NCERT Solutions for Class 12 Physics Chapter 8 – Free PDF Download

The NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves provide detailed answers to the textbook theory questions, numerical problems, worksheets and exercises. In Class 12 Physics, there are many complicated formulas and equations. In order to score good marks in the Class 12 board examination, it is important to solve the exercise questions provided at the end of each chapter using the NCERT Solutions for Class 12 Physics.

Most frequently, questions that are asked in CBSE Class 12 Physics board exams are directly taken from the NCERT textbook. Electromagnetism is one of the repeatedly asked topics in the board exam. Hence, students are suggested to refer to the NCERT Solutions for Class 12 Physics to attain a firm grip on the chapter as well as the subject. Students can download the NCERT Solutions for Class 12 Physics Chapter 8 from the link below:

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

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Class 12 Physics Chapter 8 NCERT Solutions Electromagnetic Waves Important Questions


Q 8.1) The 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 0.15A.

(a) Calculate the capacitance and the rate of change of the 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.

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves Question 1

Answer 8.1:

Given Values:

The radius of each circular plate (r) is 12 cm or 0.12 m

The distance between the plates (d) is 5 cm or 0.05 m

The charging current (I) is 0.15 A

The permittivity of free space is

\(\begin{array}{l}\varepsilon_{0} = 8.85\times 10^{-12}\; C^{2}N^{-1}m^{-2}\end{array} \)

(a) The capacitance between the two plates can be calculated as follows:

\(\begin{array}{l}C = \frac{\varepsilon _{0} A}{d}\end{array} \)

where,

A = Area of each plate =

\(\begin{array}{l}\pi r^{2}\end{array} \)
\(\begin{array}{l}C = \frac{\varepsilon _{0} \pi r^{2}}{d}\end{array} \)

=

\(\begin{array}{l}\frac{8.85\times 10^{-12}\times \pi (0.12)^{2}}{0.05}\end{array} \)

=

\(\begin{array}{l}8.0032\times 10^{-12}\; F\end{array} \)

= 80.032 pF

The charge on each plate is given by,

q = CV

where,

V is the potential difference across the plates

Differentiation on both sides with respect to time (t) gives:

\(\begin{array}{l}\frac{\mathrm{d} q}{\mathrm{d} t} = C \frac{\mathrm{d} V}{\mathrm{d} t}\end{array} \)

But,

\(\begin{array}{l}\frac{\mathrm{d} q}{\mathrm{d} t}\end{array} \)
= Current (I)

\(\begin{array}{l}∴\frac{\mathrm{d} V}{\mathrm{d} t} = \frac{I}{C}\end{array} \)

=>

\(\begin{array}{l}\frac{0.15}{80.032\times 10^{-12}} = 1.87\times 10^{9}\; V/s\end{array} \)

Therefore, the change in the potential difference between the plates is

\(\begin{array}{l}1.87\times 10^{9}\; V/s\end{array} \)
.

(b) The displacement current across the plates is the same as the conduction current. Hence, the displacement current, id is 0.15 A.

(c) Yes

Kirchhoff’s first rule is valid at each plate of the capacitor, provided that we take the sum of conduction and displacement for current.

 

Q 8.2) 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 an (angular) frequency of 300 rad s–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 3.0 cm from the axis between the plates.

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Answer 8.2:

Radius of each circular plate, R = 6.0 cm = 0.06 m

Capacitance of a parallel plate capacitor, C = 100 pF =

\(\begin{array}{l}100\times 10^{-12}\; F\end{array} \)

Supply voltage, V = 230 V

Angular frequency,

\(\begin{array}{l}\omega = 300\; rad\;s^{-1}\end{array} \)

 

(a) Rms value of conduction current, I =

\(\begin{array}{l}\frac{V}{X_{c}}\end{array} \)

Where,

\(\begin{array}{l}X_{c}\end{array} \)
= Capacitive reactance =
\(\begin{array}{l}\frac{1}{\omega C}\end{array} \)
\(\begin{array}{l}∴ I = V\times \omega C\end{array} \)

=

\(\begin{array}{l}230\times 300\times 100\times 10^{-12}\end{array} \)

=

\(\begin{array}{l}6.9\times 10^{-6} A\end{array} \)

=

\(\begin{array}{l}6.9\; \mu A\end{array} \)

Hence, the rms value of conduction current is

\(\begin{array}{l}6.9\; \mu A\end{array} \)
.

(b) Yes, conduction current is equivalent to displacement current.

(c) Magnetic field is given as:

\(\begin{array}{l}B = \frac{\mu_{0}r}{2\pi R^{2}}I_{0}\end{array} \)

Where,

\(\begin{array}{l}\mu_{0}\end{array} \)
= Permeability of free space =
\(\begin{array}{l}4\pi \times 10^{-7}\; N\;A^{-2}\end{array} \)
\(\begin{array}{l}I_{0}\end{array} \)
= Maximum value of current =
\(\begin{array}{l}\sqrt{2}\; I\end{array} \)

r = Distance between the plates from the axis = 3.0 cm = 0.03 m

\(\begin{array}{l}∴B = \frac{4\pi\times 10^{-7}\times 0.03\times \sqrt{2}\times 6.9\times 10^{-6}}{2\pi \times (0.06)^{2}}\end{array} \)

=

\(\begin{array}{l}1.63\times 10^{-11}\; T\end{array} \)

Hence, the magnetic field at the point is

\(\begin{array}{l}1.63\times 10^{-11}\; T\end{array} \)
.

Q 8.3) What physical quantity is the same for X-rays of wavelength 10–10m, the red light of wavelength 6800 Å and radiowaves of wavelength 500 m?

Answer 8.3:

The speed of light (

\(\begin{array}{l}3\times 10^{8}\end{array} \)
m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum.

 

Q 8.4) A plane electromagnetic wave travels in a vacuum along the 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?

Answer 8.4:

The electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) and the magnetic field (H) are in the x-y plane. They are mutually perpendicular.

Frequency of the wave, v = 30 MHz =

\(\begin{array}{l}30\times 10^{6}\;s^{-1}\end{array} \)

Speed of light in vacuum, C =

\(\begin{array}{l}3\times 10^{8}\end{array} \)
m/s

The wavelength of a wave is given as

\(\begin{array}{l}\lambda = \frac{c}{v}\end{array} \)

=

\(\begin{array}{l}\frac{3\times 10^{8}}{30\times 10^{6}}\end{array} \)
= 10 m

Q 8.5) A radio can tune in to any station in the 7.5 MHz to 12 MHz bands. What is the corresponding wavelength band?

Answer 8.5:

A radio can tune to a minimum frequency,

\(\begin{array}{l}v_{1} = 7.5\; MHz = 7.5\times 10^{6}\; Hz\end{array} \)

Maximum frequency,

\(\begin{array}{l}v_{2} = 12\; MHz = 12\times 10^{6}\; Hz\end{array} \)

Speed of light, c =

\(\begin{array}{l}3\times 10^{8}\; m/s\end{array} \)

Corresponding wavelength for

\(\begin{array}{l}v_{1}\end{array} \)
can be calculated as:

\(\begin{array}{l}\lambda_{1} = \frac{c}{v_{1}}\end{array} \)
\(\begin{array}{l}=\frac{3\times 10^{3}}{7.5\times 10^{6}} = 40\;m\end{array} \)

Corresponding wavelength for

\(\begin{array}{l}v_{2}\end{array} \)
can be calculated as:

\(\begin{array}{l}\lambda_{2} = \frac{c}{v_{2}}\end{array} \)
\(\begin{array}{l}=\frac{3\times 10^{3}}{12\times 10^{6}} = 25\;m\end{array} \)

Thus, the wavelength band of the radio is 40 m to 25 m.

 

Q 8.6) 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?

Answer 8.6:

The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position, i.e. 109 Hz.

 

Q 8.7) The amplitude of the magnetic field part of a harmonic electromagnetic wave in a vacuum is B0=510 nT. What is the amplitude of the electric field part of the wave?

Answer 8.7:

The amplitude of the magnetic field of an electromagnetic wave in a vacuum,

\(\begin{array}{l}B_{0} = 510\; nT = 510\times 10^{-9}\; T\end{array} \)

Speed of light in vacuum, c =

\(\begin{array}{l}3\times 10^{8}\; m/s\end{array} \)

The amplitude of the electric field of an electromagnetic wave is given by the relation,

\(\begin{array}{l}E = cB_{0} = 3\times 10^{8}\times 510\times 10^{-9} = 153\; N/C\end{array} \)

Therefore, the amplitude of the electric field part of the wave is 153 N/C.

 

Q 8.8) Suppose that the electric field amplitude of an electromagnetic wave is

\(\begin{array}{l}E_{0} = 120\; N/C\end{array} \)
and that its frequency is v = 50 MHz. (a) Determine
\(\begin{array}{l}B_{0},\; \omega,\; k\;and\; \lambda\end{array} \)
, (b) Find expressions for E and B.

Answer 8.8:

Electric field amplitude,

\(\begin{array}{l}E_{0} = 120\; N/C\end{array} \)

Frequency of source, v = 50 MHz =

\(\begin{array}{l}50\times 10^{6}\end{array} \)
Hz

Speed of light, c =

\(\begin{array}{l}3\times 10^{8}\end{array} \)
m/s

(a) Magnitude of magnetic field strength is given as:

\(\begin{array}{l}B_{0} = \frac{E_{0}}{c}\end{array} \)

=

\(\begin{array}{l}\frac{120}{3\times 10^{8}}\end{array} \)

=

\(\begin{array}{l}40\times 10^{-8}\;=400\times 10^{-9} T = 400\; nT\end{array} \)

The angular frequency of the source is given by:

\(\begin{array}{l}\omega =2\pi v=2\pi \times 50\times 10^{6}=3.14\times 10^{8}\,rads^{-1}\end{array} \)

=

\(\begin{array}{l}3.14\times 10^{8}\end{array} \)
rad/s

The propagation constant is given as:

\(\begin{array}{l}k = \frac{\omega }{c}\end{array} \)

=

\(\begin{array}{l}\frac{3.14\times 10^{8}}{3\times 10^{8}} = 1.05\; rad/m\end{array} \)

The wavelength of the wave is given by:

\(\begin{array}{l}\lambda = \frac{c}{v}\end{array} \)

=

\(\begin{array}{l}\frac{3\times 10^{8}}{50\times 10^{6}}\end{array} \)
= 6.0 m

 

(b) Suppose the wave is propagating in the positive x-direction. Then, the electric field vector will be in the positive y-direction, and the magnetic field vector will be in the positive z-direction. This is because all three vectors are mutually perpendicular.

The equation of the electric field vector is given as:

\(\begin{array}{l}\overline{E} = E_{0}\;sin(kx – \omega t)\;\widehat{j}\end{array} \)

=

\(\begin{array}{l}120\;sin[1.05x – 3.14\times 10^{8}t]\;\widehat{j}\end{array} \)

And the magnetic field vector is given as:

\(\begin{array}{l}\overline{B} = B_{0}\;sin(kx – \omega t)\;\widehat{k}\end{array} \)

 

\(\begin{array}{l}\overline{B} = (400 \times 10^{-9}) sin[1.05x – 3.14\times 10^{8}t]\;\widehat{k}\end{array} \)

 

Q 8.9) The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hν (for the energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?

Answer 8.9:

The energy of a photon is given as:

E = hv =

\(\begin{array}{l}\frac{hc}{\lambda}\end{array} \)

Where,

h = Planck’s constant =

\(\begin{array}{l}6.6\times 10^{-34}\;Js\end{array} \)

c = Speed of light =

\(\begin{array}{l}3\times 10^{8}\;m/s\end{array} \)

If the wavelength λ is in metres and the energy is in Joule, then dividing E by 1.6 × 10-19 will convert the energy into eV.

\(\begin{array}{l}E=\frac{hc}{\lambda \times 1.6\times 10^{-19}}\,eV\end{array} \)

a) For Gamma rays, the wavelength ranges from 10-10 to 10-14 m; therefore, the photon energy can be calculated as follows:

\(\begin{array}{l}E=\frac{6.62\times 10^{-34}\times 3\times 10^8}{10^{-10}\times 1.6\times 10^{-19}}=12.4\times 10^3\approx 10^4\,eV\end{array} \)
.
Therefore,
\(\begin{array}{l}\lambda =10^{-10}\,m, energy = 10^{4}\,eV\end{array} \)

 

\(\begin{array}{l}\lambda =10^{-14}\,m, energy = 10^{8}\,eV\end{array} \)

The energy for Gamma rays ranges from 104 to 108 eV.

b) The wavelength for X-rays ranges between 10-8 m to 10-13 m

For λ = 10-8,

\(\begin{array}{l}Energy = \frac{6.62\times 10^{-34}\times 3\times 10^8}{10^{-8}\times 1.6\times 10^{-19}}=124\approx 10^2\,eV\end{array} \)

For λ = 10-13 m, energy = 107 eV

c) For ultraviolet radiation, the wavelength ranges from 4 × 10-7 m to 6 × 10-7 m.

For 4 × 10-7 m,

\(\begin{array}{l}Energy = \frac{6.62\times 10^{-34}\times 3\times 10^8}{4\times 10^{-7}\times 1.6\times 10^{-19}}=3.1\approx 10^{10}\,eV\end{array} \)

For 6 × 10-7 m, the energy is equal to 103 eV.

The energy of the ultraviolet radiation varies between 1010 to 103 eV.

d) For visible light, the wavelength ranges from 4 × 10-7 m to 7 × 10-7 m.

For 4 × 10-7, the energy is the same as above, that is, 1010 eV

For 7 × 10-7 m, the energy is 100 eV

e) For infrared radiation, the wavelength ranges between 7 × 10-7 m to 7 × 10-14 m.

The energy for 7 × 10-7 m is 100 eV

The energy for 7 × 10-14 m is 10-3 eV

f) For microwaves, the wavelength ranges from 1 mm to 0.3 m.

For 1 mm, the energy is 10-3 eV.
For 0.3 m, the energy is 10-6 eV.

g) For radio waves, the wavelength ranges from 1 m to a few km.

For 1 m, the energy is 10-6 eV.

The photon energies for the different parts of the spectrum of a source indicate the spacing of the relevant energy levels of the source.

 

Q 8.10) In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude of 48 V m–1.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the B field. [ c =

\(\begin{array}{l}3\times 10^{8}\;m\;s^{-1}\end{array} \)
]

Answer 8.10:

Frequency of the electromagnetic wave, v =

\(\begin{array}{l}2\times 10^{10}\;Hz\end{array} \)

Electric field amplitude,

\(\begin{array}{l}E_{0} = 48\;V\;m^{-1}\end{array} \)

Speed of light, c =

\(\begin{array}{l}3\times 10^{8}\;m/s\end{array} \)

 

(a) Wavelength of a wave is given as:

\(\begin{array}{l}\lambda = \frac{c}{v}\end{array} \)

=

\(\begin{array}{l}\frac{3\times 10^{8}}{2\times 10^{10}} = 0.015\; m\end{array} \)

 

(b) Magnetic field strength is given as:

\(\begin{array}{l}B_{0} = \frac{E_{0}}{c}\end{array} \)

=

\(\begin{array}{l}\frac{48}{3\times 10^{8}} = 1.6\times 10^{-7}\; T\end{array} \)

 

(c) Energy density of the electric field is given as:

\(\begin{array}{l}U_{E} = \frac{1}{2}\; \epsilon _{0} \;E^{2}\end{array} \)

And the energy density of the magnetic field is given as:

\(\begin{array}{l}U_{B} = \frac{1}{2\mu_{0}}B^{2}\end{array} \)

Where,

\(\begin{array}{l}\epsilon _{0}\end{array} \)
= Permittivity of free space

 

\(\begin{array}{l}\mu_{0}\end{array} \)
= Permeability of free space

E = cB  …(1)

Where,

\(\begin{array}{l}c = \frac{1}{\sqrt{\epsilon_{0}\; \mu_{0}}}\end{array} \)
  …(2)

Putting equation (2) in equation (1), we get

\(\begin{array}{l}E = \frac{1}{\sqrt{\epsilon_{0}\; \mu_{0}}}\; B\end{array} \)

Squaring on both sides, we get

\(\begin{array}{l}E^{2} = \frac{1}{\epsilon_{0}\; \mu_{0}}\; B^{2}\end{array} \)

 

\(\begin{array}{l}\epsilon_{0}\; E^{2} = \frac{B^{2}}{\mu_{0}}\end{array} \)

 

\(\begin{array}{l}\frac{1}{2}\; \epsilon_{0}\; E^{2} = \frac{1}{2}\; \frac{B^{2}}{\mu_{0}}\end{array} \)

=>

\(\begin{array}{l}U_{E} = U_{B}\end{array} \)

Q 8.11) Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 106 rad/s)t]} ˆi.
(a) What is the direction of propagation?
(b) What is the wavelength λ?
(c) What is the frequency ν?
(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.

Answer:

(a) The direction of motion is along the negative y-direction, i.e. along -j.

(b) The given equation is compared with the equation,

E = E0 cos (ky + ωt)

⇒ k = 1.8 rad/s

ω = 5.4 x 106 rad/s

λ = 2π/k = (2 x 3.14)/1.8 = 3.492 m

(c) Frequency, ν = ω/2π =  5.4 x 106/(2 x 3.14) = 0.859  x 10Hz.

(d) Amplitude of the magnetic field, B0 = E0/c

= 3.1/(3 x 108) = 1.03 x 10-8 T= 10.3 x 10-9 T= 10.3 nT.

(e) Bz = B0 cos (ky + ωt)ˆk ={(10.3 nT) cos[(1.8 rad/m)y + (5.4 × 106 rad/s)t]} kˆ

Q 8. 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 1m from the bulb?
(b) at a distance of 10 m?
Assume that the radiation is emitted isotropically and neglects reflection.

Answer:

(a) Average intensity of the visible radiation, I = P’/4πd2

Here, the power of the visible radiation, P’ = (5/100) x 100 = 5 W

At d = 1 m

I = P’/4πd2 = 5/(4 x 3.14 x 12) = 5/12.56 = 0.39 W/m2

(b) At d = 10 m

I = P’/4πd2 = 5/(4 x 3.14 x 102)  = 5/1256 = 0.39 x 10-2 W/m2

Q 8.13) Use the formula λ m 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?

We have the equation,  λ m T = 0.29 cm K

⇒ T = (0.29/λ m )cm K

Here, T is the temperature

λ m is the maximum wavelength of the wave

For λ m = 10-4 cm

T = (0.29/10-4)cm K = 2900 K

For the visible light, λ m = 5 x 10-5 cm

T = (0.29/ 5 x 10-5 )cm K ≈ 6000 K

Note: A lower temperature will also produce wavelength but not with maximum intensity.

Q 8. 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, known as lamb shift).
(c) 2.7 K [temperature associated with the isotropic radiation filling all space-thought to be a relic of the ‘big-bang’ origin of the universe].
(d) 5890 Å – 5896 Å [double lines of sodium]
(e) 14.4 keV [energy of a particular transition in 57Fe nucleus associated with a famous high-resolution spectroscopic method (Mössbauer spectroscopy)].

Answer:

(a) Radio waves (short-wavelength end)
(b) Radio waves (short-wavelength end)
(c) Microwave
(d) Visible light (Yellow)
(e) X-rays (or soft γ-rays) region

Q 8.15) Answer the following questions:
(a) Long-distance radio broadcasts use short-wave bands. Why?
(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 the 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 an 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 on earth. What might be the basis of this prediction?

Answer:

(a) Ionosphere reflects waves in the shortwave bands.
(b) Television signals have high frequency and high energy. Therefore, it is not properly reflected by the ionosphere. Satellites are used to reflect the TV signals.
(c) Atmosphere absorbs X-rays, while visible and radio waves can penetrate it.
(d) Ozone layer absorbs the ultraviolet radiation from the sunlight and prevents it from reaching the surface of the earth and causing damage to life.
(e) If the earth did not have an atmosphere, there would be no greenhouse effect. As a result, the temperature of the earth would decrease.
(f) The smoke clouds produced by global nuclear war would perhaps cover substantial parts of the sky, preventing solar light from reaching many parts of the globe. This would cause a ‘winter’.

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

Chapter 8 Electromagnetic Waves of Class 12 Physics is prepared as per the latest CBSE Syllabus 2023-24. Electromagnetism is a physical attraction that occurs in electrically charged particles. When a capacitor is charged using an external source, there can be a potential difference between two capacitive plates. How to calculate that along with the displacement? This question will be solved using Kirchhoff’s rules. Students can make use of the NCERT Solutions for Class 12 to learn the correct methods of solving the exercise problems.

Concepts Covered in Class 12 Physics Chapter 8 Electromagnetic Waves

  1. Introduction
  2. Displacement Current
  3. Electromagnetic Waves
    1. Sources of electromagnetic waves
    2. Nature of electromagnetic waves
  4. Electromagnetic Spectrum
    1. Radio waves
    2. Microwaves
    3. Infrared waves
    4. Visible rays
    5. Ultraviolet rays
    6. X-rays
    7. Gamma rays

In these NCERT Solutions, students will be learning topics, such as how to determine the RMS value of the conduction current, the similarities between the conduction current and displacement current, and the similarities among the wavelengths of X-rays, red lights and radio waves. Besides, they will be solving questions on the wavelength of electromagnetic waves travelling in a vacuum.

Do you want to know what the frequency of electromagnetic waves produced by the oscillator is? Want to know about the electric field part of the harmonic electromagnetic wave in a vacuum? Check out the answers in the NCERT Solutions. You will be learning the photo energy of different parts of the electromagnetic spectrum and how to obtain different scales of photon energies of electromagnetic radiation.

You will be gaining knowledge on how to prove that the energy density of one field is equal to the average energy density of another field. We know that there are more fundamental forces, such as weak and strong nuclear forces and gravitational forces. You will be finding questions on them in a different chapter. The questions covered in this chapter are very common in the board exam and, if prepared thoroughly, will definitely help you understand the concept of electromagnetism easily and score well in the board exam.

Also Access
NCERT Exemplar for Class 12 Maths Chapter 8
CBSE Notes for Class 12 Maths Chapter 8

BYJU’S provides class-wise NCERT Solutions, along with study materials, notes, books, assignments and sample papers prepared by top-notch subject experts of the country who have been involved in teaching the CBSE Syllabus for years.

Disclaimer – 

Dropped Topics – 

Example 8.1
8.3.2 Nature of Electromagnetic Waves (delete only about ether and page 277)
Examples 8.4 and 8.5
Exercises 8.11–8.15

Frequently Asked Questions on NCERT Solutions for Class 12 Physics Chapter 8

Q1

Do NCERT Solutions for Class 12 Physics Chapter 8 have answers for all the textbook questions?

The NCERT Solutions for Class 12 Physics Chapter 8 is available in PDF, designed by the subject-matter experts. These solutions are completely based on the latest CBSE Syllabus 2023-24 and cover all the important concepts in the textbook. The textbook problems are solved in a stepwise manner as per the marks weightage in the CBSE  exam. Both chapter-wise and exercise-wise PDF links are available on BYJU’S website, which can be accessed by the students to get their doubts clarified instantly.
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How can students score full marks in NCERT Solutions for Class 12 Physics Chapter 8?

The NCERT Solutions for Class 12 Physics Chapter 8 are designed by experts at BYJU’S after conducting vast research on each concept. Every minute detail is explained in a comprehensive manner to help students score well in class tests as well as in board exams. It also helps students in completing the assignments given to them on time without any difficulty.
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Are NCERT Solutions for Class 12 Physics Chapter 8 PDF enough to score well in the board exams?

NCERT Solutions for Class 12 Physics Chapter 8 are available in PDF format, which can be downloaded and used by the students without any time constraints. The solutions are created by the highly experienced faculty at BYJU’S, based on the latest CBSE Syllabus and its guidelines. The exercise-wise solutions help students to gain an overall idea about the concepts which are important for the board exams. Practising these questions on a regular basis will improve the time management and problem-solving abilities of students.
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