A parallel beam of the light is incident normally on a plane surface absorbing 40% of the light and reflecting the rest. If the incident beam carries 10 watt power, find the force exerted by it on the surface.

Explanation

To solve the problem step-by-step, we will follow these calculations:

### Step 1: Determine the power absorbed and reflected
The total power of the incident beam is given as \( P = 10 \) watts. The surface absorbs 40% of the light and reflects the remaining 60%.

- Power absorbed \( P_a = 0.40 \times P = 0.40 \times 10 = 4 \) watts
- Power reflected \( P_r = 0.60 \times P = 0.60 \times 10 = 6 \) watts

### Step 2: Calculate the momentum change due to absorption and reflection
The momentum of light can be calculated using the formula:
\[ \text{Momentum per second} = \frac{\text{Power}}{c} \]
where \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \) m/s.

#### For absorbed power:
- Momentum change due to absorption \( P_a \):
\[ \Delta p_a = \frac{P_a}{c} = \frac{4}{3 \times 10^8} \]

#### For reflected power:
- Momentum change due to reflection \( P_r \):
\[ \Delta p_r = \frac{P_r}{c} = \frac{6}{3 \times 10^8} \]
However, since the light is reflected, the momentum change is twice that of the reflected power:
\[ \Delta p_r = 2 \times \frac{6}{3 \times 10^8} \]

### Step 3: Calculate the total momentum change
The total momentum change \( \Delta p \) is the sum of the momentum change due to absorption and twice the momentum change due to reflection:
\[ \Delta p = \Delta p_a + \Delta p_r = \frac{4}{3 \times 10^8} + 2 \times \frac{6}{3 \times 10^8} \]
\[ \Delta p = \frac{4 + 12}{3 \times 10^8} = \frac{16}{3 \times 10^8} \]

### Step 4: Calculate the force exerted by the light
Force is defined as the rate of change of momentum:
\[ F = \Delta p \]
Since the power is constant, the force exerted by the light on the surface is:
\[ F = \frac{16}{3 \times 10^8} \text{ N} \]

### Step 5: Calculate the numerical value of the force
Calculating the numerical value:
\[ F \approx \frac{16}{3 \times 10^8} \approx 5.33 \times 10^{-8} \text{ N} \]

### Final Answer
The force exerted by the light on the surface is approximately \( 5.33 \times 10^{-8} \) N.

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