A `"50 W m"^(-2)` energy density of sunlight is incident normally 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)ms^(-1))`

Options

A. `25xx10^(-8)N`
B. `20xx10^(-8)N`
C. `10xx10^(-8)N`
D. `35xx10^(-8)N`

Answer

To solve the problem, we need to determine the force exerted on a solar panel due to the incident sunlight. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the incident energy density and the absorbed energy
The energy density of sunlight incident on the solar panel is given as \( I = 50 \, \text{W/m}^2 \). Since 25% of this energy is reflected, the absorbed energy density can be calculated as follows:

\[ \text{Reflected energy density} = 0.25 \times I = 0.25 \times 50 = 12.5 \, \text{W/m}^2 \]

\[ \text{Absorbed energy density} = I - \text{Reflected energy density} = 50 - 12.5 = 37.5 \, \text{W/m}^2 \]

### Step 2: Calculate the momentum change due to the absorbed and reflected energy
The momentum associated with the incident energy can be expressed as:

\[ p_1 = \frac{E_1}{c} \]

Where \( E_1 \) is the incident energy and \( c \) is the speed of light (\( c = 3 \times 10^8 \, \text{m/s} \)).

For the absorbed energy, the momentum change is:

\[ p_{\text{absorbed}} = \frac{E_{\text{absorbed}}}{c} = \frac{37.5 \, \text{W/m}^2}{c} \]

For the reflected energy, the momentum change is:

\[ p_{\text{reflected}} = \frac{E_{\text{reflected}}}{c} = \frac{12.5 \, \text{W/m}^2}{c} \]

### Step 3: Calculate the net momentum change
The net momentum change \( \Delta p \) is given by the momentum of the absorbed energy minus the momentum of the reflected energy:

\[ \Delta p = p_{\text{absorbed}} + p_{\text{reflected}} = \frac{37.5}{c} + \frac{12.5}{c} = \frac{50}{c} \]

### Step 4: Calculate the force exerted on the solar panel
The force \( F \) exerted on the solar panel can be calculated using the relation:

\[ F = \frac{\Delta p}{\Delta t} \]

Assuming \( \Delta t = 1 \, \text{s} \) for simplicity, we have:

\[ F = \Delta p = \frac{50}{c} \]

### Step 5: Substitute the value of \( c \)
Substituting \( c = 3 \times 10^8 \, \text{m/s} \):

\[ F = \frac{50}{3 \times 10^8} \]

Calculating this gives:

\[ F \approx 1.67 \times 10^{-7} \, \text{N} \]

### Final Result
Thus, the force exerted on a \( 1 \, \text{m}^2 \) surface area of the solar panel will be approximately:

\[ F \approx 1.67 \times 10^{-7} \, \text{N} \]