To solve the problem, we need to find the force exerted by light on a surface when light of a certain intensity is incident on it. Let's break down the solution step by step.
### Step 1: Understand the given data
- Intensity of light, \( I_0 = 50 \, \text{W/m}^2 \)
- Area of the surface, \( A = 1 \, \text{m}^2 \)
- Percentage of light reflected, \( R = 25\% = 0.25 \)
- Therefore, the percentage of light absorbed is \( 1 - R = 0.75 \).
### Step 2: Calculate the total power incident on the surface
The power \( P \) incident on the surface can be calculated using the formula:
\[
P = I_0 \times A
\]
Substituting the values:
\[
P = 50 \, \text{W/m}^2 \times 1 \, \text{m}^2 = 50 \, \text{W}
\]
### Step 3: Calculate the power reflected and absorbed
- Power reflected, \( P_r = R \times P = 0.25 \times 50 \, \text{W} = 12.5 \, \text{W} \)
- Power absorbed, \( P_a = (1 - R) \times P = 0.75 \times 50 \, \text{W} = 37.5 \, \text{W} \)
### Step 4: Calculate the radiation pressure
The radiation pressure \( P \) is given by the formula:
\[
P = \frac{I}{c} \times (1 + R)
\]
Where \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \).
Substituting the values:
\[
P = \frac{50 \, \text{W/m}^2}{3 \times 10^8 \, \text{m/s}} \times (1 + 0.25)
\]
\[
P = \frac{50}{3 \times 10^8} \times 1.25
\]
\[
P = \frac{62.5}{3 \times 10^8} \, \text{N/m}^2
\]
### Step 5: Calculate the force exerted by the light on the surface
The force \( F \) exerted by the light can be calculated using:
\[
F = P \times A
\]
Substituting the area:
\[
F = \left(\frac{62.5}{3 \times 10^8}\right) \times 1
\]
\[
F = \frac{62.5}{3 \times 10^8} \, \text{N}
\]
Calculating the value:
\[
F \approx 20.83 \times 10^{-8} \, \text{N}
\]
### Final Answer
The force exerted by the light on the surface is approximately:
\[
F \approx 20.83 \times 10^{-8} \, \text{N}
\]
