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

[Given: The speed of light in vacuum, $$c = 3 \times 10^8$$ ms$$^{-1}$$]

The electric field is given as
$$\vec{E}(z,t)=30\,(2\hat{x}+\hat{y})\, \sin\!\Bigl[2\pi\Bigl(5\times10^{14}\,t-\dfrac{10^{7}}{3}\,z\Bigr)\Bigr] \;{\rm V\,m^{-1}}$$

The argument of the sine has the standard form $$2\pi(ft-kz)$$, so

$$f = 5\times10^{14}\ {\rm Hz} \quad -(1)$$
$$k = 2\pi\!\left(\dfrac{10^{7}}{3}\right)=\dfrac{2\pi\times10^{7}}{3}\ {\rm rad\,m^{-1}} \quad -(2)$$

Because the positive coefficient of $$z$$ appears with a minus sign in the phase, the wave propagates along $$+z$$.

From $$k=\dfrac{2\pi}{\lambda}$$, the wavelength is
$$\lambda=\dfrac{2\pi}{k}= \dfrac{2\pi}{2\pi\times10^{7}/3}= \dfrac{3}{10^{7}} =3\times10^{-7}\ {\rm m} \quad -(3)$$

The speed of the wave in the medium is
$$v = f\lambda = (5\times10^{14})(3\times10^{-7}) =1.5\times10^{8}\ {\rm m\,s^{-1}} \quad -(4)$$

Hence the refractive index is
$$n = \dfrac{c}{v} = \dfrac{3\times10^{8}}{1.5\times10^{8}} = 2 \quad -(5)$$
So Option D is correct.

The amplitude of the electric field is
$$\vec{E}_0 = 30\,(2\hat{x}+\hat{y}) = 60\hat{x}+30\hat{y}\;{\rm V\,m^{-1}} \quad -(6)$$

For a plane electromagnetic wave moving along $$\hat{z}$$, the magnetic field amplitude is related by
$$\vec{B}_0 = \dfrac{1}{v}\,\hat{z}\times\vec{E}_0 \quad -(7)$$

Using $$\hat{z}\times\hat{x}= \hat{y}$$ and $$\hat{z}\times\hat{y} = -\hat{x}$$:

$$\hat{z}\times\vec{E}_0 = 60(\hat{z}\times\hat{x})+30(\hat{z}\times\hat{y}) = 60\hat{y}-30\hat{x} \quad -(8)$$

Therefore
$$\vec{B}_0 = \dfrac{60\hat{y}-30\hat{x}}{1.5\times10^{8}} = -2\times10^{-7}\hat{x}+4\times10^{-7}\hat{y}\ {\rm T} \quad -(9)$$

Incorporating the same space-time factor as in $$\vec{E}$$ gives
$$\vec{B}(z,t)=\left[-2\times10^{-7}\hat{x}+4\times10^{-7}\hat{y}\right] \sin\!\Bigl[2\pi\Bigl(5\times10^{14}t-\dfrac{10^{7}}{3}z\Bigr)\Bigr] \;{\rm T} \quad -(10)$$

The $$x$$-component matches Option A, while Option B lists the $$y$$-component with the wrong magnitude (it should be $$4\times10^{-7}{\rm \,T}$$, not $$2\times10^{-7}{\rm \,T}$$). Thus Option B is incorrect.

The electric field lies entirely in the $$xy$$-plane along the direction $$2\hat{x}+\hat{y}$$. The angle with the $$x$$-axis is
$$\tan\theta=\dfrac{1}{2}\;\Longrightarrow\;\theta\approx26.6^{\circ} \quad -(11)$$
not $$30^{\circ}$$, so Option C is also incorrect.

Hence the correct options are:
Option A and Option D.