A parallel beam of light is incident normally on a plane surface absorbing 40% of the light and reflecting the rest. If the incident beam carries 60 W of power, the force exerted by it on the surface is

Options

A. `3.2 xx 10^(-8) N`
B. `3.2 xx 10^(-7) N`
C. `5.12 xx 10^(-7) N`
D. `5.12 xx 10^(-8) N`

Answer

To solve the problem, we need to calculate the force exerted by a parallel beam of light on a plane surface, given that 40% of the light is absorbed and the incident beam carries a power of 60 W.

### Step-by-Step Solution:

1. **Identify the Power of the Incident Beam:**
The power of the incident beam is given as \( P = 60 \, \text{W} \).

2. **Calculate the Power Absorbed and Reflected:**
- The power absorbed by the surface is \( 40\% \) of the incident power:
\[ P_{\text{absorbed}} = 0.4 \times P = 0.4 \times 60 \, \text{W} = 24 \, \text{W} \]
- The power reflected by the surface is \( 60\% \) of the incident power:
\[ P_{\text{reflected}} = 0.6 \times P = 0.6 \times 60 \, \text{W} = 36 \, \text{W} \]

3. **Calculate the Force Due to Absorption:**
The force due to the absorbed power can be calculated using the formula:
\[ F_{\text{absorbed}} = \frac{P_{\text{absorbed}}}{c} \]
where \( c \) is the speed of light (\( c \approx 3 \times 10^8 \, \text{m/s} \)).
Substituting the values:
\[ F_{\text{absorbed}} = \frac{24}{3 \times 10^8} = 8 \times 10^{-8} \, \text{N} \]

4. **Calculate the Force Due to Reflection:**
The force due to the reflected power can be calculated using the formula:
\[ F_{\text{reflected}} = \frac{2 \times P_{\text{reflected}}}{c} \]
Substituting the values:
\[ F_{\text{reflected}} = \frac{2 \times 36}{3 \times 10^8} = \frac{72}{3 \times 10^8} = 2.4 \times 10^{-7} \, \text{N} \]

5. **Calculate the Total Force:**
The total force exerted by the light on the surface is the sum of the forces due to absorption and reflection:
\[ F_{\text{total}} = F_{\text{absorbed}} + F_{\text{reflected}} = 8 \times 10^{-8} + 2.4 \times 10^{-7} \]
Converting \( 8 \times 10^{-8} \) to the same power of ten:
\[ F_{\text{total}} = 8 \times 10^{-8} + 24 \times 10^{-8} = 32 \times 10^{-8} = 3.2 \times 10^{-7} \, \text{N} \]

### Final Answer:
The total force exerted by the beam of light on the surface is:
\[ F_{\text{total}} = 3.2 \times 10^{-7} \, \text{N} \]