`50 W//m^(2)` energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy (25%) is reflected from the surface and the rest is absorbed. The force exerted on `1m^(2)` surface area will be close to `(c=3xx10^(8) m//s)`:

Options

A. `15xx10^(-8)N`
B. `20xx10^(-8) N`
C. `10xx10^(-8) N`
D. `35xx10^(-8) N`

Answer

To solve the problem, we need to calculate the total force exerted on a solar panel of area \(1 \, m^2\) when sunlight with an intensity of \(50 \, W/m^2\) is incident on it. We are given that 25% of the incident energy is reflected, and the remaining 75% is absorbed.

### Step-by-Step Solution:

1. **Identify the intensity of sunlight**:
The intensity \(I\) of the sunlight is given as:
\[ I = 50 \, W/m^2 \]

2. **Calculate the reflected energy**:
Since 25% of the incident energy is reflected, the intensity of the reflected energy \(I_r\) is:
\[ I_r = 0.25 \times I = 0.25 \times 50 = 12.5 \, W/m^2 \]

3. **Calculate the absorbed energy**:
The remaining 75% of the incident energy is absorbed, so the intensity of the absorbed energy \(I_a\) is:
\[ I_a = 0.75 \times I = 0.75 \times 50 = 37.5 \, W/m^2 \]

4. **Calculate the force due to reflected energy**:
The force \(F_r\) exerted by the reflected light can be calculated using the formula:
\[ F_r = \frac{2 I_r}{c} \]
Substituting the values:
\[ F_r = \frac{2 \times 12.5}{3 \times 10^8} = \frac{25}{3 \times 10^8} \, N \]

5. **Calculate the force due to absorbed energy**:
The force \(F_a\) exerted by the absorbed light can be calculated using the formula:
\[ F_a = \frac{I_a}{c} \]
Substituting the values:
\[ F_a = \frac{37.5}{3 \times 10^8} \, N \]

6. **Calculate the total force**:
The total force \(F\) exerted on the solar panel is the sum of the forces due to reflected and absorbed light:
\[ F = F_r + F_a = \frac{25}{3 \times 10^8} + \frac{37.5}{3 \times 10^8} \]
\[ F = \frac{25 + 37.5}{3 \times 10^8} = \frac{62.5}{3 \times 10^8} \, N \]

7. **Final calculation**:
To express the force in a more manageable form:
\[ F = \frac{62.5}{3} \times 10^{-8} \, N \approx 20.83 \times 10^{-8} \, N \]

### Conclusion:
The force exerted on the \(1 \, m^2\) surface area of the solar panel is approximately:
\[ F \approx 20.83 \times 10^{-8} \, N \]