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

Options

A. `3.2xx10^(-8)N`
B. `3.2xx10^(-7)N`
C. `5.12xx10^(-7)N`
D. `5.12xx10^(-8)N`

Explanation

To solve the problem, we need to determine the force exerted by the light beam on the surface. Here's a step-by-step breakdown of the solution:

### Step 1: Calculate the Power Absorbed and Reflected
The total power of the incident beam is given as 60 W. The surface absorbs 40% of this power.

\[ \text{Power absorbed} = 60 \, \text{W} \times 0.40 = 24 \, \text{W} \]

The remaining power is reflected:

\[ \text{Power reflected} = 60 \, \text{W} - 24 \, \text{W} = 36 \, \text{W} \]

### Step 2: Calculate the Change in Momentum
When light is absorbed, it transfers momentum to the surface. The momentum change due to the absorbed power can be calculated using the formula:

\[ \text{Momentum change per second} = \frac{\text{Power absorbed}}{c} \]

Where \( c \) is the speed of light (\( c \approx 3 \times 10^8 \, \text{m/s} \)).

For the absorbed power:

\[ \text{Momentum change (absorbed)} = \frac{24 \, \text{W}}{3 \times 10^8 \, \text{m/s}} = 8 \times 10^{-8} \, \text{kg m/s} \]

For the reflected power, the momentum change is twice the power reflected (because the light reverses direction):

\[ \text{Momentum change (reflected)} = 2 \times \frac{36 \, \text{W}}{3 \times 10^8 \, \text{m/s}} = 2 \times 12 \times 10^{-8} = 24 \times 10^{-8} \, \text{kg m/s} \]

### Step 3: Total Change in Momentum
The total change in momentum per second (which is also the force exerted on the surface) is the sum of the momentum changes due to absorption and reflection:

\[ \text{Total momentum change} = \text{Momentum change (absorbed)} + \text{Momentum change (reflected)} \]

\[ \text{Total momentum change} = 8 \times 10^{-8} + 24 \times 10^{-8} = 32 \times 10^{-8} \, \text{kg m/s} \]

### Step 4: Calculate the Force
The force exerted by the light on the surface is equal to the total change in momentum per second:

\[ F = 32 \times 10^{-8} \, \text{N} = 3.2 \times 10^{-7} \, \text{N} \]

### Final Answer
The force exerted by the light beam on the surface is:

\[ F = 3.2 \times 10^{-7} \, \text{N} \]

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