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Electromagnetic Waves question

2021 · 25 Jul · Shift 1 · Q59
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  5. /2021 · 25 Jul · Shift 1 · Q59

Electromagnetic Waves question

2021 · 25 Jul · Shift 1 · Q59

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
A linearly polarized electromagnetic wave in vacuum is E=3.1cos⁡[(1.8)z−(5.4×106)t]i^N/CE = 3.1\cos \left[ {(1.8)z - (5.4 \times {{10}^6})t} \right]\widehat iN/CE=3.1cos[(1.8)z−(5.4×106)t]iN/C is incident normally on a perfectly reflecting wall at z = a. Choose the correct option
  1. A
    The wavelength is 5.4 m
  2. B
    The frequency of electromagnetic wave is 54 ×\times× 104 Hz.
  3. C
    The transmitted wave will be 3.1cos⁡[(1.8)z−(5.4×106)t]i^N/C3.1\cos \left[ {(1.8)z - (5.4 \times {{10}^6})t} \right]\widehat iN/C3.1cos[(1.8)z−(5.4×106)t]iN/C
  4. D
    The reflected wave will be 3.1cos⁡[(1.8)z+(5.4×106)t]i^N/C3.1\cos \left[ {(1.8)z + (5.4 \times {{10}^6})t} \right]\widehat iN/C3.1cos[(1.8)z+(5.4×106)t]iN/C
View written solutionFree

Correct answer: D

  1. Identify the wave parameters

The given incident electric field is

E⃗i=3.1cos⁡[(1.8)z−(5.4×106)t] i^  N/C\vec E_i = 3.1\cos\big[(1.8)z-(5.4\times 10^6)t\big] \, \hat i \; \text{N/C}Ei​=3.1cos[(1.8)z−(5.4×106)t]i^N/C

Compare with the standard form:

E⃗=E0cos⁡(kz−ωt)\vec E = E_0\cos(kz-\omega t)E=E0​cos(kz−ωt)

So,

  • Wave number: k=1.8 rad/mk=1.8\ \text{rad/m}k=1.8 rad/m
  • Angular frequency: ω=5.4×106 rad/s\omega=5.4\times 10^6\ \text{rad/s}ω=5.4×106 rad/s

Since the phase is kz−ωtkz-\omega tkz−ωt, the wave is traveling in the +z direction.

  1. Check option A: wavelength

We know

λ=2πk\lambda = \frac{2\pi}{k}λ=k2π​

Thus,

λ=2π1.8≈3.49 m\lambda = \frac{2\pi}{1.8} \approx 3.49\ \text{m}λ=1.82π​≈3.49 m

This is not 5.4 m5.4\,\text{m}5.4m.

So, A is false.

  1. Check option B: frequency

We know

f=ω2πf = \frac{\omega}{2\pi}f=2πω​

Therefore,

f=5.4×1062πf = \frac{5.4\times 10^6}{2\pi}f=2π5.4×106​ f≈5.4×1066.283≈8.59×105 Hzf \approx \frac{5.4\times 10^6}{6.283} \approx 8.59\times 10^5\ \text{Hz}f≈6.2835.4×106​≈8.59×105 Hz

Option B says:

54×104=5.4×105 Hz54\times 10^4 = 5.4\times 10^5\ \text{Hz}54×104=5.4×105 Hz

which is incorrect.

So, B is false.

  1. Check option C: transmitted wave

The wave is incident on a perfectly reflecting wall at z=az=az=a.

For a perfectly reflecting wall, there is no transmitted wave beyond the wall.

Hence option C is false.

  1. Check option D: reflected wave

For reflection from a perfectly reflecting wall, the reflected wave must travel in the -z direction.

A wave traveling in the −z-z−z direction has the form

cos⁡(kz+ωt)\cos(kz+\omega t)cos(kz+ωt)

or equivalently cos⁡(−kz−ωt)\cos(-kz-\omega t)cos(−kz−ωt) etc., depending on phase constants.

Thus the reflected wave must have the sign of ttt reversed relative to the incident wave, i.e. of the form

E⃗r=E0cos⁡(kz+ωt+ϕ) i^\vec E_r = E_0\cos(kz+\omega t+\phi)\,\hat iEr​=E0​cos(kz+ωt+ϕ)i^

Option D gives

3.1cos⁡[(1.8)z+(5.4×106)t]i^ N/C3.1\cos\left[(1.8)z+(5.4\times 10^6)t\right]\hat i\ \text{N/C}3.1cos[(1.8)z+(5.4×106)t]i^ N/C

which indeed represents a wave propagating in the -z direction.

Hence D is the correct option.

Note: At a perfectly reflecting wall, there can be an additional phase factor depending on boundary conditions at z=az=az=a, but among the given options, D is the correct reflected-wave form.

  1. Final conclusion
  • A: False
  • B: False
  • C: False
  • D: True

Therefore, the correct answer is:

D\boxed{D}D​
  1. Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.

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