NCERT Solutions For Class 12 Physics Chapter 8: Electromagnetic Waves

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NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves are provided in this article. The waves that are produced when an electric field comes into contact with a magnetic field are known as Electromagnetic Waves or EM waves. The concept of EM waves can also be understood by visualizing the waves in which there are sinusoidal variations of vectors of the magnetic and electric field which are perpendicular to each other and also at right angles to the wave propagation direction.

Unit 5 - Electromagnetic Waves along with Unit 6 - Optics has a weightage of 18 marks in the CBSE Class 12 Physics Examination. The NCERT Solutions for Class 12 Physics Chapter 8 covers concepts of displacement current, electromagnetic spectrum, Maxwell Equations, etc.

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NCERT Solutions for Class 12 Physics Chapter 8

The NCERT solutions for class 12 physics chapter 8: Electromagnetic waves are given below:

CBSE Class 12 Physics Chapter 8 Important Topics

Electromagnetic Waves are the waves produced when an electric field comes in contact with a magnetic field.

The direction of the propagation of EM waves is given by a vector cross product of the electric and magnetic fields as

$\overrightarrow E \times \overrightarrow B$

Equation of Speed of EM Wave: The relation between the speed of light, permittivity constant, and permeability constant is as follows:

$c = {1 \over \mu_0 \epsilon_0}$

The wavelength for the various spectrum of light is as follows -
- Radio Waves > 0.1 m
- Microwave - 0.1m to 1 mm
- Infrared waves - 1 mm to 700 nm
- Visible light - 700 nm to 400 nm
- Ultraviolet - 400 nm to 1nm
- X-rays - 1nm to 10^-3 nm
- Gamma rays < 10^-3 nm

According to their wavelength or frequency, EM waves can be classified as Electromagnetic Spectrum that ranges from 400 nm to 700 nm.

Applications of Electromagnetic Waves: Electromagnetic Waves play an important role in communication technology.

The infrared radiations are used in the security cameras for night vision. UV rays can be helped to detect forged banknotes.

Also Read:

Chapter 8 Related Articles
Electromagnetic Radiations	Electromagnetic Pulse	Wave Nature of Electromagnetic Radiation
Electromagnetism	Wavelength formula	Electromagnetic Field
Faraday’s Laws of Induction	Electromagnetic Induction	Electromagnetic Spectrum
X-Rays	Radio waves	Infrared radiation
Gamma radiation	Types of radiation	Impedance of free space

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CBSE CLASS XII Related Questions

1.
Two air-filled capacitors of capacitances $C_1$ and $C_2$ are connected in parallel with a dc battery. After the capacitors are fully charged, a slab of dielectric constant K is inserted between the plates of each capacitor. How will the (i) charge on each capacitor and (ii) energy stored in the capacitor affected after the slab is introduced.

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2.
With the help of a labelled diagram, explain the principle, construction and working of an a.c. generator.

View Solution

3.
Consider a cylindrical conductor of length $ l $ and area of cross-section $ A $. Current $ I $ is maintained in the conductor and electrons drift with velocity $ \vec{v}_d \, (|\vec{v}_d| = \frac{eE}{m} \tau) $, where symbols have their usual meanings. Show that the conductivity of the material of the conductor is given by \[ \sigma = \frac{n e^2 \tau}{m}. \]

View Solution

4.
Read the following paragraph and answer the questions that follow.
A p-type or n-type semiconductor can be converted into a p-n junction by doping it with suitable impurity. The motion of majority charge carriers causes diffusion current across the junction while the barrier electric field causes motion of minority carriers for drift current. In case of unbiased diode, the diffusion and drift currents are equal. This equilibrium is disturbed by the biasing batteries. Diodes, therefore, allow currents in one direction. This property of diode is used in making rectifiers.

5.
The resistance of a metal wire at $ 20^\circ \text{C} $ is $ 1.05 \, \Omega $ and at $ 100^\circ \text{C} $ is $ 1.38 \, \Omega $. Determine the temperature coefficient of resistivity of this metal.

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6.
Read the following paragraph and answer the questions that follow.
In an experiment with convex lens of focal length f, the screen is fixed at a distance D from the object. A student slowly moves the lens away from the object towards the screen and finds that she is able to form sharp image of the object for two positions of the lens. The distance between these two positions of the lens is d.

NCERT Solutions For Class 12 Physics Chapter 8: Electromagnetic Waves

NCERT Solutions for Class 12 Physics Chapter 8

CBSE Class 12 Physics Chapter 8 Important Topics

CBSE CLASS XII Related Questions

2.
With the help of a labelled diagram, explain the principle, construction and working of an a.c. generator.

5.
The resistance of a metal wire at \( 20^\circ \text{C} \) is \( 1.05 \, \Omega \) and at \( 100^\circ \text{C} \) is \( 1.38 \, \Omega \). Determine the temperature coefficient of resistivity of this metal.

Similar Physics Concepts

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NCERT Solutions For Class 12 Physics Chapter 8: Electromagnetic Waves

NCERT Solutions for Class 12 Physics Chapter 8

CBSE Class 12 Physics Chapter 8 Important Topics

CBSE CLASS XII Related Questions

2.With the help of a labelled diagram, explain the principle, construction and working of an a.c. generator.

5. The resistance of a metal wire at \( 20^\circ \text{C} \) is \( 1.05 \, \Omega \) and at \( 100^\circ \text{C} \) is \( 1.38 \, \Omega \). Determine the temperature coefficient of resistivity of this metal.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
With the help of a labelled diagram, explain the principle, construction and working of an a.c. generator.

5.
The resistance of a metal wire at \( 20^\circ \text{C} \) is \( 1.05 \, \Omega \) and at \( 100^\circ \text{C} \) is \( 1.38 \, \Omega \). Determine the temperature coefficient of resistivity of this metal.