NCERT Questions for Class 12 Physics Chapter 8 – Electromagnetic Waves

Important Questions with Solutions of Class 12 Physics Chapter 8 – Electromagnetic Waves

Short Answer Questions

1) What is the role of the ozone layer in the atmosphere?

Ans – The ozone layer in the atmosphere mainly prevents harmful ultraviolet rays by acting as an absorber of those rays and saves us from harm caused by them.

2) Would the earth’s average surface temperature be higher or lower than it is currently if it didn’t have an atmosphere?

Ans – The globe warms as a result of the greenhouse effect, which traps infrared energy inside the atmosphere. The earth’s average temperature would have been low as a result.

3) Why aren’t TV broadcasts sent via sky waves? Provide two strategies for extending the TV transmission range.

Ans – Sky waves are not used to broadcast television signals because they are not reflected by the ionosphere.

Techniques for extending TV transmission range
(1) Elevated antenna
(2) Satellites that are geostationary

4) An electromagnetic wave propagates across a vacuum in the z-direction. What can you state regarding the orientations of its electric and magnetic field vectors? What is the wavelength of a wave at a frequency of 30 MHz?

Ans – Since the electromagnetic wave is traveling along the z-direction, the electric field (E) and magnetic field (B) vectors must be perpendicular to the z-direction and also perpendicular to each other.

Wave frequency, f = 30 MHz = 30 × 10⁶ s⁻¹

In vacuum speed of light will be, c = 3 × 10⁸ m/s

Wavelength of the wave,

5) A radio may receive any station within the 7.5MHz to 12MHz frequency range. What is the associated wavelength range?

Ans – To find the wavelength corresponding to a given frequency band, we use

where c = 3 × 10⁸ m/s

We need to calculate minimum and maximum frequency

For minimum frequency,

For maximum frequency,

The wavelength range of the radio is from 40m to 25m.

3 Mark Questions

1) The electric field in a plane electromagnetic wave oscillates sinusoidally with an amplitude of 48V/m and a frequency of 2 x 10¹⁶ Hz.

a) What is the electromagnetic wave’s wavelength?
b) Determine the oscillating magnetic field’s amplitude.
c) Determine the electromagnetic field’s average energy density at the wave.

Ans – Given that,

Frequency of the electric field, f = 2 × 10¹⁶ Hz

Amplitude of the electric field, E₀ = 48 V/m

Speed of light will be, c = 3 × 10⁸ m/s

a) The wavelength of an electromagnetic wave in a vacuum is given by,

Substituting the values,

= 1.5 × 10⁻⁸ m = 15 nm

b) The correlation between the amplitudes of the electric field E₀ and the magnetic field B₀ in an electromagnetic wave is defined by

Substituting the values,

= 1.6 × 10⁻⁷ T

c) The total energy density 𝑢 in an electromagnetic wave is the sum of the energy densities of the electric and magnetic fields. The average energy density 𝑢 is given by,

where ε₀​ = 8.85 × 10⁻¹² F/m (the permittivity of free space)

= 1.02 × 10⁻⁸ J/m³

2) Determine the wavelength of 5 × 10¹⁸ Hz electromagnetic waves in open space. List its two uses.

Ans – We can find the wavelength by using the formula,

Uses for gamma rays:

(i) Atomic structural information is obtained using these rays.

(ii) They are employed for detection because of their extremely strong penetrating power.

3) Utilise the formula λ_xT = 0.29 cmK to derive the characteristic temperature ranges for various segments of the electromagnetic spectrum. What information do the obtained numbers convey?

Ans – A body at a specific temperature generates a continuous spectrum of wavelengths. Planck’s law identifies the wavelength at which a black body emits radiation with maximum intensity. The relationship can be used to determine it.

The temperature corresponding to various wavelengths can be calculated as

For λ_s = 10⁻⁴ cm,

For λ_n = 5 x 10⁻⁵ cm,

For λ_m =10⁻⁶ cm,

The results also show that temperature ranges are required to reduce radiation in different parts of the electromagnetic spectrum. As the wavelength decreases, the temperature increases.

5 Marks Questions

1) Answer the following:

(a) Short-wave bands are used for long-distance radio transmissions. Why?

Ans – The ionosphere can only refractively bend shortwave frequencies, which is why they are utilised for long-distance radio transmissions.

(b) Long-distance TV broadcasting requires the use of satellites. Why?

Ans – Satellites are used for long distance TV transmission since the television signal consists of high frequencies and amplitude. Therefore, such signals are not reflected by the ionosphere. Therefore, satellites contribute towards reflection of television signals.

(c) While radio and optical telescopes are constructed on the ground, only satellites in orbit around the earth can perform gamma-ray astronomy. Why?

Ans – According to gamma-ray astronomy, the atmosphere absorbs gamma-rays. However, it can be penetrated by radio and visible wavelengths. Consequently, radio and optical telescopes are built on the ground, but only satellites in orbit around the Earth can make X-ray astronomy possible.

(d) Human survival depends on the thin layer of ozone that sits on top the stratosphere. Why?

Ans – This is because the ozone layer at the top of the atmosphere helps block harmful UV-rays from reaching the Earth’s surface, and hence essential for human survival.

(e) Some scientists predict that a global nuclear war would mean a catastrophic ‘nuclear winter’ where life on Earth will be wiped out. What might this forecast be based on?

Ans – A nuclear war would indeed prove disastrous for this world to be seen on Earth’s surface. Earth will have a hard winter after a nuclear war, due to the reason that smoke clouds from the war will fill out most of the sky, and thus sunlight cannot reach the atmosphere. Apart from that, it will contribute towards ozone layer depletion.

2) Let the electric field amplitude of an electromagnetic wave be E₀ = 120 N/C and its frequency be v = 50.0 MHz.

(a) Identify B₀, ω, k, and λ.

(b) Derive expressions for E and B.

Ans – Given that ,

Electric field amplitude, E₀ = 120 N/C

Frequency of source, v = 50.0 MHz = 50 × 10⁶ Hz

Speed of light, c = 3 × 10⁸ m/s

(a) The magnetic field strength’s magnitude is provided as

= 4 x 10^-7 T

Angular frequency ω is given as,

ω = 2πv

ω = 2π x 50 × 10⁶

= 3.14 x 10⁸ rad/s

Wavelength λ is given as,

(b) The general expression for the electric field in an electromagnetic wave traveling in the z-direction is,

E(z,t) = E₀ cos(kz − ωt)

We know E₀ = 120 N/C, k = 1.05 rad/m, and ω = 3.14 x 10⁸ rad/s

E(z,t) = 120 cos(1.05z − 3.14 x 10⁸t) N/C

The general expression for the magnetic field is given by,

B(z,t) = B₀ cos(kz − ωt)

We know B₀ = 4 x 10^-7 T, k = 1.05 rad/m, and ω = 3.14 x 10⁸ rad/s

B(z,t) = 4 x 10^-7 cos(1.05z − 3.14 x 10⁸t) T