The incident intensity on a horizontal surface at sea level from the sun is about `1kW m^(-2)`.
Assuming that 50 per cent of this intensity is reflected and 50 per cent is absorbed, determine the radiation pressure on this horizontal surface (in pascals).

Options

A. `8.2xx10^(-2)`
B. `5xx10^(-6)`
C. `6xx10^(-12)`
D. `8xx10^(-11)`

Explanation

To solve the problem of determining the radiation pressure on a horizontal surface due to sunlight, we can follow these steps:

### Step 1: Understand the Given Data
The intensity of sunlight incident on the surface is given as:
- \( S = 1 \, \text{kW/m}^2 = 1000 \, \text{W/m}^2 \)

It is stated that:
- 50% of this intensity is reflected.
- 50% of this intensity is absorbed.

### Step 2: Calculate the Pressure from Absorbed Light
The pressure exerted by the absorbed light can be calculated using the formula:
\[ P_{\text{absorbed}} = \frac{S_{\text{absorbed}}}{c} \]
Where:
- \( S_{\text{absorbed}} = 0.5 \times S = 0.5 \times 1000 \, \text{W/m}^2 = 500 \, \text{W/m}^2 \)
- \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \).

Thus, the pressure from absorbed light is:
\[ P_{\text{absorbed}} = \frac{500}{3 \times 10^8} \, \text{Pa} \]

### Step 3: Calculate the Pressure from Reflected Light
The pressure exerted by the reflected light is given by:
\[ P_{\text{reflected}} = \frac{2S_{\text{reflected}}}{c} \]
Where:
- \( S_{\text{reflected}} = 0.5 \times S = 500 \, \text{W/m}^2 \)

Thus, the pressure from reflected light is:
\[ P_{\text{reflected}} = \frac{2 \times 500}{3 \times 10^8} \, \text{Pa} = \frac{1000}{3 \times 10^8} \, \text{Pa} \]

### Step 4: Calculate Total Radiation Pressure
The total radiation pressure \( P_{\text{total}} \) on the surface is the sum of the pressures from absorbed and reflected light:
\[ P_{\text{total}} = P_{\text{absorbed}} + P_{\text{reflected}} \]
Substituting the values we calculated:
\[ P_{\text{total}} = \frac{500}{3 \times 10^8} + \frac{1000}{3 \times 10^8} \]
\[ P_{\text{total}} = \frac{1500}{3 \times 10^8} \, \text{Pa} \]
\[ P_{\text{total}} = \frac{1500}{3} \times 10^{-8} \, \text{Pa} = 500 \times 10^{-8} \, \text{Pa} = 5 \times 10^{-6} \, \text{Pa} \]

### Final Answer
The radiation pressure on the horizontal surface is:
\[ P_{\text{total}} = 5 \times 10^{-6} \, \text{Pa} \]