Light with an energy flux of `18 W//cm^2` falls on a non-reflecting surface at normal incidence. If the surface has an area of `20 cm^2`, find the average force exerted on the surface during a `30` minute time span.

Options

A. `2.1xx10^(-6)N`
B. `1.2xx10^(-6)N`
C. `1.2xx10^(6)N`
D. `2.1xx10^(6)N`

Explanation

To solve the problem, we need to calculate the average force exerted on a non-reflecting surface by light over a specified time period. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the given data
- Energy flux (intensity) of light, \( I = 18 \, \text{W/cm}^2 \)
- Area of the surface, \( A = 20 \, \text{cm}^2 \)
- Time duration, \( \Delta t = 30 \, \text{minutes} = 30 \times 60 = 1800 \, \text{seconds} \)

### Step 2: Calculate the total power absorbed by the surface
The total power \( P \) absorbed by the surface can be calculated using the formula:
\[ P = I \times A \]
Substituting the values:
\[ P = 18 \, \text{W/cm}^2 \times 20 \, \text{cm}^2 = 360 \, \text{W} \]

### Step 3: Relate power to momentum change
The average force \( F_{\text{avg}} \) exerted on the surface can be expressed in terms of the change in momentum over time:
\[ F_{\text{avg}} = \frac{\Delta P}{\Delta t} \]
Where \( \Delta P \) is the change in momentum. For light, the momentum change can be related to power by the equation:
\[ \Delta P = \frac{P}{c} \times \Delta t \]
Where \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \).

### Step 4: Calculate the change in momentum
Substituting the values into the equation for \( \Delta P \):
\[ \Delta P = \frac{360 \, \text{W}}{3 \times 10^8 \, \text{m/s}} \times 1800 \, \text{s} \]
Calculating \( \Delta P \):
\[ \Delta P = \frac{360}{3 \times 10^8} \times 1800 = \frac{648000}{3 \times 10^8} = 2.16 \times 10^{-3} \, \text{kg m/s} \]

### Step 5: Calculate the average force
Now substituting \( \Delta P \) back into the equation for \( F_{\text{avg}} \):
\[ F_{\text{avg}} = \frac{2.16 \times 10^{-3} \, \text{kg m/s}}{1800 \, \text{s}} = 1.2 \times 10^{-6} \, \text{N} \]

### Step 6: Conclusion
Since the force exerted by light on a non-reflecting surface is in the direction of the light, we can express the average force as:
\[ F_{\text{avg}} = 1.2 \times 10^{-6} \, \text{N} \]

### Final Answer:
The average force exerted on the surface during a 30-minute time span is \( 1.2 \times 10^{-6} \, \text{N} \).