Q.A linearly polarized electromagnetic wave given as $\mathbf{E} = E_0\hat{i}\cos(kz - \omega t)$ is incident normally on a perfectly reflecting infinite wall at $z = a$. Assuming that the material of the wall is optically inactive, the reflected wave will be given as

The wall at $z=a$ is a perfect reflector, so the tangential electric field must vanish at its surface. The reflected wave keeps the same amplitude and frequency but travels back along $-z$, and its electric field is reversed (a phase change of $\pi$).

• Incident wave: $\mathbf{E}_i = E_0\hat{i}\cos(kz-\omega t)$, propagating along $+z$.
• Reflected wave: same $\hat{i}$ polarisation, propagation reversed to $-z$ (argument becomes $kz+\omega t$), with a sign flip from the $\pi$ phase shift.

Principle. Inside a perfect conductor the electromagnetic field is zero. At the surface of the wall the tangential component of the total electric field (incident + reflected) must therefore vanish for all times $t$. This single boundary condition fixes the reflected wave.

Step 1 — Direction of the reflected wave. The incident wave $\mathbf{E}_i = E_0\hat{i}\cos(kz-\omega t)$ moves along $+z$. On reflection the wave must move along $-z$, so its spatial-temporal argument changes from $kz-\omega t$ to $kz+\omega t$.

Step 2 — Polarisation. The wall is optically inactive, so it does not rotate the plane of polarisation. The reflected electric field stays along $\hat{i}$ (the $x$-direction), with the same amplitude $E_0$.

Step 3 — Apply the boundary condition (phase reversal). For a perfect conductor the tangential electric field is continuous and equals zero at the surface, which requires the reflected electric field to be exactly out of phase with the incident one at the wall. This is the standard $\pi$ phase change of the electric field on reflection from a denser/perfectly-reflecting medium, giving an overall minus sign:

Step 4 — Consistency check. The magnetic field does not undergo a $\pi$ shift on reflection; the tangential magnetic field at the surface doubles rather than cancels. Only the electric field is reversed, which is exactly what the accepted NCERT Exemplar option states.

Method: Finding the Reflected Wave from a Perfect Reflector

Use this whenever an EM wave is normally incident on a perfectly reflecting (conducting) surface and you must write the reflected wave's field expression.

Steps

Step 1: Identify the physical boundary condition

Inside/at the surface of a perfect conductor, the total tangential electric field (incident + reflected) must be zero at every instant — this single condition determines everything about the reflected wave.

Step 2: Reverse the direction of propagation in the wave's argument

If the incident wave travels in $+z$ with argument $(kz-\omega t)$, the reflected wave travels in $-z$, so its argument becomes $(kz+\omega t)$ (same $k,\omega$, since the medium and frequency are unchanged).

Step 3: Keep the same polarisation direction (for an optically inactive medium)

The field stays along the same unit vector ($\hat i$, etc.) with the same amplitude $E_0$ — reflection off an optically inactive wall does not rotate the plane of polarisation.

Step 4: Apply the $\pi$ phase reversal to the electric field only

To make the tangential $E$ vanish at the wall for all $t$, the reflected electric field must carry an overall minus sign relative to the incident one:

This reversal applies to $\vec E$ only — the tangential magnetic field does not undergo this phase flip (it doubles, rather than cancels, at a perfectly conducting surface), so never apply the same sign flip to $\vec B$.