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. `36xx10^4 kg-m//s`
C. `1.08 xx 10^4 kg -m//s`
D. `1.08xx10^7 kg-m//s`

Explanation

To solve the problem step by step, we will follow the principles of energy flux and momentum transfer due to light.

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

### Step 2: Convert units
1. Convert the area from cm² to m²:
\[ A = 30 \, \text{cm}^2 = 30 \times 10^{-4} \, \text{m}^2 = 3 \times 10^{-3} \, \text{m}^2 \]

2. Convert time from minutes to seconds:
\[ t = 30 \, \text{minutes} = 30 \times 60 \, \text{seconds} = 1800 \, \text{s} \]

### Step 3: Calculate the total energy absorbed
The total energy \( E \) absorbed by the surface can be calculated using the formula:
\[ E = I \times A \times t \]
Substituting the values:
\[ E = 20 \, \text{W/cm}^2 \times 30 \, \text{cm}^2 \times 1800 \, \text{s} \]
Calculating \( E \):
\[ E = 20 \times 30 \times 1800 \, \text{W} \cdot \text{s} = 1080000 \, \text{J} = 10.8 \times 10^5 \, \text{J} \]
Since \( 1 \, \text{W} = 1 \, \text{J/s} \), we can also express this in terms of \( \text{W/cm}^2 \):
\[ E = 20 \times 30 \times 1800 \times 10^{-4} \, \text{J} = 10.8 \, \text{J} \]

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

Substituting the values:
\[ p = \frac{10.8 \, \text{J}}{3 \times 10^8 \, \text{m/s}} = 3.6 \times 10^{-8} \, \text{kg m/s} \]

### Step 5: Final result
Thus, the total momentum delivered during 30 minutes is:
\[ p = 3.6 \times 10^{-8} \, \text{kg m/s} \]