Consider an electromagnetic wave propagating in vacuum. Choose the correct statement:

For a monochromatic plane electromagnetic wave travelling through vacuum we always have three experimentally verified geometric relations:

1. The wave-vector $$\vec k$$ (hence the direction of propagation), the electric field $$\vec E$$ and the magnetic field $$\vec B$$ are mutually perpendicular. Mathematically

$$\vec k\cdot\vec E = 0,\qquad \vec k\cdot\vec B = 0,\qquad \vec E\cdot\vec B = 0.$$

2. The right-hand rule fixes their relative orientation. Stating it formally, the Poynting vector is

$$\vec S=\frac1{\mu_0}\,\vec E\times\vec B,$$

and this $$\vec S$$ points in the same direction as $$\vec k$$.

3. Their magnitudes are related by $$|\vec B|=\dfrac{|\vec E|}{c}$$, but for the present multiple-choice problem only the directions matter.

Now we test every option against the two purely geometrical requirements written in (1) and (2).

Option A
Propagation is specified as $$+y$$, i.e. $$\vec k=+\hat y$$. The proposal is

$$\vec E=\frac1{\sqrt2}E_{yz}(x,t)\,\hat z,\qquad \vec B=\frac1{\sqrt2}B_{z}(x,t)\,\hat y.$$

Immediately we notice $$\vec B$$ is parallel to $$\vec k$$ because both point along $$\hat y$$, so $$\vec k\cdot\vec B\neq 0$$. This violates $$\vec k\cdot\vec B=0$$, therefore Option A is incorrect.

Option B
Again the propagation direction is $$+y$$ but here

$$\vec E=\frac1{\sqrt2}E_{yz}(x,t)\,\hat y,\qquad \vec B=\frac1{\sqrt2}B_{yx}(x,t)\,\hat z.$$

Now $$\vec E$$ itself is parallel to $$\vec k$$, giving $$\vec k\cdot\vec E\neq 0$$, contradicting $$\vec k\cdot\vec E=0$$. Hence Option B is also wrong.

Option C
Propagation is along $$+x$$, so $$\vec k=+\hat x$$. The fields are given as

$$\vec E=\frac1{\sqrt2}E_{yz}(y,z,t)\,(\hat y+\hat z),\qquad \vec B=\frac1{\sqrt2}E_{yz}(y,z,t)\,(\hat y+\hat z).$$

Although each vector lies in the $$yz$$‐plane and is therefore perpendicular to $$\vec k$$ (good so far), notice that $$\vec E$$ and $$\vec B$$ are identical. Their dot product equals their magnitudes squared:

$$\vec E\cdot\vec B=|\vec E|\,|\vec B|\neq 0,$$

so they are not perpendicular, violating $$\vec E\cdot\vec B=0$$. Consequently Option C is ruled out.

Option D
Propagation is again along $$+x$$, i.e. $$\vec k=+\hat x$$. Here we are told

$$\vec E=\frac1{\sqrt2}E_{yz}(x,t)\,(\hat y-\hat z),\qquad \vec B=\frac1{\sqrt2}B_{yz}(x,t)\,(\hat y+\hat z).$$

First, both $$\vec E$$ and $$\vec B$$ lie wholly in the $$yz$$‐plane, so each is automatically perpendicular to $$\vec k$$; thus $$\vec k\cdot\vec E=0$$ and $$\vec k\cdot\vec B=0$$ are satisfied.

Next we check their mutual perpendicularity:

$$\vec E\cdot\vec B=\frac12(\hat y-\hat z)\cdot(\hat y+\hat z) =\frac12(1-1)=0,$$

so $$\vec E$$ and $$\vec B$$ are indeed perpendicular.

Finally we verify the right-hand rule by computing $$\vec E\times\vec B$$:

$$$ \begin{aligned} \vec E\times\vec B &=\frac12(\hat y-\hat z)\times(\hat y+\hat z)\\[4pt] &=\frac12\bigl[\hat y\times\hat y+\hat y\times\hat z-\hat z\times\hat y-\hat z\times\hat z\bigr]\\[4pt] &=\frac12\bigl[0+\hat x-(-\hat x)+0\bigr]\\[4pt] &=\frac12(2\hat x)=\hat x. \end{aligned} $$$

The cross product points along $$+\hat x$$, exactly the specified direction of propagation. Therefore Option D satisfies all the requirements imposed by Maxwell’s equations.

Hence, the correct answer is Option D.