A linearly polarised electromagnetic wave in vacuum is $$$E = 3.1\cos 1.8z - 5.4 \times 10^6 t \hat{i}$$$ N C$$^{-1}$$ is incident normally on a perfectly reflecting wall at $$z = a$$. Choose the correct option.

We have been given an incident electric field in free space

$$$\vec E_i(z,t)=3.1\cos(1.8\,z-5.4\times10^{6}\,t)\,\hat i\;{\rm N\,C^{-1}}.$$$

This is of the standard form $$\vec E_i=E_0\cos(kz-\omega t)\hat i,$$ so by simple comparison

$$$k=1.8\;{\rm rad\,m^{-1}},\qquad \omega=5.4\times10^{6}\;{\rm rad\,s^{-1}},\qquad E_0=3.1\;{\rm N\,C^{-1}}.$$$

First, let us find the wavelength. The relation between the propagation constant and wavelength in vacuum is

$$k=\frac{2\pi}{\lambda}\;\Longrightarrow\;\lambda=\frac{2\pi}{k}.$$

Substituting the numerical value

$$\lambda=\frac{2\pi}{1.8}= \frac{6.2832}{1.8}\approx3.49\;{\rm m}.$$

This is clearly not equal to $$5.4\;{\rm m}$$, so statement A (“The wavelength is 5.4 m”) is wrong.

Next, let us calculate the ordinary frequency $$f$$. The angular frequency and ordinary frequency are related by

$$\omega=2\pi f\;\Longrightarrow\;f=\frac{\omega}{2\pi}.$$

Hence

$$$f=\frac{5.4\times10^{6}}{2\pi}= \frac{5.4\times10^{6}}{6.2832}\approx8.6\times10^{5}\;{\rm Hz}.$$$

The option claims $$54\times10^{4}\;{\rm Hz}=5.4\times10^{5}\;{\rm Hz}$$, which is different from $$8.6\times10^{5}\;{\rm Hz}$$. Therefore statement B is also wrong.

Because the wall is perfectly reflecting (perfect conductor), no electromagnetic field can be transmitted into it; the tangential component of $$\vec E$$ must vanish at the surface. Hence there is no transmitted wave. Statement C therefore cannot be true.

We now construct the reflected wave. For normal incidence, the reflected field must travel in the $$-z$$ direction and can be written in the general form

$$\vec E_r(z,t)=E_0\cos(kz+\omega t+\phi)\,\hat i,$$

where $$\phi$$ is a phase constant to be fixed by the boundary condition at the wall situated at $$z=a$$.

The total tangential electric field on the wall must be zero, so we impose

$$\vec E_i(a,t)+\vec E_r(a,t)=0.$$

That gives

$$$E_0\cos(k a-\omega t)+E_0\cos(k a+\omega t+\phi)=0\qquad\forall\;t.$$$

Dividing through by $$E_0$$ and using the identity

$$\cos C+\cos D=2\cos\frac{C+D}{2}\,\cos\frac{C-D}{2},$$

we obtain

$$$2\cos\!\Bigl(k a+\frac{\phi}{2}\Bigr)\cos\!\Bigl(-\omega t-\frac{\phi}{2}\Bigr)=0\qquad\forall\;t.$$$

For the product of cosines to vanish at every instant, the first cosine must be zero:

$$\cos\!\Bigl(k a+\tfrac{\phi}{2}\Bigr)=0.$$

This condition is satisfied if

$$k a+\frac{\phi}{2}=\frac{(2n+1)\pi}{2},\qquad n=0,1,2,\dots$$

The simplest choice is to keep $$\phi=0$$ and take

$$k a=\frac{(2n+1)\pi}{2}\;\;(n=0,1,2,\dots).$$

With this choice, the reflected field becomes

$$$\boxed{\;\vec E_r(z,t)=3.1\cos(1.8\,z+5.4\times10^{6}\,t)\,\hat i\;{\rm N\,C^{-1}}\;},$$$

which is exactly the expression listed in option D.

We have therefore shown

• Options A and B disagree with the numerical values of wavelength and frequency.
• Option C is impossible because a perfectly reflecting wall allows no transmission.
• Option D is consistent with the required boundary condition and is therefore correct.

Hence, the correct answer is Option D.