CBSE Class 12 Physics Electromagnetic Waves Assignment Set 02

Important Questions for NCERT Class 12 Physics Electromagnetic Waves

Question. An em wave is propagating in a medium with a velocity v- = v ^i . The instantaneous oscillating electric field of this em wave is along +y axis. Then the direction of oscillating magnetic field of the em wave will be along

(a) –z direction
(b) +z direction
(c) –y direction
(d) –x direction

Answer B


Question. In an electromagnetic wave in free space the root mean square value of the electric field is Erms = 6 V m–1. The peak value of the magnetic field is
(a) 2.83 × 10^–8 T
(b) 0.70 × 10^–8 T
(c) 4.23 × 10^–8 T
(d) 1.41 × 10^–8 T

Answer A
 

Question. Which of the following are not electromagnetic waves? 
(a) cosmic rays
(b) g-rays
(c) b-rays
(d) X-rays

Answer C


Question. Out of the following options which one can be used to produce a propagating electromagnetic wave ?
(a) A chargeless particle
(b) An accelerating charge
(c) A charge moving at constant velocity
(d) A stationary charge 

Answer B

Question. Light with an energy flux of 25 × 104 W m–2 falls on a perfectly reflecting surface at normal incidence. If the surface area is 15 cm2, the average force exerted on the surface is
(a) 1.25 × 10^–6 N
(b) 2.50 × 10^–6 N
(c) 1.20 × 10^–6 N
(d) 3.0 × 10^–6 N 

Answer B


Question. An electromagnetic wave of frequency u = 3.0 MHz passes from vacuum into a dielectric medium with relative permittivity e = 4.0. Then 
(a) Wavelength is doubled and frequency becomes half.
(b) Wavelength is halved and frequency remains unchanged.
(c) Wavelength and frequency both remain unchanged.
(d) Wavelength is doubled and frequency unchanged

Answer B
 

Question. The electric field associated with an em wave in vacuum is given by [E = i ^ 40 cos (kz – 6 × 108t)]; where E, z and t are in volt/m, meter and seconds respectively. The value of wave vector k is
(a) 2 m^–1
(b) 0.5 m^–1
(c) 6 m^–1
(d) 3 m^–1 

Answer A
 

Question. Which of the following electromagnetic radiations have the smaller wavelength?   
(a) X-rays
(b) g-rays
(c) UV waves
(d) microwaves 

Answer  B
 

Question. According to Maxwell’s hypothesis, changing of electric field give rise to :    
(a) magnetic field
(b) pressure gradient
(c) charge
(d) voltage

Answer  A


Question. Frequency of infrared wave is approximately:   
(a) 1018 Hz
(b) 1014 Hz 
(c) 109 Hz
(d) 1016 Hz

Answer  B

ELECTROMAGNETIC WAVES

TOPICS TO BE COVERED

Displacement current, electromagnetic waves and their characteristics (qualitative ideas only).
Transverse nature of electromagnetic waves. Electromagnetic spectrum (radio-waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays) including elementary facts about their uses.

KEY POINTS

EM waves are produced by accelerated (only by the change in speed) charged particles.
\(\vec{E}\) and \(\vec{B}\) vectors oscillate with the frequency of oscillating charged particles.
Properties of em waves :
(i) Transverse nature
(ii) Can travel through vacuum.
(iii) \(E_0 / B_0 = E / B = v\) where \(v\) = Speed of EM waves.
(iv) Speed = \(3 \times 10^8\) m/s in vacuum.
(v) In any medium \(v = \frac{1}{\sqrt{\mu \epsilon}}\)
Where \(\mu = \mu_r \mu_0, \epsilon = \epsilon_r \epsilon_0\)
\(n =\) refractive index of medium \(\sqrt{\epsilon_r} = n\)
Also \(v = c / n\)
(vi) Wave intensity equals average of Poynting vector \(I = |\vec{S}|_{\text{av}} = \frac{B_0 E_0}{2\mu_0}\)
(vii) Average electric and average magnetic energy densities are equal.
In an em spectrum, different waves have different frequency and wavelengths.
Penetration power of em waves depends on frequency. Higher the frequency larger the penetration power.
Wavelength \(\lambda\) and frequency \(\nu\) are related with each other \(v = \lambda \nu\). Here \(v\) is the wave velocity.
A wave travelling along \(+x\) axis is represented by
\(E_y = E_{0y} \cos(\omega t - kx)\)
\(B_z = B_{0z} \cos(\omega t - kx)\)
\(\omega = \frac{2\pi}{T} = 2\pi\nu\)
\(\frac{\omega}{k} = \lambda\nu = v\) wave speed
\(k = \frac{2\pi}{\lambda} = 2\pi\bar{\nu}\)
\(\nu \rightarrow\) frequency
\(\bar{\nu} = \frac{1}{\lambda} \rightarrow\) wave number.

QUESTIONS

VERY SHORT ANSWER QUESTIONS (1 Mark)

Question 1. Every EM wave has certain frequency. Name two parameters of an em wave that oscillate with this frequency.
Answer: Electric field vector \((\vec{E})\) and magnetic field vector \((\vec{B})\).
In simple words: The electric and magnetic fields oscillate at the same frequency as the wave.

Exam Tip: Remember that both fields are perpendicular to each other and to the direction of wave propagation.

Question 2. What is the phase difference between electric and magnetic field vectors in an em wave?
Answer: Zero phase difference. Both electric and magnetic fields reach their maxima and minima at the same time and place.
In simple words: They are completely in phase with each other.

Exam Tip: Phase difference is zero, unlike alternating current circuits where voltage and current can have a phase difference.

Question 3. Name em radiations used for detecting fake currency notes.
Answer: Ultraviolet (UV) radiations.
In simple words: UV rays cause certain materials in real currency to fluoresce.

Exam Tip: Associate UV rays with fluorescence, water purification, and fake currency detection.

Question 4. Give any two uses of microwaves.
Answer:
1. In RADAR systems for aircraft navigation.
2. In microwave ovens for cooking.
In simple words: Microwaves are used for radar communication and heating food.

Exam Tip: Mention specific practical applications like RADAR and microwave cooking to score full marks.

Question 5. Name the phenomenon which justifies the transverse nature of em waves.
Answer: Polarization.
In simple words: Only transverse waves can be polarized.

Exam Tip: Polarization is the definitive test for transverse wave nature.

Question 6. Arrange the following em waves in descending order of wavelengths : \(\gamma\) ray, microwaves, UV radiations.
Answer: Microwaves > UV radiations > \(\gamma\) rays.
In simple words: Microwaves have the longest wavelength among these, while gamma rays have the shortest.

Question 7. Which component \(\vec{E}\) or \(\vec{B}\) of an em wave is responsible for visible effect?
Answer: Electric field vector \((\vec{E})\).
In simple words: The electric field vector exerts force on electrons in the retina, producing vision.

Exam Tip: The magnitude of the electric field is much larger than the magnetic field when expressed in SI units \((E = cB)\), making the electric force dominant on charged particles.

Question 8. Write expression for speed of em waves in a medium of electrical permittivity \(\epsilon\) and magnetic permeability \(\mu\).
Answer:
\(v = \frac{1}{\sqrt{\mu \epsilon}}\)
In simple words: Speed depends inversely on the square root of the medium's permeability and permittivity.

Question 9. Which of the following has longest penetration power? UV radiation, X-ray, Microwaves.
Answer: X-rays (since they have the highest frequency among the given options).
In simple words: Higher frequency means higher energy and greater penetration power.

Exam Tip: Penetration power is directly proportional to frequency and inversely proportional to wavelength.

Question 10. Which of the following has least frequency? IR radiations, visible radiation, radio waves.
Answer: Radio waves.
In simple words: Radio waves are at the low-frequency end of the spectrum.

Question 11. Which physical quantity is the same for microwaves of wavelength 1 mm and UV radiations of 1600 A° in vacuum?
Answer: Speed of propagation in vacuum \((c = 3 \times 10^8 \text{ m/s})\).
In simple words: All electromagnetic waves travel at the same speed in a vacuum.

Exam Tip: Never confuse speed in vacuum with frequency or wavelength; speed is a universal constant for all EM waves in vacuum.

Question 12. Name two physical quantities which are imparted by an em wave to a surface on which it falls.
Answer: Momentum and Energy.
In simple words: EM waves exert radiation pressure, transferring both energy and momentum.

Question 13. Name the physical quantity with unit same as that of \(\left[ I_d = \epsilon_0 \frac{d\phi_e}{dt} \right]\)
Answer: Electric current (Ampere).
In simple words: Displacement current has the exact same units as conduction current.

Exam Tip: Displacement current is not a flow of charge, but it has the dimensions and unit of current.

Question 14. What is the source of energy associated with propagating em waves?
Answer: Oscillating or accelerated electric charges.
In simple words: Energy comes from the accelerating charge that creates the fields.

Question 15. What is the wavelength range of em waves that were produced and observed by J.C. Bose?
Answer: From 25 mm to 20 mm.
In simple words: J.C. Bose produced millimeter-range electromagnetic waves in his laboratory.

Question 16. Name the device used for producing microwaves.
Answer: Klystron valve, Magnetron, or Gunn diode.
In simple words: Special electronic vacuum tubes or semiconductor diodes generate microwaves.

Question 17. Name the em radiations which are detected using Gieger tube.
Answer: X-rays and Gamma rays.
In simple words: Geiger counters detect high-energy ionizing radiation.

Question 18. Relative electric permittivity of a medium is 8 and relative permeability is close to unity. What is the speed of em waves in the medium?
Answer:
Given: Relative permittivity \(\epsilon_r = 8\), Relative permeability \(\mu_r \approx 1\)
Refractive index \(n = \sqrt{\mu_r \epsilon_r} = \sqrt{1 \times 8} = \sqrt{8} = 2\sqrt{2}\)
Speed \(v = \frac{c}{n} = \frac{3 \times 10^8}{2\sqrt{2}} = 1.06 \times 10^8 \text{ m/s}\)
In simple words: Use the formula \(v = \frac{c}{\sqrt{\epsilon_r}}\) to find the speed.

Exam Tip: Always state the formula \(v = \frac{c}{\sqrt{\mu_r \epsilon_r}}\) before substituting values.

Question 19. Identify the part of the electromagnetic spectrum to which the following wavelengths belong :
(i) \(10^{-1}\) m
(ii) \(10^{-12}\) m
Answer:
(i) \(10^{-1}\) m: Microwaves (or short radio waves)
(ii) \(10^{-12}\) m: Gamma rays
In simple words: Check the standard wavelength ranges for the em spectrum.

Exam Tip: Keep order-of-magnitude ranges of the em spectrum memorized.

Question 20. Name the part of the electromagnetic spectrum of wavelength \(10^{-2}\) m and mention its one application.
Answer:
Wavelength \(10^{-2}\) m corresponds to Microwaves.
Application: Used in microwave ovens for cooking / RADAR.
In simple words: Microwave radiation is used in radar and cooking.

Exam Tip: State both the wave type and its corresponding application clearly.

Question 21. Which of the following, if any, can act as a source of electromagnetic waves?
(i) A charge moving with a constant velocity.
(ii) A charge moving in a circular orbit.
(iii) A charge at rest.
Answer:
(ii) A charge moving in a circular orbit.
In simple words: Only an accelerated charge emits electromagnetic waves, and circular motion involves continuous acceleration (centripetal acceleration).

Exam Tip: Constant velocity means zero acceleration (no em waves); stationary charge creates only a static electric field.

Question 22. Mention the pair of space and time varying E and B fields which would generate a plane em wave trevelling in Z-direction.
Answer: Electric field \(E_x\) (or \(E_y\)) along the X-axis (or Y-axis) and magnetic field \(B_y\) (or \(B_x\)) along the Y-axis (or X-axis) varying sinusoidally with time and position \(z\).
In simple words: The fields must oscillate perpendicular to the z-direction and to each other.

Exam Tip: Direction of propagation is given by the cross product \(\vec{E} \times \vec{B}\).

Question 23. The charging current for a capacitor is 0.2A. What is the displacement Current?
Answer:
Displacement current \(I_d\) is always equal to the conduction (charging) current \(I_c\) in magnitude to maintain continuity of current.
\(I_d = 0.2 \text{ A}\)
In simple words: Displacement current equals charging current.

Exam Tip: Remember Maxwell's modified Ampere's circuital law where \(I_c = I_d\).

Question 24. Give the ratio of Velocities of light waves of wavelengths \(4000\text{ A}^\circ\) and \(8000\text{ A}^\circ\) in Vaccum.
Answer:
Ratio = \(1 : 1\)
In simple words: All electromagnetic waves travel at the exact same speed \(c\) in a vacuum, regardless of their wavelength.

Exam Tip: Do not get tricked by different wavelengths in a vacuum; speed remains constant.

Question 25. Which physical quantity, if any has the same value for waves belonging to the different parts of the electromagnetic spectrum?
Answer: Speed in vacuum \((c = 3 \times 10^8 \text{ m/s})\).
In simple words: Their speed in vacuum is universal.

Exam Tip: Speed in vacuum is the invariant constant for all electromagnetic waves.

SHORT ANSWER QUESTIONS (2 Marks)

Question 1. Give one use of each of the following (i) UV ray (ii) \(\gamma\)-ray
Answer:
(i) UV ray: Used for sterilizing surgical instruments / water purification.
(ii) \(\gamma\)-ray: Used in radiotherapy for cancer treatment / destroying tumor cells.
In simple words: UV sterilizes things, while gamma rays treat cancer.

Exam Tip: Mention clear medical or industrial applications.

Question 2. Represent EM waves propagating along the \(x\)-axis. In which electric and magnetic fields are along \(y\)-axis and \(z\)-axis respectively.
Answer:
Equations representing the wave:
\(E_y = E_0 \sin(\omega t - kx)\)
\(B_z = B_0 \sin(\omega t - kx)\)
In simple words: The electric field oscillates along y and the magnetic field along z, while the wave moves along x.

Exam Tip: Ensure the phase term \((\omega t - kx)\) is identical for both field equations.

Question 3. State the principles of production of EM waves. An EM wave of wavelength \(\lambda\) goes from vacuum to a medium of refractive index \(n\). What will be the frequency of wave in the medium?
Answer:
Principle: An accelerated electric charge produces electromagnetic waves.
Frequency remains unchanged when a wave enters from vacuum into a medium because frequency depends only on the source.
Frequency in medium \(\nu' = \nu\)
In simple words: Frequency never changes upon refraction; only wavelength and speed change.

Exam Tip: Always state that frequency is a source property and remains constant across media.

Question 4. An EM wave has amplitude of electric field \(E_0\) and amplitude of magnetic field is \(B_0\) the electric field at some instant become \(\frac{3}{4} E_0\). What will be magnetic field at this instant? (Wave is travelling in vacuum).
Answer:
At any instant in an electromagnetic wave, the ratio of electric field to magnetic field is constant: \(\frac{E}{B} = \frac{E_0}{B_0} = c\)
Therefore, when \(E = \frac{3}{4} E_0\):
\(B = \frac{E}{c} = \frac{\frac{3}{4} E_0}{c} = \frac{3}{4} \left(\frac{E_0}{c}\right) = \frac{3}{4} B_0\)
In simple words: The magnetic field decreases in the same proportion as the electric field.

Exam Tip: Electric and magnetic fields in an EM wave vary in step (are in phase).

Question 5. State two applications of infrared radiations.
Answer:
1. In physical therapy to treat muscular strain.
2. In remote controls of TVs and electronic devices.
In simple words: IR rays are used in TV remotes and heat therapy.

Exam Tip: Infrared waves are also known as heat waves.

Question 6. State two applications of ultraviolet radiations.
Answer:
1. In water purifiers to kill germs and bacteria.
2. In the detection of fake currency and legal documents.
In simple words: UV rays purify water and detect forged currency.

Exam Tip: UV radiation causes fluorescence in certain substances.

Question 7. State two applications of x-rays.
Answer:
1. In medical diagnostics for detecting bone fractures.
2. In engineering for detecting internal cracks or flaws in metal structures.
In simple words: X-rays are used for medical bone scans and checking metal flaws.

Question 8. Show that the average energy density of the electric field \(\vec{E}\) equals the average energy density of the magnetic fields \(\vec{B}\)?
Answer:
Average electric energy density \(u_E = \frac{1}{2} \epsilon_0 E_{\text{rms}}^2 = \frac{1}{4} \epsilon_0 E_0^2\)
Since \(E_0 = c B_0\) and \(c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}\), we get \(c^2 = \frac{1}{\mu_0 \epsilon_0}\)
Substituting \(E_0 = \frac{B_0}{\sqrt{\mu_0 \epsilon_0}}\) into \(u_E\):
\(u_E = \frac{1}{4} \epsilon_0 \left(\frac{B_0^2}{\mu_0 \epsilon_0}\right) = \frac{B_0^2}{4\mu_0}\)
Average magnetic energy density \(u_B = \frac{1}{2} \frac{B_{\text{rms}}^2}{\mu_0} = \frac{B_0^2}{4\mu_0}\)
Thus, \(u_E = u_B\).
In simple words: Total energy in an electromagnetic wave is shared equally between electric and magnetic fields.

Exam Tip: This equipartition of energy is a fundamental property of electromagnetic waves.

SHORT ANSWER QUESTIONS (3 Marks)

Question 1. Name EM radiations used (i) in the treatment of cancer. (ii) For detecting flaw in pipes carrying oil. (iii) In sterilizing surgical instruments.
Answer:
(i) Gamma rays \((\gamma\text{-rays})\)
(ii) X-rays
(iii) Ultraviolet (UV) radiations
In simple words: Gamma rays treat cancer, X-rays check pipes, and UV rays sterilize tools.

Exam Tip: Match each industrial/medical application to its correct ray type.

Question 2. How would you experimentally show that EM waves are transverse in nature?
Answer:
By passing an unpolarized EM wave (like light) through a polarizer (e.g., a Tourmaline crystal or Polaroid). When the polarizer is rotated, the transmitted intensity varies from maximum to zero. This phenomenon of polarization proves that electric and magnetic field vectors oscillate perpendicular to the direction of wave propagation, confirming the transverse nature of EM waves.
In simple words: Using a Polaroid filter, the intensity changes upon rotation, proving transverse vibration.

Exam Tip: Mention polarization and the variation of intensity with rotation angle.

Question 3. List any three properties of EM waves.
Answer:
1. They are transverse in nature.
2. They do not require any material medium for propagation and can travel through vacuum.
3. In vacuum, their speed is given by \(c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} = 3 \times 10^8 \text{ m/s}\).
In simple words: EM waves are transverse, need no medium, and travel at the speed of light in vacuum.

Exam Tip: Bullet points are ideal for listing properties.

Question 4. Find the wavelength of electromagnetic waves of frequency \(5 \times 10^{19}\) Hz in free space. Give its two applications
Answer:
Given frequency \(\nu = 5 \times 10^{19}\) Hz, speed \(c = 3 \times 10^8 \text{ m/s}\)
Wavelength \(\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{5 \times 10^{19}} = 0.6 \times 10^{-11} \text{ m} = 6 \times 10^{-12} \text{ m}\) (Gamma rays)
Applications of Gamma rays:
1. In radiotherapy for treating cancer.
2. To produce nuclear reactions.
In simple words: Calculate wavelength using \(\lambda = \frac{c}{\nu}\) and list gamma ray uses.

Exam Tip: State the wave type based on the calculated wavelength.

NUMERICALS

Question 1. The refractive index of medium is 1.5. A beam of light of wavelength \(6000\text{ A}^\circ\) enters in the medium from air. Find wavelength and frequency of light in the medium.
Answer:
Given: Refractive index \(n = 1.5\), Wavelength in air \(\lambda_0 = 6000\text{ A}^\circ = 6 \times 10^{-7} \text{ m}\)
Frequency in air \(\nu = \frac{c}{\lambda_0} = \frac{3 \times 10^8}{6 \times 10^{-7}} = 5 \times 10^{14} \text{ Hz}\)
Frequency in medium remains unchanged: \(\nu' = 5 \times 10^{14} \text{ Hz}\)
Wavelength in medium \(\lambda_m = \frac{\lambda_0}{n} = \frac{6000\text{ A}^\circ}{1.5} = 4000\text{ A}^\circ = 4 \times 10^{-7} \text{ m}\)
In simple words: Frequency stays the same, while wavelength decreases by a factor of 1.5.

Exam Tip: Remember that frequency never changes when light enters a new medium.

Question 2. An EM wave is travelling in vaccum. Amplitude of the electric field vector is \(5 \times 10^4\) V/m. Calculate amplitude of magnetic field vector.
Answer:
Given: \(E_0 = 5 \times 10^4 \text{ V/m}\), \(c = 3 \times 10^8 \text{ m/s}\)
\(B_0 = \frac{E_0}{c} = \frac{5 \times 10^4}{3 \times 10^8} = 1.67 \times 10^{-4} \text{ T}\)
In simple words: Divide electric field amplitude by the speed of light to get magnetic field amplitude.

Exam Tip: Include proper SI units (Tesla for magnetic field).

Question 3. Suppose the electric field amplitude of an em wave is \(E_0 = 120 \text{ NC}^{-1}\) and that its frequency is \(\nu = 50.0 \text{ MHz}\).
(a) Determine \(B_0\), \(\omega\), \(k\) and \(\lambda\)
(b) Find expressions for E and B.
Answer:
Given: \(E_0 = 120 \text{ NC}^{-1}\), \(\nu = 50 \text{ MHz} = 50 \times 10^6 \text{ Hz}\)
(a) Magnetic field amplitude \(B_0 = \frac{E_0}{c} = \frac{120}{3 \times 10^8} = 4 \times 10^{-7} \text{ T}\)
Angular frequency \(\omega = 2\pi\nu = 2 \times 3.14 \times 50 \times 10^6 = 3.14 \times 10^8 \text{ rad/s}\)
Wavelength \(\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{50 \times 10^6} = 6 \text{ m}\)
Propagation constant \(k = \frac{2\pi}{\lambda} = \frac{2 \times 3.14}{6} = 1.05 \text{ rad/m}\)
(b) Assuming wave propagates along X-axis with electric field along Y-axis:
\(E_y = 120 \sin(3.14 \times 10^8 t - 1.05 x) \text{ N/C}\)
\(B_z = 4 \times 10^{-7} \sin(3.14 \times 10^8 t - 1.05 x) \text{ T}\)
In simple words: Calculate each wave parameter step-by-step using standard formulas.

Exam Tip: Be careful with powers of 10 when calculating frequency and angular frequency.

Question 4. A radio can tune into any station of frequency band 7.5 MHz to 10 MHz. Find the corresponding wave length range.
Answer:
Maximum frequency \(\nu_1 = 10 \text{ MHz} = 10 \times 10^6 \text{ Hz}\)
Minimum frequency \(\nu_2 = 7.5 \text{ MHz} = 7.5 \times 10^6 \text{ Hz}\)
Minimum wavelength \(\lambda_1 = \frac{c}{\nu_1} = \frac{3 \times 10^8}{10 \times 10^6} = 30 \text{ m}\)
Maximum wavelength \(\lambda_2 = \frac{c}{\nu_2} = \frac{3 \times 10^8}{7.5 \times 10^6} = 40 \text{ m}\)
Wavelength range is from 30 m to 40 m.
In simple words: Higher frequency corresponds to shorter wavelength, so the range is inverted.

Exam Tip: Note that maximum frequency gives minimum wavelength and vice versa.

Question 5. The amplitude of the magnetic field vector of an electromagnetic wave travelling in vacuum is \(2.4\text{ mT}\). Frequency of the wave is \(16 \text{ MHz}\). Find :
(i) Amplitude of electric field vector and
(ii) Wavelength of the wave.
Answer:
Given: \(B_0 = 2.4 \text{ mT} = 2.4 \times 10^{-3} \text{ T}\), \(\nu = 16 \text{ MHz} = 16 \times 10^6 \text{ Hz}\)
(i) Electric field amplitude \(E_0 = c B_0 = 3 \times 10^8 \times 2.4 \times 10^{-3} = 7.2 \times 10^5 \text{ V/m}\)
(ii) Wavelength \(\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{16 \times 10^6} = \frac{300}{16} = 18.75 \text{ m}\)
In simple words: Multiply magnetic field by \(c\) to find electric field, and divide \(c\) by frequency to find wavelength.

Exam Tip: Convert milli-Tesla (mT) to Tesla before doing calculations.

Question 6. An EM wave travelling through a medium has electric field vector \(E_y = 4 \times 10^5 \cos(3.14 \times 10^8 t - 1.57 x)\) N/C. Here \(x\) is in \(m\) and \(t\) in \(s\). Then find :
(i) Wavelength
(ii) Frequency
(iii) Direction of propagation
(iv) Speed of wave
(v) Refractive index of medium
(vi) Amplitude of magnetic field vector.
Answer:
Comparing given equation \(E_y = 4 \times 10^5 \cos(3.14 \times 10^8 t - 1.57 x)\) with standard equation \(E_y = E_0 \cos(\omega t - kx)\):
Amplitude \(E_0 = 4 \times 10^5 \text{ N/C}\)
Angular frequency \(\omega = 3.14 \times 10^8 \text{ rad/s}\)
Propagation constant \(k = 1.57 \text{ rad/m}\)
(i) Wavelength \(\lambda = \frac{2\pi}{k} = \frac{2 \times 3.14}{1.57} = 4 \text{ m}\)
(ii) Frequency \(\nu = \frac{\omega}{2\pi} = \frac{3.14 \times 10^8}{2 \times 3.14} = 5 \times 10^7 \text{ Hz}\)
(iii) Direction of propagation: Along positive X-axis \((+x)\)
(iv) Speed of wave \(v = \nu\lambda = 5 \times 10^7 \times 4 = 2 \times 10^8 \text{ m/s}\)
(v) Refractive index \(n = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5\)
(vi) Amplitude of magnetic field vector \(B_0 = \frac{E_0}{v} = \frac{4 \times 10^5}{2 \times 10^8} = 2 \times 10^{-3} \text{ T}\)
In simple words: Compare the wave equation coefficients with standard wave formulas to extract all required quantities.

Exam Tip: Note that speed in medium uses wave velocity \(v\), not \(c\), when calculating refractive index.