A plane light wave of intensity `I=0.20Wcm^-2` falls on a plane mirror surface with reflection coefficient `rho=0.8`. The angle of incidence is `45^circ`. In terms of corpuscular theory, find the magnitude of the normal pressure exerted on that surface.

Options

A. `1.2N cm^-2`
B. `0.2Ncm^-2`
C. `2.6Ncm^-2`
D. `0.5N cm^-2`

Explanation

To find the magnitude of the normal pressure exerted on the surface of a plane mirror by a plane light wave, we can follow these steps:

### Step 1: Understand the Problem
We have a plane light wave with intensity \( I = 0.20 \, \text{W/cm}^2 \) incident on a plane mirror at an angle of \( 45^\circ \). The reflection coefficient \( \rho = 0.8 \). We need to find the normal pressure exerted on the mirror surface.

### Step 2: Calculate the Momentum Change Due to Incident Photons
The momentum change per unit time for the incident photons normal to the surface can be expressed as:
\[ \frac{dP_{\text{incident}}}{dt} = I \cdot dA \cdot \cos^2 \theta \]
Where:
- \( I \) is the intensity of the light,
- \( dA \) is the area,
- \( \theta \) is the angle of incidence.

Substituting the values:
\[ \theta = 45^\circ \Rightarrow \cos^2(45^\circ) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]
Thus,
\[ \frac{dP_{\text{incident}}}{dt} = I \cdot dA \cdot \frac{1}{2} = 0.20 \cdot dA \cdot \frac{1}{2} = 0.10 \, dA \]

### Step 3: Calculate the Momentum Change Due to Reflected Photons
The momentum change per unit time for the reflected photons is given by:
\[ \frac{dP_{\text{reflected}}}{dt} = -I \cdot dA \cdot \rho \cdot \cos^2 \theta \]
Substituting the reflection coefficient \( \rho = 0.8 \):
\[ \frac{dP_{\text{reflected}}}{dt} = -0.20 \cdot dA \cdot 0.8 \cdot \frac{1}{2} = -0.08 \, dA \]

### Step 4: Calculate the Total Rate of Change of Momentum
The total rate of change of momentum is the sum of the momentum change due to incident and reflected photons:
\[ \frac{dP}{dt} = \frac{dP_{\text{incident}}}{dt} + \frac{dP_{\text{reflected}}}{dt} = 0.10 \, dA - 0.08 \, dA = 0.02 \, dA \]

### Step 5: Relate Rate of Change of Momentum to Force
According to Newton's second law, the force \( F \) exerted on the surface is equal to the rate of change of momentum:
\[ F = \frac{dP}{dt} \]
Thus,
\[ F = 0.02 \, dA \]

### Step 6: Calculate the Pressure
Pressure \( P \) is defined as force per unit area:
\[ P = \frac{F}{dA} = \frac{0.02 \, dA}{dA} = 0.02 \, \text{N/cm}^2 \]

### Step 7: Convert to Standard Units
To express this in standard units:
\[ P = 0.02 \, \text{N/cm}^2 = 0.02 \times 10^4 \, \text{N/m}^2 = 200 \, \text{Pa} \]

### Final Answer
The magnitude of the normal pressure exerted on the surface is:
\[ P = 0.02 \, \text{N/cm}^2 \text{ or } 200 \, \text{Pa} \]

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