# NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

Q. 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. 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. What physical quantity is the same for X-rays of wavelength 10–10 m, red light of wavelength 6800 Å and radio waves of wavelength 30 cm?
Ans. The speed of light (3 × 108 m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum.
Q. Use the formula λm 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 A body at a particular temperature produces a continuous spectrum of wavelengths. In case of a black body, the wavelength corresponding to maximum intensity of radiation is given according to Planck’s law. It can be given by the relation,

$$\lambda_m=\frac{0.29}{T}\text{cm}\space K\\\text{where},\lambda_m=\text{Maximum wavelength}\\\text{T}=\text{Temperature}\\\text{Thus, the temperature for different wavelength can be obtained as :}\\\text{For}\space\lambda_m=10^{-4}\text{cm};\text{T}=\frac{0.29}{10^{-4}}=2900K\\\text{For}\space\lambda_m=5×10^{-5}\text{cm};\text{T}=\frac{0.29}{5×10^{-5}}=5800K\\\text{For}\space\lambda_m=10^{-6}\text{cm};\text{T}=\frac{0.29}{10^{-6}}=290000K\text{and so on.}$$

The numbers obtained tell us that temperature ranges are required for obtaining radiations in different parts of an electromagnetic spectrum.
As the wavelength decreases, the corresponding temperature increases.

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

• (i) 21 cm (wavelength emitted by atomic hydrogen in interstellar spaces).
• (ii) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift).
• (iii) 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].
• (iv) 5890 Å – 5896 Å [double lines of sodium].
• (v) 14.4 keV [energy of a particular transition in 57Fe nucleus associated with a famous high resolution spectroscopic method (Mössbauer spectroscopy)].
• Ans(i) Radio waves; short wavelength end of the electromagnetic spectrum.
• (ii) Radio waves; short wavelength end.
• (iii) Temperature, T = 2.7 K λm is given by Planck’s law by

$$\lambda_m=\frac{0.29}{2.7}=0.11\space\text{cm}\\\text{This wavelength corresponds to microwaves.}$$

• (iv) This is the yellow light of the visible spectrum.
• (v) Transition energy is given by the relation,

$$\text{E}=hv\\\text{Here},\space h=\text{Planck’s constant} \\= 6.6 × 10^{\normalsize–34}\space Js\\\text{v= Frequency of radiation}\\\text{Energy, E = 14.4 keV}\\\therefore\space\text{v}=\frac{E}{h}\\=\frac{14.4×10^{3}×1.6×10^{-19}}{6.6×10^{-34}}\\= 3.4 × 10^{18} Hz.\\\text{This corresponds to X-rays.}$$

Q. If the surface has an area of 20 cm2, find the average force exerted on the surface during a 30 minute time span.
• Ans. The total energy falling at the surface
• U = (18 W/cm2) × (20 cm2) × (30 × 60)
• = 6.48 × 105 J
• Thus, the total momentum delivered (for complete absorption) will be

$$\text{p}=\frac{U}{c}=\frac{6.48×10^{5}J}{3×10^{8}m/s}\\=2.16 × 10^{–3}\text{kg m}/\text{s}\\\text{The average force exerted on the surface will be}\\\text{F}=\frac{p}{t}=\frac{2.16×10^{\normalsize-3}}{0.18×10^{4}}=1.2×10^{-6}N$$

Q. 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 2.5% and it is a point source.

Ans. The bulb (point source) radiates light in all directions uniformly. At distance of 3 m, the surface area of the surrounding sphere will be

$$\text{A}=4\pi r^{2}=4\pi(3)^{2}=113 m^{2}\\\text{The intensity at this distance will be}\\\text{I}=\frac{\text{Power}}{\text{Area}}=\frac{100 W×2.5\%}{113m^{2}}\\=0.022 W/m^{2}\\\text{Half of this intensity is provided by the electric field and half by the magnetic field.}\\\frac{1}{2}\text{I}=\frac{1}{2}(\epsilon_0E^{2}_{rms}c)\\=\frac{1}{2}(0.022 W/m^{2})\\=\text{E}_\text{{rms}}\sqrt{\frac{0.022}{(8.85×10^{-12})(3×10^{8})}}V/m\\=2.9\space V/m\\\text{The peak electric field, E}_0 is,\\\text{E}_0=\sqrt{2E_{rms}}\\=\sqrt{2}×2.9V/m=4.07V/m\\\text{Compare it with electric field strength of TV or FM waves, which is of the order of a few microvolts per metre.}\\\text{Let us calculate the strength of the magnetic field. It is}\\\text{B}_{rms}=\frac{E_{rms}}{c}=\frac{2.9×Vm^{-1}}{3×10^{8} ms^{-1}}\\=9.6×10^{-9}\text{T}\\\text{The peak magnetic field is B}_0 =\sqrt{2}B_{rms}=1.4×10^{-8}\text{T}.$$

Note that although the energy in the magnetic field is equal to the energy in the electric field, the magnetic field strength is evidently very weak.

Q. 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]}
• (i) What is the direction of propagation ?
• (ii) What is the wavelength λ ?
• (iii) What is the frequency v ?
• (iv) What is the amplitude of the magnetic field part of the wave ?
• (v) Write an expression for the magnetic field part of the wave.

Ans. (i) From the given electric field vector, it can be understood that the electric field is directed along the negative X-direction. Hence, the direction of motion is along the negative

$$\text{Y-direction}\space i.e.,(-\hat{j}).\\\text{(ii) From question}\\\vec{\text{E}}=3.2\space\text{cos}\\\space[(1.8\space\text{rad/m})y+(5.4×10^{8}\text{rad/s})t]\hat{i}\space\text{…(i)}\\\text{The general equation for the electric field vector in the positive x-direction is given by}\\\vec{\text{E}}=\text{E}_0\text{sin}(kx-\omega t)\hat{i}\space…(ii)\\\text{From equations (i) and (ii)}\\\text{Electric field amplitude,}\text{E}_0=3.1\text{N/C}\\\text{Angular frequency},\omega=5.4×10^{8}\text{rad/s}\\\text{Wave number, k}=1.8\space\text{rad/m}\\\text{Wavelength,}\space\lambda=\frac{2\pi}{1.8}=3.490\space\text{m}\\\text{(iii) Frequency of wave is given as :}\\ \text{v}=\frac{\omega}{2\pi}=\frac{5.4×10^{8}}{2\pi}\\=8.6×10^{7}\text{Hz}\\\text{(iv) Magnetic field strength is given by}\\\text{B}_0=\frac{\text{E}_0}{c}\\\text{where},c=\text{Speed of light}=3×10^{8}\text{m/s}\\\therefore\text{B}_0=\frac{3.1}{3×10^{8}}=1.03×10^{-8}\text{T}\\\text{(v) From given vector field, it can be inferred that the magnetic field vector is directed along the negative Z-direction.}\\\text{ Hence, the general equation for the magnetic field vector is written as :}\\\vec{\text{B}}=\text{B}_0\text{cos}(ky+\omega t)\hat{k}\\=\lbrace(1.03×10^{-7}\text{T})\\\text{cos}[(1.8\space\text{rad/m})y+(5.4×10^{6}\space\text{rad/s})]\rbrace\hat{\text{K}}$$

• Q. Answer the following questions :