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← JEE Advanced 2023 · Paper 2
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JEE Advanced 2023 · Paper 2, Physics question 7

The electric field associated with an electromagnetic wave propagating in a dielectric medium is given by E⃗=30(2x^+y^)sin⁡[2π(5×1014 t−1073z)] V m−1\vec E = 30(2\hat x + \hat y)\sin\left[2\pi\left(5\times10^{14}\,t - \frac{10^7}{3}z\right)\right]\ \mathrm{V\,m^{-1}}. Which of the following option(s) is(are) correct?

[Given: The speed of light in vacuum, c=3×108 m s−1c = 3\times10^8\ \mathrm{m\,s^{-1}}]

Choose one or more options
Answer and worked solution

Official answer: A, D

Wave parameters. The wave travels along +z+z with
f=5×1014 Hz,1λ=1073 m−1.f=5\times10^{14}\ \text{Hz},\qquad\frac1\lambda=\frac{10^7}3\ \text{m}^{-1}.

(D) True. v=fλ=5×1014×3×10−7=1.5×108v=f\lambda=5\times10^{14}\times3\times10^{-7}=1.5\times10^8 m/s, so n=c/v=2n=c/v=2.

Magnetic field. B⃗=1vz^×E⃗\vec B=\frac1v\hat z\times\vec E, and z^×(2x^+y^)=2y^−x^\hat z\times(2\hat x+\hat y)=2\hat y-\hat x. So
B⃗=301.5×108(−x^+2y^)sin⁡[⋯ ]=2×10−7(−x^+2y^)sin⁡[⋯ ].\vec B=\frac{30}{1.5\times10^8}(-\hat x+2\hat y)\sin[\cdots]=2\times10^{-7}(-\hat x+2\hat y)\sin[\cdots].

  • (A) True: Bx=−2×10−7sin⁡[⋯ ]B_x=-2\times10^{-7}\sin[\cdots].
  • (B) False: By=4×10−7sin⁡[⋯ ]B_y=4\times10^{-7}\sin[\cdots].

(C) False. E⃗\vec E is along 2x^+y^2\hat x+\hat y, at tan⁡−112≈26.6∘\tan^{-1}\frac12\approx26.6^\circ to the xx-axis, not 30∘30^\circ.

Topics:Properties of EM wavesPolarisation and Brewster's law
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