Light is incident normally on a completely absorbing surface with an energy flux of `25 Wcm^(-2)` If the surface has an area of `25 cm^(2)` the momentum transferred to the surface in 40 min time duration will be:

Options

A. `6.3 xx 10^(-4) Ns`
B. `5.0 xx 10^(-3) Ns`
C. `1.4 xx 10^(-6) Ns`
D. `3.5 xx 10^(-6) Ns`

Explanation

To solve the problem step by step, we need to calculate the momentum transferred to a completely absorbing surface when light is incident on it.

### Step 1: Understand the Given Data
- Energy flux (intensity) \( I = 25 \, \text{W/cm}^2 \)
- Area of the surface \( A = 25 \, \text{cm}^2 \)
- Time duration \( t = 40 \, \text{minutes} \)

### Step 2: Convert Units
1. Convert energy flux from \( \text{W/cm}^2 \) to \( \text{W/m}^2 \):
\[ I = 25 \, \text{W/cm}^2 = 25 \times 10^4 \, \text{W/m}^2 \]
2. Convert area from \( \text{cm}^2 \) to \( \text{m}^2 \):
\[ A = 25 \, \text{cm}^2 = 25 \times 10^{-4} \, \text{m}^2 \]
3. Convert time from minutes to seconds:
\[ t = 40 \, \text{minutes} = 40 \times 60 = 2400 \, \text{seconds} \]

### Step 3: Calculate the Momentum Transferred
The momentum transferred to the surface can be calculated using the formula:
\[ \Delta p = \frac{I \cdot A \cdot t}{c} \]
where:
- \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \).

Substituting the values:
\[ \Delta p = \frac{(25 \times 10^4) \cdot (25 \times 10^{-4}) \cdot (2400)}{3 \times 10^8} \]

### Step 4: Simplify the Expression
1. Calculate the numerator:
\[ (25 \times 10^4) \cdot (25 \times 10^{-4}) = 625 \]
Then multiply by time:
\[ 625 \cdot 2400 = 1500000 \]

2. Now, substitute back into the momentum equation:
\[ \Delta p = \frac{1500000}{3 \times 10^8} \]

### Step 5: Final Calculation
\[ \Delta p = 5 \times 10^{-3} \, \text{Ns} \]

### Conclusion
The momentum transferred to the surface in 40 minutes is \( 5 \times 10^{-3} \, \text{Ns} \).

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