A beam of white light is incident normaly on a plane surface absorbing 70% of the light and reflecting the rest. If the incident beam carries 10 w of power, find the force exerated by it on the surface.

Explanation

To solve the problem, we need to find the force exerted by the beam of light on the surface. We can use the concept of momentum change due to the reflection and absorption of light.

### Step-by-Step Solution:

1. **Identify the Power of the Incident Beam**:
The power of the incident beam is given as \( P = 10 \, \text{W} \).

2. **Determine the Fraction of Light Absorbed and Reflected**:
The problem states that 70% of the light is absorbed and 30% is reflected.
- Absorbed Power: \( P_{absorbed} = 0.7 \times P = 0.7 \times 10 \, \text{W} = 7 \, \text{W} \)
- Reflected Power: \( P_{reflected} = 0.3 \times P = 0.3 \times 10 \, \text{W} = 3 \, \text{W} \)

3. **Calculate the Change in Momentum**:
- For absorbed light, the change in momentum is given by:
\[ \Delta p_{absorbed} = \frac{P_{absorbed}}{c} = \frac{7 \, \text{W}}{c} \]
- For reflected light, the change in momentum is:
\[ \Delta p_{reflected} = \frac{P_{reflected}}{c} \times 2 = \frac{3 \, \text{W}}{c} \times 2 = \frac{6 \, \text{W}}{c} \]

4. **Total Change in Momentum**:
The total change in momentum per unit time (force) is:
\[ F = \Delta p_{absorbed} + \Delta p_{reflected} = \frac{7}{c} + \frac{6}{c} = \frac{13}{c} \]

5. **Substituting the Speed of Light**:
The speed of light \( c \) is approximately \( 3 \times 10^8 \, \text{m/s} \). Therefore:
\[ F = \frac{13 \, \text{W}}{3 \times 10^8 \, \text{m/s}} = \frac{13}{3 \times 10^8} \, \text{N} \]

6. **Calculating the Force**:
\[ F \approx 4.33 \times 10^{-8} \, \text{N} \]

### Final Answer:
The force exerted by the beam of light on the surface is approximately \( 4.33 \times 10^{-8} \, \text{N} \).