A linearly polarised electromagnetic wave given as `E=E_(0)haticos(kz-omegat)` is incident normally on a perfectly reflecting wall `z=a`. Assuming that the material of the optically inactive, the reflected wave will be give as

Options

A. `vec(E)_(r)=-E_(0)hat(i)cos(kx-omegat)`
B. `vec(E)_(r)=E_(0)hat(i)cos(kz+omegat)`
C. `vec(E)_(r)=-E_(0)hat(i)cos(kz+omegat)`
D. `vec(E)_(r)=E_(0) hat(i)sin(kz-omegat)`

Explanation

To solve the problem of finding the equation for the reflected wave when a linearly polarized electromagnetic wave is incident normally on a perfectly reflecting wall, we can follow these steps:

### Step 1: Understand the Incident Wave
The incident wave is given by:
\[ E = E_0 \hat{i} \cos(kz - \omega t) \]
This wave is linearly polarized in the direction of \(\hat{i}\) and travels in the positive \(z\) direction.

### Step 2: Reflection at the Wall
When the wave hits a perfectly reflecting wall at \(z = a\), it reflects back in the opposite direction. The reflection occurs normally, meaning the angle of incidence is equal to the angle of reflection.

### Step 3: Change of Direction
Since the wave reflects back, the \(z\) component in the wave equation will change sign. Thus, the wave equation for the reflected wave can be written as:
\[ E = E_0 \hat{i} \cos(k(-z) - \omega t) \]
This simplifies to:
\[ E = E_0 \hat{i} \cos(-kz - \omega t) \]

### Step 4: Phase Change upon Reflection
When an electromagnetic wave reflects off a denser medium (like a perfectly reflecting wall), there is a phase change of \(\pi\) (180 degrees). This means we need to adjust our wave equation to account for this phase change.

### Step 5: Incorporate the Phase Change
The phase change of \(\pi\) can be incorporated into the equation:
\[ E = E_0 \hat{i} \cos(-kz - \omega t + \pi) \]
Using the property of cosine, \(\cos(\theta + \pi) = -\cos(\theta)\), we can rewrite the equation:
\[ E = -E_0 \hat{i} \cos(-kz - \omega t) \]

### Step 6: Simplify the Equation
Now, we can simplify the equation using the property of cosine:
\[ E = -E_0 \hat{i} \cos(kz + \omega t) \]
Since \(\cos(-\theta) = \cos(\theta)\), we can express it as:
\[ E = E_0 \hat{i} \cos(kz + \omega t) \]

### Final Answer
Thus, the equation for the reflected wave is:
\[ E = E_0 \hat{i} \cos(kz + \omega t) \]

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