CBSE Class 12 Physics Electromagnetic Waves Assignment Set 01

Answer: D

Question. A parallel plate capacitor of capacitance 20 mF is being charged by a voltage source whose potential is changing at the rate of 3 V/s. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively 
(a) zero, zero
(b) zero, 60 mA
(c) 60 mA, 60 mA
(d) 60 mA, zero

Answer: C

Question. A 100 W resistance and a capacitor of 100 W reactance are connected in series across a 220 V source. When the capacitor is 50% charged, the peak value of the displacement current is
(a) 2.2 A
(b) 11 A
(c) 4.4 A
(d) 11√2 A

Answer: A

Question. Light with an average flux of 20 W/cm2 falls on a non-reflecting surface at normal incidence having surface area 20 cm2. The energy received by the surface during time span of 1 minute is
(a) 10 × 10³ J
(b) 12 × 10³ J
(c) 24 × 10³ J
(d) 48 × 10³ J 

Answer: C

Question. The ratio of contributions made by the electric field and magnetic field components to the intensity of an electromagnetic wave is (c = speed of electromagnetic waves)
(a) c : 1
(b) 1 : 1
(c) 1 : c
(d) 1 : c2 

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. For a transparent medium relative permeability and permittivity, μr and er are 1.0 and 1.44 respectively. The velocity of light in this medium would be
(a) 2.5 × 10⁸ m/s
(b) 3 × 10⁸ m/s
(c) 2.08 × 10⁸ m/s
(d) 4.32 × 10⁸ m/s

Answer: A

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: Both the electric and magnetic fields oscillate at the same frequency as the electromagnetic wave.

Exam Tip: Remember that \(\vec{E}\) and \(\vec{B}\) are mutually perpendicular to each other and to the direction of propagation.

Question 2. What is the phase difference between electric and magnetic field vectors in an em wave?
Answer: Zero phase difference (\(0\) or in phase).
In simple words: Both electric and magnetic fields reach their maximum and minimum values at the same time and place.

Exam Tip: In a plane electromagnetic wave, \(E\) and \(B\) attain their zero and peak values simultaneously.

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

Exam Tip: UV rays are also used for water purification and in LASIK eye surgery.

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 high-frequency radio waves useful for communication and heating food molecules.

Exam Tip: Mention concise and standard applications like RADAR and telecommunications.

Question 5. Name the phenomenon which justifies the transverse nature of em waves.
Answer: Polarization.
In simple words: Only transverse waves can be polarized because their oscillations are perpendicular to the direction of travel.

Exam Tip: Longitudinal waves like sound waves cannot be polarized.

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

Exam Tip: Recall the full electromagnetic spectrum order: Radio > Microwave > Infrared > Visible > UV > X-rays > Gamma.

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 vector exerts a force on electrons in our eyes, producing the sensation of vision.

Exam Tip: Optical effects and biological interactions are predominantly due to the electric field vector.

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

Exam Tip: For a vacuum, substitute \(\mu_0\) and \(\epsilon_0\) to get \(c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}\).

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

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 beginning of the electromagnetic spectrum and have the lowest frequency.

Exam Tip: Lowest frequency corresponds to the longest wavelength.

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

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: Electromagnetic waves exert radiation pressure because they carry both energy and momentum.

Exam Tip: Momentum \(p = \frac{U}{c}\), where \(U\) is energy absorbed.

Question 13. Name the physical quantity with unit same as that of \(I_d = \epsilon_0 \frac{d\phi_e}{dt}\) where \(\phi_e \to\) electric flux.
Answer: Electric current (Ampere).
In simple words: This expression defines displacement current, which has the exact same units and dimensions as conduction current.

Exam Tip: Displacement current maintains continuity of current through capacitors.

Question 14. What is the source of energy associated with propagating em waves?
Answer: An accelerated (or oscillating) charged particle.
In simple words: Accelerated charges radiate energy in the form of electromagnetic waves.

Exam Tip: Stationary charges produce only electric fields; uniform motion produces magnetic fields; accelerated charges produce EM waves.

Question 15. What is the wavelength range of em waves that were produced and observed by J.C. Bose?
Answer: From \(25\text{ mm}\) to \(5\text{ mm}\) (Microwaves).
In simple words: J.C. Bose produced short wavelength radio waves (microwaves) in his laboratory.

Exam Tip: Memorize historical contributions: J.C. Bose (25 mm to 5 mm) and Marconi (several kilometers).

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

Exam Tip: Magnetron is the standard source used in common microwave ovens.

Question 17. Name the em radiations which are detected using Gieger tube.
Answer: X-rays and \(\gamma\)-rays.
In simple words: Geiger-Muller counters are designed to detect ionizing radiations like X-rays and gamma rays.

Exam Tip: Ionization detectors work by gas ionization caused by high-energy photons or particles.

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 \(\mu_r \approx 1\)
Speed \(v = \frac{c}{\sqrt{\mu_r \epsilon_r}}\)
\(v = \frac{3 \times 10^8}{\sqrt{1 \times 8}} = \frac{3 \times 10^8}{2\sqrt{2}} = 1.06 \times 10^8\text{ m/s}\)
In simple words: The speed of light in a medium decreases by a factor equal to the square root of its dielectric constant.

Exam Tip: Use the formula \(v = \frac{c}{n}\) where refractive index \(n = \sqrt{\mu_r \epsilon_r}\).

Question 19. Identify the part of the electromagnetic spectrum to which the following wavelengths belong: (i) \(10^{-1}\text{ m}\) (ii) \(10^{-12}\text{ m}\)
Answer:
(i) \(10^{-1}\text{ m} \to\) Microwaves (or short radio waves)
(ii) \(10^{-12}\text{ m} \to\) Gamma rays
In simple words: Wavelength of \(10^{-1}\text{ m}\) is in the microwave region, and \(10^{-12}\text{ m}\) is extremely short, belonging to gamma rays.

Exam Tip: Keep standard wavelength intervals of the spectrum handy for quick identification.

Question 20. Name the part of the electromagnetic spectrum of wavelength \(10^{-2}\text{ m}\) and mention its one application.
Answer:
Wavelength \(10^{-2}\text{ m}\) belongs to Microwaves.
Application: Used in RADAR systems for aircraft navigation.
In simple words: Microwaves of centimetre wavelength are ideal for directional wireless communication and radar.

Exam Tip: Microwaves span wavelengths from \(0.1\text{ m}\) down to \(1\text{ mm}\) (\(10^{-3}\text{ m}\)).

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: A charge in a circular orbit undergoes centripetal acceleration, and accelerated charges always radiate electromagnetic waves.

Question 22. Mention the pair of space and time varying E and B fields which would generate a plane em wave travelling in Z-direction.
Answer: Electric field along X-axis (\(E_x\)) and magnetic field along Y-axis (\(B_y\)) (or vice versa).
In simple words: The direction of wave propagation is given by the cross product of the electric and magnetic fields (\(\vec{E} \times \vec{B}\)).

Exam Tip: If wave travels along Z (\(\hat{k}\)), fields must oscillate along X (\(\hat{i}\)) and Y (\(\hat{j}\)) since \(\hat{i} \times \hat{j} = \hat{k}\).

Question 23. The charging current for a capacitor is \(0.2\text{ A}\). What is the displacement Current?
Answer: Displacement current \(I_d = 0.2\text{ A}\).
In simple words: Displacement current in the gap between capacitor plates is exactly equal to the conduction current in the connecting wires.

Exam Tip: Due to continuity of current, \(I_c = I_d\).

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

Exam Tip: Do not get distracted by different wavelengths when velocity in a vacuum is asked.

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: Speed of propagation in free space is identical for all electromagnetic waves.

Exam Tip: Frequency and wavelength vary widely across the spectrum, but speed in vacuum remains constant.

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 or water purifiers.
(ii) \(\gamma\)-ray: Used in radiotherapy for treatment of malignant tumors (cancer).
In simple words: UV rays kill bacteria, while high-energy gamma rays destroy cancer cells.

Exam Tip: Clearly bullet both parts for full credit.

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:
The equations representing the electric and magnetic fields are:
\(E_y = E_0 \sin(\omega t - kx)\)
\(B_z = B_0 \sin(\omega t - kx)\)
In simple words: The wave travels in the x-direction with electric oscillations along y and magnetic oscillations along z.

Exam Tip: Ensure phase terms \((\omega t - kx)\) are identical for both fields.

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 charge is the source of electromagnetic waves. When a charge oscillates, it produces oscillating electric and magnetic fields that sustain each other as an EM wave.
Frequency in medium: The frequency remains unchanged (\(\nu\) remains same).
In simple words: Frequency is a characteristic of the source and does not change when entering a medium, though wavelength and speed decrease.

Exam Tip: Frequency never changes upon refraction; only speed and wavelength change (\(v = \frac{c}{n}\), \(\lambda' = \frac{\lambda}{n}\)).

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:
In an electromagnetic wave, the ratio of electric field to magnetic field at any instant is constant and equal to the speed of light \(c\):
\(\frac{E}{B} = \frac{E_0}{B_0} = c\)
Therefore, magnetic field \(B\) is proportional to electric field \(E\):
\(B = \frac{E}{c} = \frac{\frac{3}{4}E_0}{c} = \frac{3}{4}B_0\)
In simple words: Since electric and magnetic fields oscillate in phase, if the electric field becomes \(3/4\) of its maximum value, the magnetic field also becomes \(3/4\) of its maximum value.

Exam Tip: Use linear proportionality between \(E\) and \(B\) in in-phase wave propagation.

Question 5. State two applications of infrared radiations.
Answer:
(1) Used in remote controls of TVs and home appliances.
(2) Used in physical therapy to treat muscular strains.
In simple words: Infrared waves carry heat energy and are widely used in remote controllers and thermal imaging.

Exam Tip: Also mention their use in greenhouse effect maintenance.

Question 6. State two applications of ultraviolet radiations.
Answer:
(1) Used in water purifiers to kill harmful bacteria and germs.
(2) Used in detection of forged documents and fake currency notes.
In simple words: UV light has germicidal properties and causes fluorescence.

Question 7. State two applications of x-rays.
Answer:
(1) Used in medical diagnostics to detect bone fractures.
(2) Used in security scanners at airports to check baggage.
In simple words: X-rays can pass through soft tissues but are absorbed by dense bones, making them useful for medical imaging.

Exam Tip: Mention medical diagnosis and engineering flaw detection.

Question 8. Show that the average energy density of the electric field \(\vec{E}\) equals the average energy density of the magnetics fields \(\vec{B}\)?
Answer:
Energy density of electric field: \(u_E = \frac{1}{2}\epsilon_0 E^2\)
For a sinusoidal wave, root mean square electric field is \(E_{rms} = \frac{E_0}{\sqrt{2}}\), so average electric energy density is:
\(u_{E(avg)} = \frac{1}{2}\epsilon_0 E_{rms}^2 = \frac{1}{4}\epsilon_0 E_0^2\)
Energy density of magnetic field: \(u_B = \frac{1}{2\mu_0} B^2\)
Average magnetic energy density is:
\(u_{B(avg)} = \frac{1}{2\mu_0} B_{rms}^2 = \frac{1}{4\mu_0} B_0^2\)
Since \(E_0 = c B_0\) and \(c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}\), substituting \(c^2 = \frac{1}{\mu_0 \epsilon_0}\):
\(u_{E(avg)} = \frac{1}{4}\epsilon_0 (c B_0)^2 = \frac{1}{4}\epsilon_0 \left(\frac{1}{\mu_0 \epsilon_0}\right) B_0^2 = \frac{1}{4\mu_0} B_0^2 = u_{B(avg)}\)
Hence, average electric energy density equals average magnetic energy density.
In simple words: Electromagnetic wave energy is shared equally between the electric and magnetic fields.

Exam Tip: Show substitution of \(E_0 = cB_0\) clearly to prove equality.

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 (\(\gamma\)-rays)
(ii) X-rays
(iii) Ultraviolet (UV) radiations
In simple words: Gamma rays destroy cancer cells, X-rays check for internal pipeline cracks, and UV rays kill bacteria on medical tools.

Exam Tip: Match each industrial/medical application to the correct high-energy EM band.

Question 2. How would you experimentally show that EM waves are transverse in nature?
Answer:
We can demonstrate the transverse nature of electromagnetic waves using the phenomenon of polarization:
(1) Take a source of light (or microwaves) and pass it through a polarizer (e.g., a tourmaline crystal or grid).
(2) Rotate a second identical crystal, known as an analyzer, placed in the path of the beam.
(3) It is observed that the intensity of transmitted light varies from maximum to zero as the analyzer is rotated through \(90^\circ\).
(4) This asymmetric behavior proves that electromagnetic oscillations are strictly perpendicular to the direction of wave propagation, confirming their transverse nature.
In simple words: Passing light through polarizing filters blocks waves vibrating in specific directions, proving they vibrate sideways (transversely) rather than back-and-forth longitudinally.

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 produced by accelerated charges and do not require any material medium for propagation.
(2) In a vacuum, they travel with a constant speed of \(c = 3 \times 10^8\text{ m/s}\).
(3) They are transverse in nature, with electric and magnetic field vectors oscillating mutually perpendicular to each other and to the direction of propagation.
In simple words: EM waves travel through empty space at the speed of light and consist of perpendicular electric and magnetic vibrations.

Question 4. Find the wavelength of electromagnetic waves of frequency \(5 \times 10^{19}\text{ Hz}\) in free space. Give its two applications
Answer:
Frequency \(\nu = 5 \times 10^{19}\text{ Hz}\)
Speed of light \(c = 3 \times 10^8\text{ m/s}\)
Wavelength \(\lambda = \frac{c}{\nu}\)
\(\lambda = \frac{3 \times 10^8}{5 \times 10^{19}} = 0.6 \times 10^{-11}\text{ m} = 6 \times 10^{-12}\text{ m}\) (Gamma rays)
Applications of gamma rays:
(1) Used in radiotherapy for cancer treatment.
(2) Used to induce mutations in agriculture for crop improvement.
In simple words: Using the wave formula \(\lambda = c/\nu\), we find the wavelength falls in the gamma ray region.

Exam Tip: Always state the formula \(\lambda = c/\nu\) before substituting values.

NUMERICALS

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

Exam Tip: Remember that frequency never changes during refraction.

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

Exam Tip: Use relation \(B_0 = E_0/c\) with proper units (Tesla).

Question 3. Suppose the electric field amplitude of an em wave is \(E_0 = 120\text{ NC}^{-1}\) and its frequency is \(\nu = 50.0\text{ MHz}\). (a) Determine \(B_0, \omega, k\) and \(\lambda\) (b) Find expressions for E and B.
Answer:
(a) Calculations:
Electric field amplitude \(E_0 = 120\text{ V/m}\)
Frequency \(\nu = 50.0\text{ MHz} = 50 \times 10^6\text{ Hz}\)
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 \(k = \frac{2\pi}{\lambda} = \frac{2 \times 3.14}{6} = 1.05\text{ m}^{-1}\)

(b) Expressions for E and B (assuming wave travels along x-axis with E along y-axis):
\(E_y(x,t) = 120 \sin(3.14 \times 10^8 t - 1.05 x)\text{ V/m}\)
\(B_z(x,t) = 4 \times 10^{-7} \sin(3.14 \times 10^8 t - 1.05 x)\text{ T}\)
In simple words: Calculate each wave parameter step-by-step using standard wave formulas and substitute them into the general sine wave equation.

Exam Tip: Be careful with unit conversions for MHz (\(10^6\) Hz).

Question 4. A radio can tune into any station of frequency band \(7.5\text{ MHz}\) to \(10\text{ MHz}\). Find the corresponding wave length range.
Answer:
Speed of light \(c = 3 \times 10^8\text{ m/s}\)
For lower frequency \(\nu_1 = 7.5\text{ MHz} = 7.5 \times 10^6\text{ Hz}\):
\(\lambda_1 = \frac{c}{\nu_1} = \frac{3 \times 10^8}{7.5 \times 10^6} = 40\text{ m}\)
For upper frequency \(\nu_2 = 10\text{ MHz} = 10 \times 10^6\text{ Hz}\):
\(\lambda_2 = \frac{c}{\nu_2} = \frac{3 \times 10^8}{10 \times 10^6} = 30\text{ m}\)
Wavelength range is from \(30\text{ m}\) to \(40\text{ m}\).
In simple words: Since wavelength is inversely proportional to frequency, the lower frequency corresponds to the longer wavelength (\(40\text{ m}\)) 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 magnetic field amplitude \(B_0 = 2.4\text{ mT} = 2.4 \times 10^{-3}\text{ T}\)
Frequency \(\nu = 16\text{ MHz} = 16 \times 10^6\text{ Hz}\)
(i) Amplitude of electric field vector:\br /> \(E_0 = c B_0 = (3 \times 10^8) \times (2.4 \times 10^{-3}) = 7.2 \times 10^5\text{ V/m}\)
(ii) Wavelength of the wave:
\(\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{16 \times 10^6} = \frac{300}{16} = 18.75\text{ m}\)
In simple words: Multiply magnetic field by \(c\) to find the electric field, and divide \(c\) by frequency to find wavelength.

Exam Tip: Convert milli-Tesla (mT) to Tesla (\(10^{-3}\) T) before calculating.

Question 6. An EM wave travelling through a medium has electric field vector. \(E_y = 4 \times 10^5 \cos(3.14 \times 10^8 t - 1.57 x)\text{ N/C}\). 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 the given equation \(E_y = 4 \times 10^5 \cos(3.14 \times 10^8 t - 1.57 x)\) with standard wave equation \(E_y = E_0 \cos(\omega t - kx)\):
Angular frequency \(\omega = 3.14 \times 10^8\text{ rad/s}\)
Propagation constant \(k = 1.57\text{ m}^{-1}\)
Electric field amplitude \(E_0 = 4 \times 10^5\text{ N/C}\)

(i) Wavelength \(\lambda = \frac{2\pi}{k} = \frac{2 \times 3.14}{1.57} = 4\text{ m}\)
(ii) Frequency \(\nu = \frac{\omega}{2\pi} = \frac{3.14 \times 10^8}{2 \times 3.14} = 0.5 \times 10^8\text{ Hz} = 5 \times 10^7\text{ Hz}\)
(iii) Direction of propagation: Positive x-axis (indicated by \(-kx\))
(iv) Speed of wave \(v = \frac{\omega}{k} = \frac{3.14 \times 10^8}{1.57} = 2 \times 10^8\text{ m/s}\)
(v) Refractive index of medium \(n = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5\)
(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: By comparing the wave equation coefficients with standard wave parameters, we can directly extract frequency, wavelength, speed, refractive index, and field amplitudes.

Exam Tip: Speed in a medium is given by \(v = \omega/k\), and refractive index is \(n = c/v\).