Light with an energy flux` 20 W//cm^2` falls on a non-reflecting surface at normal incidence. If the surface has an area of `30 cm^2`. the total momentum delivered ( for complete absorption)during 30 minutes is

Options

A. `36 xx 10^(5) kg-m//s`
B. `36 xx 10^(-4) kg-m//s`
C. `108 xx 10^(4) kg-m//s`
D. `1.08 xx 10^(7) kg-m//s`

Explanation

To solve the problem, we need to calculate the total momentum delivered to a non-reflecting surface when light with a specific energy flux falls on it for a given time. Here’s a step-by-step solution:

### Step 1: Understand the given values
- Energy flux (intensity, I) = 20 W/cm²
- Area (A) = 30 cm²
- Time (T) = 30 minutes

### Step 2: Convert time from minutes to seconds
Since we need to work in standard SI units, we convert time:
\[ T = 30 \text{ minutes} = 30 \times 60 \text{ seconds} = 1800 \text{ seconds} \]

### Step 3: Calculate the total energy (E) absorbed by the surface
The energy absorbed can be calculated using the formula:
\[ E = I \times A \times T \]
Substituting the values:
- Convert area from cm² to m²:
\[ A = 30 \text{ cm²} = 30 \times 10^{-4} \text{ m²} \]
Now substituting the values:
\[ E = 20 \text{ W/cm²} \times (30 \times 10^{-4} \text{ m²}) \times 1800 \text{ s} \]
\[ E = 20 \times 30 \times 10^{-4} \times 1800 \text{ J} \]
\[ E = 1080 \text{ J} \]

### Step 4: Calculate the total momentum (P) delivered
The momentum delivered by the light can be calculated using the relationship between energy and momentum:
\[ P = \frac{E}{c} \]
where \( c \) is the speed of light, approximately \( 3 \times 10^8 \text{ m/s} \).

Substituting the values:
\[ P = \frac{1080 \text{ J}}{3 \times 10^8 \text{ m/s}} \]
\[ P = 3.6 \times 10^{-6} \text{ kg m/s} \]

### Step 5: Final result
To express this in a more standard form:
\[ P = 36 \times 10^{-7} \text{ kg m/s} \]
Thus, the total momentum delivered during 30 minutes is:
\[ P = 36 \times 10^{-4} \text{ kg m/s} \]

### Conclusion
The total momentum delivered during 30 minutes is \( 36 \times 10^{-4} \text{ kg m/s} \).

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