}{\text{ms}}^{-1}\right)}\\ {\text{= 1.6 ×10}}^{\text{-7}}\text{T\hspace{0.17em}}\\ \left(\text{c}\right){\text{In vacuum, the average energy density of electric field (U\hspace{0.17em}}}_{\text{E}}\text{),}\\ {\text{U\hspace{0.17em}}}_{\text{E}}\text{=}\frac{\text{1}}{\text{2}}{\text{\hspace{0.17em}ε}}_{\text{0\hspace{0.17em}}}{\text{E}}^{\text{2}}\\ {\text{whereas, the average energy density of magnetic field (U\hspace{0.17em}}}_{\text{B}}\text{),}\\ {\text{ U\hspace{0.17em}}}_{\text{B}}\text{=}\frac{\text{1}}{\text{2}}\frac{{\text{B}}^{\text{2}}}{{\text{µ}}_{\text{0}}}\\ \text{Thus}\\ \frac{{\text{U\hspace{0.17em}}}_{\text{E}}}{{\text{U}}_{\text{\hspace{0.17em}B}}}\text{=}\frac{{\text{µ}}_{\text{0}}{\text{\hspace{0.17em}ε}}_{\text{0}}{\text{\hspace{0.17em}E}}^{\text{2}}\text{}}{{\text{B}}^{\text{2}}}\\ \text{=}\frac{{\text{µ}}_{\text{0\hspace{0.17em}}}{\text{ε}}_{\text{0}}}{{\text{µ}}_{\text{0\hspace{0.17em}}}{\text{ε}}_{\text{0}}}\text{= 1}\\ \text{as,}\frac{\text{E}}{\text{B}}\text{=}\frac{\text{1}}{\sqrt{{\text{µ}}_{\text{0\hspace{0.17em}}}{\text{ε}}_{\text{0}}\text{}}}\\ {\text{Hence U\hspace{0.17em}}}_{\text{E}}{\text{= U\hspace{0.17em}}}_{\text{B}}\end{array}$

Q.11 Suppose that electric field part of an electromagnetic wave in vacuum is

$\stackrel{\to }{\text{E}}{\text{= 3.1 \hspace{0.17em}N\hspace{0.17em}C}}^{\text{-1}}\text{cos\hspace{0.17em}}\left[{\text{(1.8 \hspace{0.17em}radm}}^{\text{-1}}{\text{)\hspace{0.17em}y + (5.4 × 10}}^{\text{8}}{\text{\hspace{0.17em}rads}}^{\text{-1}}\text{)\hspace{0.17em}t}\right]\text{\hspace{0.17em}}\stackrel{^}{\mathrm{i}}$

(a) What is the direction of propagation?
(b) What is the wavelength λ?
(c) What is the frequency

$\mathrm{\upsilon }$

?
(d) What is the amplitude of the magnetic field part of the wave?
(e) Write an expression for the magnetic field part of the wave.

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{From the given equation, it is clear that the direction of propagation of the wave is}\\ \text{along negative y-direction}\left(\text{i.e}\right)\text{–j, because coefficient of y is positive.}\\ \left(\text{b}\right)\text{Comparing the given equation with the equation,}\\ {\text{E = E}}_{\text{0}}{\text{\hspace{0.17em}cos\hspace{0.17em}(k\hspace{0.17em}y+ω\hspace{0.17em}t), we get the value of k = 1.8 \hspace{0.17em}radm}}^{\text{-1}}\text{,}\\ {\text{ω = 5.4 × 10}}^{\text{8}}{\text{\hspace{0.17em}rad\hspace{0.17em}s}}^{\text{-1}}\text{,}\\ {\text{E}}_{\text{0}}{\text{= 3.1\hspace{0.17em}NC}}^{\text{-1}}\text{ then}\\ \text{wavelength is given by}\\ \text{ λ =}\frac{\text{2\hspace{0.17em}π}}{\text{k}}\\ \text{=}\frac{\text{2\hspace{0.17em}π}}{\text{1.8}}\text{= 3.5 m\hspace{0.17em}}\\ \left(\text{c}\right)\text{As, }\mathrm{\nu }\text{=}\frac{\text{ω}}{\text{2\hspace{0.17em}π}}\text{=}\frac{{\text{5.4×10}}^{\text{6}}}{\text{2\hspace{0.17em}π}}\\ \text{= 0.86 MHz\hspace{0.17em}}\\ \left(\text{d}\right)\text{c =}\frac{{\text{E}}_{\text{0}}}{{\text{B}}_{\text{0}}}{\text{ thus, B}}_{\text{0}}\text{=}\frac{{\text{E}}_{\text{0}}}{\text{c}}\\ \text{=}\frac{\text{3.1}}{\left({\text{3×10}}^{\text{8}}\right)}{\text{=1.03×10}}^{\text{-8}}\text{=10.3 \hspace{0.17em}nT\hspace{0.17em}}\\ \left(\text{e}\right)\text{Using B =}\left({\text{B}}_{\text{0}}\text{\hspace{0.17em}cos}\left(\text{k\hspace{0.17em}y+ ω\hspace{0.17em}t}\right)\right)\text{, we get}\\ \text{B =}\left(\text{10.03 nT}\right)\text{cos\hspace{0.17em}}\left[{\text{(1.8\hspace{0.17em}rad\hspace{0.17em}m}}^{\text{-1}}{\text{)\hspace{0.17em}y+(5.4×10}}^{\text{8}}{\text{\hspace{0.17em}rad\hspace{0.17em}s}}^{\text{-1}}\text{)\hspace{0.17em}t}\right]\text{}\stackrel{^}{\mathrm{k}}\text{\hspace{0.17em}}\end{array}$

Q.12 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.

Ans.

$\begin{array}{l}\text{Power rating of the bulb, P = 100 W}\\ \text{It is given that 5 % of its power is convered into visible radiation.}\\ \therefore \text{Power of visible radiation, P’ =}\frac{\text{5}}{\text{100}}\text{× 100 = 5 W}\\ \left(\text{a}\right)\text{As average intensity of radiation at 1 m}\\ \text{distance from the bulb, I =}\frac{\text{Power}}{{\text{4\hspace{0.17em}π\hspace{0.17em}r}}^{\text{2}}}\\ \text{Thus, intensity, I =}\frac{\text{5 W}}{{\text{4×π×1 m}}^{\text{2}}}{{\text{= 0.4 Wm}}^{\text{–}}}^{\text{2}}\text{.}\\ \left(\text{b}\right)\text{Average intensity of radiation at 10 m}\\ \text{Distance from the bulb,I’ =}\frac{\text{5 W}}{{\text{4×π×10}}^{\text{2}}{\text{m}}^{\text{2}}}{\text{= 0.004 Wm}}^{\text{-2}}\end{array}$

Q.13 Use the formula

${\mathrm{\lambda }}_{\text{m}}\text{T}$

= 0.29 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?

Ans.

$\begin{array}{l}{\text{According to\hspace{0.17em}Wein’s law, λ}}_{\text{m}}\text{T = 0.29\hspace{0.17em}cm K, thus T =}\frac{\text{0.29 cm K}}{{\text{λ}}_{\text{m}}}\text{ }\\ {\text{which implies that, T = 0.29 K for λ}}_{\text{m}}\text{= 1\hspace{0.17em}cm\hspace{0.17em}.}\\ \left(\text{a}\right){\text{For λ}}_{\text{m}}{\text{= 10}}^{\text{-10}}\text{cm we get,}\\ \text{T =}\frac{\text{0.29 cm K}}{{\text{10}}^{\text{-10}}\text{cm}}\\ {\text{= 2.9×10}}^{\text{9}}\text{K\hspace{0.17em}}\\ \left(\text{b}\right){\text{For λ}}_{\text{m}}{\text{=10}}^{\text{-6}}\text{\hspace{0.17em}m}\\ {\text{=10}}^{\text{-4}}\text{cm}\\ \text{we get,}\\ \text{T =}\frac{\text{0.29 cmK}}{{\text{10}}^{\text{-4}}\text{cm}}\\ \text{= 2900 \hspace{0.17em}K}\\ \text{Similarly, temperatures for other wavelengths can be found. These numbers tell us the temperature ranges required for obtaining radiations in different}\\ {\text{parts of the electromagnetic spectrum. Thus to obtain visible radiation of wavelength 5 x 10}}^{\text{-7}}\text{m, the source should have a temperature of about}\\ \text{T =}\frac{\text{0.29 cmK}}{{\text{5×10}}^{\text{-5}}\text{cm}}\\ \text{= 5800 K}\\ \text{T}\simeq \text{6000 K.}\end{array}$

Q.14 Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs:

(a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space)

(b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen)

(c) 2.7K ( temperature associated with the isotropic radiation filling all space – thought to be a relic of the big-bang origin of the universe)

(d) 5890 A – 5896 A (double lines of sodium)

(e) 14.4 keV (energy of a particular transition in Fe nucleus associated with a famous high resolution spectroscopic method.

Ans.

$\begin{array}{l}\left(\text{a}\right){\text{Short radio wave as wave length is of the order of 10}}^{\text{-2}}\text{m.}\\ \left(\text{b}\right){\text{Short radio wave as frequency is of the order of 10}}^{\text{-9}}\text{Hz.}\\ \left(\text{c}\right){\text{As λ}}_{\text{m}}\text{T = 0.29 cm K, thus,}\\ {\text{λ}}_{\text{m}}\text{=}\frac{\text{0.29}}{\text{T}}\text{=}\frac{\text{0.29 cm K}}{\text{2.7 K}}\\ \text{= 0.11 cm}\\ \text{Thus wavelength corresponds to microwave region of}\\ \text{e.m. spectrum\hspace{0.17em}.}\\ \left(\text{d}\right)\text{Visible radiations}\left(\text{Yellow light}\right){\text{as given wavelengths are of the order of 10}}^{\text{-7}}\text{m.}\\ \left(\text{e}\right){\text{E = 14.4\hspace{0.17em}keV = 14.4 × 10}}^{\text{3}}{\text{× 1.6 × 10}}^{\text{-19}}\text{\hspace{0.17em}J}\\ \text{Using ,}\\ \text{E = hν}\\ \text{ν =}\frac{\text{E}}{\text{h}}\\ \text{=}\frac{\left({\text{14.4 × 10}}^{\text{3}}{\text{× 1.6 x 10}}^{\text{-19}}\text{J}\right)}{\left({\text{6.63 ×10}}^{\text{-34}}\mathrm{Js}\right)}\\ {\text{= 3.47 × 10}}^{\text{18}}\text{\hspace{0.17em}Hz\hspace{0.17em}.}\\ \text{This corresponds to X rays.}\end{array}$

(b) It is necessary to use satellites for long distance TV transmission. Why?
(c) Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting earth. Why?
(d) The small ozone layer on top of the stratosphere is crucial for human survival. Why?
(e) If the earth did not have its atmosphere, would its average surface temperature be higher or lower than what it is now?
(f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a devastating effect on life of earth. What might be the basis of this prediction?

Ans.

(a) As long distance radio broadcasts make use of sky waves where Ionosphere of earth’s atmosphere reflects the radiations of this range.

(b) For every long distance transmission of TV signals, a very high frequency is required. Waves of this frequency just pass through the ionosphere, without being reflected, so a satellite is required to return the signals to the earth.

(c) As X-rays has smaller wavelength, it can be absorbed by the earth. Hence X ray astronomy is possible only from satellites orbiting the earth but visible waves and radio waves can pass through the atmosphere therefore we can work with visible waves and radio waves on earth’s surface.

(d) The small ozone layer on top of the stratosphere is crucial for human survival as it absorbs harmful radiations (ultraviolet radiations) present in the sunlight and prevents it from reaching the surface of the earth. Ultraviolet radiations are harmful for the life on earth.

(e) In this case, there will be no green house effect. So the earth will be at lower temperature.

(f) In the case of worldwide nuclear war the sky may get overcast with clouds. These clouds will prevent sunlight from reaching many parts of the globe. Thus earth will be as cool as in winter.

## 1. Which essential concepts are in NCERT Solutions Class 12 Physics Chapter 8?

NCERT Solutions Class 12 Mathematics Chapter 1 covers solutions of all questions from the NCERT books based on topics like Displacement Current, Electromagnetic Waves, and Sources of electromagnetic waves. Students also gain detailed information about the Electromagnetic Spectrum and different types of electromagnetic waves like Radio Waves, Microwaves, Infrared waves, Visible rays, Ultraviolet (UV) rays, X-rays, and Gamma rays. These NCERT solutions give clear and precise information about these topics.

## 2. Which chapters are included in the NCERT Exemplar Class 12 Physics?

Extramarks has covered all essential topics and provided well-structured answers of all the chapters in the NCERT Solutions Class 12 Physics, which are as follows:

Chapter 1 Electric Charges and Fields

Chapter 2 Electrostatic Potential and Capacitance

Chapter 3 Current Electricity

Chapter 4 Moving Charges and Magnetism

Chapter 5 Magnetism and Matter

Chapter 6 Electromagnetic Induction

Chapter 7 Alternating Current

Chapter 8 Electromagnetic Waves

Chapter 9 Ray Optics and Optical Instruments

Chapter 10 Wave Optics

Chapter 11 Dual Nature of Radiation and Matter

Chapter 12 Atoms

Chapter 13 Nuclei

Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits

Chapter 15 Communication Systems