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

aNS is 5.33 x 10 -8 N

To determine the force exerted by a beam of light on a surface, we can use the relationship between power, momentum, and force. When light strikes a surface, it can either be absorbed or reflected, and each of these interactions contributes to the momentum transfer, which in turn relates to the force exerted on the surface.

Understanding the Power Distribution

In this scenario, we have a parallel beam of light with a total power of 10 W. The problem states that 40% of this light is absorbed by the surface, while the remaining 60% is reflected. Let's break this down:

• Power absorbed: 40% of 10 W = 0.4 × 10 W = 4 W
• Power reflected: 60% of 10 W = 0.6 × 10 W = 6 W

Calculating Momentum Change

Light carries momentum, and when it interacts with a surface, it changes momentum, which is what creates a force. The momentum \( p \) of light can be calculated using the formula:

p = \frac{P}{c}

where \( P \) is the power of the light and \( c \) is the speed of light in a vacuum (approximately \( 3 \times 10^8 \) m/s).

Momentum from Absorbed Light

For the absorbed power (4 W), the momentum change is:

p_{absorbed} = \frac{4 \text{ W}}{c} = \frac{4}{3 \times 10^8} \text{ kg m/s}

Momentum from Reflected Light

For the reflected power (6 W), the light changes direction, effectively doubling the momentum change because it reverses direction:

p_{reflected} = 2 \times \frac{6 \text{ W}}{c} = 2 \times \frac{6}{3 \times 10^8} \text{ kg m/s}

Total Momentum Change

The total momentum change \( \Delta p \) is the sum of the momentum changes from both absorbed and reflected light:

\Delta p = p_{absorbed} + p_{reflected}

Substituting the values:

\Delta p = \frac{4}{3 \times 10^8} + 2 \times \frac{6}{3 \times 10^8} = \frac{4 + 12}{3 \times 10^8} = \frac{16}{3 \times 10^8} \text{ kg m/s}

Calculating Force

Force can be defined as the rate of change of momentum over time. If we assume the light beam is continuous and steady, we can express force \( F \) as:

F = \frac{\Delta p}{\Delta t}

Since we are considering a steady beam, we can relate the power to force directly. The force exerted by the light can be calculated using the total power \( P \) and the speed of light \( c \):

F = \frac{P}{c}

In this case, we need to consider only the effective power that contributes to the momentum change, which is the total power (10 W):

F = \frac{10 \text{ W}}{3 \times 10^8 \text{ m/s}} = \frac{10}{3 \times 10^8} \text{ N}

Calculating this gives:

F ≈ 3.33 \times 10^{-8} \text{ N}

However, since we have both absorbed and reflected components, we need to consider the effective force from both interactions:

F_{total} = \frac{16}{3 \times 10^8} \text{ N} = 5.33 \times 10^{-8} \text{ N}

Final Result

Thus, the total force exerted by the beam of light on the surface is approximately 5.33 x 10^-8 N. This calculation illustrates how light, despite being intangible, can exert measurable forces through its interactions with surfaces.