A beam of white light is incident normally on a plane surface absorbing `70%` of the light and reflecting the rest. If the incident beam carries `30 W` of power, find the force exerted by it on the surface.

Explanation

To find the force exerted by a beam of white light on a plane surface, we can follow these steps:

### Step 1: Determine the power absorbed and reflected
The total power of the incident beam is given as \( P = 30 \, \text{W} \). Since \( 70\% \) of the light is absorbed, the power absorbed \( P_a \) is:
\[ P_a = 0.7 \times P = 0.7 \times 30 \, \text{W} = 21 \, \text{W} \]
The power reflected \( P_r \) is:
\[ P_r = 0.3 \times P = 0.3 \times 30 \, \text{W} = 9 \, \text{W} \]

### Step 2: Calculate the change in momentum for absorbed light
When light is absorbed, it transfers its momentum to the surface. The momentum \( p \) of light can be expressed in terms of power \( P \) and the speed of light \( c \):
\[ p = \frac{P}{c} \]
For the absorbed power:
\[ p_a = \frac{P_a}{c} = \frac{21 \, \text{W}}{3 \times 10^8 \, \text{m/s}} = 7 \times 10^{-8} \, \text{kg m/s} \]

### Step 3: Calculate the change in momentum for reflected light
For the reflected light, the momentum change is twice the momentum of the reflected power because it reverses direction:
\[ p_r = 2 \times \frac{P_r}{c} = 2 \times \frac{9 \, \text{W}}{3 \times 10^8 \, \text{m/s}} = 6 \times 10^{-8} \, \text{kg m/s} \]

### Step 4: Total change in momentum
The total change in momentum \( \Delta p \) is the sum of the momentum changes for absorbed and reflected light:
\[ \Delta p = p_a + p_r = 7 \times 10^{-8} \, \text{kg m/s} + 6 \times 10^{-8} \, \text{kg m/s} = 1.3 \times 10^{-7} \, \text{kg m/s} \]

### Step 5: Calculate the force exerted on the surface
The force \( F \) exerted by the light on the surface is the rate of change of momentum. Since power is the rate of energy transfer, we can relate it to momentum change over time. The force can be calculated as:
\[ F = \frac{\Delta p}{\Delta t} \]
Since we know the power and the speed of light, we can express the force in terms of power:
\[ F = \frac{P}{c} \]
Substituting the total power \( P = 30 \, \text{W} \):
\[ F = \frac{30 \, \text{W}}{3 \times 10^8 \, \text{m/s}} = 1 \times 10^{-7} \, \text{N} \]

### Final Answer
The force exerted by the beam of white light on the surface is:
\[ F = 1 \times 10^{-7} \, \text{N} \]