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.

Download PDF: NCERT Solutions for Class 12 Physics Chapter 8

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.
A deuteron contains a proton and a neutron and has a mass of \( 2.01355 \, \text{u} \). Calculate the mass defect for it in u and its energy equivalence in MeV. (\( m_p = 1.007277 \, \text{u} \), \( m_n = 1.008665 \, \text{u} \), \( 1 \, \text{u} = 931.5 \, \text{MeV/c}^2 \))

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2.
Two point charges \( 5 \, \mu C \) and \( -1 \, \mu C \) are placed at points \( (-3 \, \text{cm}, 0, 0) \) and \( (3 \, \text{cm}, 0, 0) \), respectively. An external electric field \( \vec{E} = \frac{A}{r^2} \hat{r} \) where \( A = 3 \times 10^5 \, \text{V m} \) is switched on in the region. Calculate the change in electrostatic energy of the system due to the electric field.

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3.
(i) Write Biot-Savart's law in vector form.
(ii) Two identical circular coils A and B, each of radius \( R \), carrying currents \( I \) and \( \sqrt{3} I \) respectively, are placed concentrically in XY and YZ planes respectively. Find the magnitude and direction of the net magnetic field at their common centre.

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4.
In a compound microscope, an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. The eyepiece has a focal length of 5 cm. The final image is formed at infinity. Calculate the distance between the objective and the eyepiece.

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5.
In Young's double slit experiment, the separation between the two slits is 1.0 mm and the screen is 1.0 m away from the slits. A beam of light consisting of two wavelengths, 500 nm and 600 nm, is used to obtain interference fringes. Calculate:
(a) The distance between the first maxima for the two wavelengths.
(b) The least distance from the central maximum, where the bright fringes due to both the wavelengths coincide.

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6.
A resistor of 400 \( \Omega \), an inductor of \( \frac{5}{\pi} \) H, and a capacitor of \( \frac{50}{\pi} \) µF are joined in series across an AC source \( v = 140 \sin (100 \pi t) \) V. Find the rms voltages across these three circuit elements. The algebraic sum of these voltages is more than the rms voltage of source. Explain.

View Solution

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

1.
A deuteron contains a proton and a neutron and has a mass of \( 2.01355 \, \text{u} \). Calculate the mass defect for it in u and its energy equivalence in MeV. (\( m_p = 1.007277 \, \text{u} \), \( m_n = 1.008665 \, \text{u} \), \( 1 \, \text{u} = 931.5 \, \text{MeV/c}^2 \))

4.
In a compound microscope, an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. The eyepiece has a focal length of 5 cm. The final image is formed at infinity. Calculate the distance between the objective and the eyepiece.

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

1.A deuteron contains a proton and a neutron and has a mass of \( 2.01355 \, \text{u} \). Calculate the mass defect for it in u and its energy equivalence in MeV. (\( m_p = 1.007277 \, \text{u} \), \( m_n = 1.008665 \, \text{u} \), \( 1 \, \text{u} = 931.5 \, \text{MeV/c}^2 \))

4.In a compound microscope, an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. The eyepiece has a focal length of 5 cm. The final image is formed at infinity. Calculate the distance between the objective and the eyepiece.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
A deuteron contains a proton and a neutron and has a mass of \( 2.01355 \, \text{u} \). Calculate the mass defect for it in u and its energy equivalence in MeV. (\( m_p = 1.007277 \, \text{u} \), \( m_n = 1.008665 \, \text{u} \), \( 1 \, \text{u} = 931.5 \, \text{MeV/c}^2 \))

4.
In a compound microscope, an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. The eyepiece has a focal length of 5 cm. The final image is formed at infinity. Calculate the distance between the objective and the eyepiece.