To determine which option(s) are correct, we analyze the given electric field equation:
E ⃗ = 30 ( 2 x ^ + y ^ ) sin [ 2 π ( 5 × 10 14 t − 10 7 3 z ) ] V m − 1 \vec{E} = 30(2\hat{x} + \hat{y})\sin\left[2\pi\left(5 \times 10^{14}t - \frac{10^7}{3}z\right)\right] \text{ V m}^{-1} E = 30 ( 2 x ^ + y ^ ) sin [ 2 π ( 5 × 1 0 14 t − 3 1 0 7 z ) ] V m − 1
1. Speed of the Wave and Refractive Index (Checking Option D) Comparing the phase of the wave with the standard form sin ( ω t − k z ) \sin(\omega t - kz) sin ( ω t − k z )
Angular frequency, ω = 2 π × ( 5 × 10 14 ) rad s − 1 = 10 15 π rad s − 1 \omega = 2\pi \times \left(5 \times 10^{14}\right) \text{ rad s}^{-1} = 10^{15}\pi \text{ rad s}^{-1} ω = 2 π × ( 5 × 1 0 14 ) rad s − 1 = 1 0 15 π rad s − 1
Wave number, k = 2 π × 10 7 3 m − 1 = 2 π × 10 7 3 m − 1 k = 2\pi \times \frac{10^7}{3} \text{ m}^{-1} = \frac{2\pi \times 10^7}{3} \text{ m}^{-1} k = 2 π × 3 1 0 7 m − 1 = 3 2 π × 1 0 7 m − 1
The phase velocity v v v v = ω k = 5 × 10 14 10 7 3 = 1.5 × 10 8 m s − 1 v = \frac{\omega}{k} = \frac{5 \times 10^{14}}{\frac{10^7}{3}} = 1.5 \times 10^8 \text{ m s}^{-1} v = k ω = 3 1 0 7 5 × 1 0 14 = 1.5 × 1 0 8 m s − 1
The refractive index n n n n = c v = 3 × 10 8 m s − 1 1.5 × 10 8 m s − 1 = 2 n = \frac{c}{v} = \frac{3 \times 10^8 \text{ m s}^{-1}}{1.5 \times 10^8 \text{ m s}^{-1}} = 2 n = v c = 1.5 × 1 0 8 m s − 1 3 × 1 0 8 m s − 1 = 2
Thus, Option D is correct.
2. Angle of Polarization (Checking Option C) The electric field vector oscillates along E ⃗ 0 = 30 ( 2 x ^ + y ^ ) = 60 x ^ + 30 y ^ V m − 1 \vec{E}_0 = 30(2\hat{x} + \hat{y}) = 60\hat{x} + 30\hat{y} \text{ V m}^{-1} E 0 = 30 ( 2 x ^ + y ^ ) = 60 x ^ + 30 y ^ V m − 1 x y xy x y
The angle of polarization θ \theta θ x x x tan θ = E 0 y E 0 x = 30 60 = 1 2 = 0.5 \tan\theta = \frac{E_{0y}}{E_{0x}} = \frac{30}{60} = \frac{1}{2} = 0.5 tan θ = E 0 x E 0 y = 60 30 = 2 1 = 0.5
For a 30 ∘ 30^\circ 3 0 ∘ tan 30 ∘ = 1 3 ≈ 0.577 \tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577 tan 3 0 ∘ = 3 1 ≈ 0.577 tan θ = 0.5 ⟹ θ = arctan ( 0.5 ) ≈ 26.57 ∘ ≠ 30 ∘ \tan\theta = 0.5 \implies \theta = \arctan(0.5) \approx 26.57^\circ \neq 30^\circ tan θ = 0.5 ⟹ θ = arctan ( 0.5 ) ≈ 26.5 7 ∘ = 3 0 ∘
Thus, Option C is incorrect.
3. Magnetic Field Vector B ⃗ \vec{B} B The direction of wave propagation is along the + z +z + z k ^ = z ^ \hat{k} = \hat{z} k ^ = z ^
The magnetic field vector B ⃗ \vec{B} B E ⃗ \vec{E} E B ⃗ = 1 v ( k ^ × E ⃗ ) \vec{B} = \frac{1}{v} (\hat{k} \times \vec{E}) B = v 1 ( k ^ × E )
Calculating k ^ × E ⃗ \hat{k} \times \vec{E} k ^ × E z ^ × ( 60 x ^ + 30 y ^ ) = 60 ( z ^ × x ^ ) + 30 ( z ^ × y ^ ) = 60 y ^ − 30 x ^ = − 30 x ^ + 60 y ^ \hat{z} \times (60\hat{x} + 30\hat{y}) = 60(\hat{z} \times \hat{x}) + 30(\hat{z} \times \hat{y}) = 60\hat{y} - 30\hat{x} = -30\hat{x} + 60\hat{y} z ^ × ( 60 x ^ + 30 y ^ ) = 60 ( z ^ × x ^ ) + 30 ( z ^ × y ^ ) = 60 y ^ − 30 x ^ = − 30 x ^ + 60 y ^
Therefore:
B ⃗ = − 30 x ^ + 60 y ^ 1.5 × 10 8 sin [ 2 π ( 5 × 10 14 t − 10 7 3 z ) ] \vec{B} = \frac{-30\hat{x} + 60\hat{y}}{1.5 \times 10^8} \sin\left[2\pi\left(5 \times 10^{14}t - \frac{10^7}{3}z\right)\right] B = 1.5 × 1 0 8 − 30 x ^ + 60 y ^ sin [ 2 π ( 5 × 1 0 14 t − 3 1 0 7 z ) ] B ⃗ = ( − 2 × 10 − 7 x ^ + 4 × 10 − 7 y ^ ) sin [ 2 π ( 5 × 10 14 t − 10 7 3 z ) ] Wb m − 2 \vec{B} = \left(-2 \times 10^{-7}\hat{x} + 4 \times 10^{-7}\hat{y}\right) \sin\left[2\pi\left(5 \times 10^{14}t - \frac{10^7}{3}z\right)\right] \text{ Wb m}^{-2} B = ( − 2 × 1 0 − 7 x ^ + 4 × 1 0 − 7 y ^ ) sin [ 2 π ( 5 × 1 0 14 t − 3 1 0 7 z ) ] Wb m − 2
From this, the components of the magnetic field are:
B x = − 2 × 10 − 7 sin [ 2 π ( 5 × 10 14 t − 10 7 3 z ) ] Wb m − 2 B_x = -2 \times 10^{-7} \sin\left[2\pi\left(5 \times 10^{14}t - \frac{10^7}{3}z\right)\right] \text{ Wb m}^{-2} B x = − 2 × 1 0 − 7 sin [ 2 π ( 5 × 1 0 14 t − 3 1 0 7 z ) ] Wb m − 2 B y = 4 × 10 − 7 sin [ 2 π ( 5 × 10 14 t − 10 7 3 z ) ] Wb m − 2 B_y = 4 \times 10^{-7} \sin\left[2\pi\left(5 \times 10^{14}t - \frac{10^7}{3}z\right)\right] \text{ Wb m}^{-2} B y = 4 × 1 0 − 7 sin [ 2 π ( 5 × 1 0 14 t − 3 1 0 7 z ) ] Wb m − 2
Evaluating the options:
Option B states B y = 2 × 10 − 7 sin [ 2 π ( 5 × 10 14 t − 10 7 3 z ) ] B_y = 2 \times 10^{-7} \sin\left[2\pi\left(5 \times 10^{14}t - \frac{10^7}{3}z\right)\right] B y = 2 × 1 0 − 7 sin [ 2 π ( 5 × 1 0 14 t − 3 1 0 7 z ) ] 2 × 10 − 7 2 \times 10^{-7} 2 × 1 0 − 7 4 × 10 − 7 4 \times 10^{-7} 4 × 1 0 − 7 Option B is incorrect. Option A as given in the problem statement lacks the factor of 2 π 2\pi 2 π sin ( 5 × 10 14 t − 10 7 3 z ) \sin\left(5 \times 10^{14}t - \frac{10^7}{3}z\right) sin ( 5 × 1 0 14 t − 3 1 0 7 z ) Option A incorrect.
Conclusion The correct option is D .