A plane electromagnetic wave of wave intensity `6 W//m^(2)` strikes a small mirror of area `40cm(2)`, held perpendicular to the approaching wave. The momentum transferred by the wave to the mirror each second will be

Options

A. `6.4 xx10^(-7) kg ms^(-1)`
B. `4.8 xx10^(-8) kg ms^(-1)`
C. `3.2 xx10^(-9) kg ms^(-1)`
D. `1.6 xx10^(-10) kg ms^(-1)`

Answer

To solve the problem of determining the momentum transferred by the electromagnetic wave to the mirror each second, we can follow these steps:

### Step 1: Understand the relationship between intensity and momentum
The momentum \( P \) transferred by an electromagnetic wave can be calculated using the formula:
\[ P = \frac{2 \cdot S \cdot A}{c} \]
where:
- \( S \) is the intensity of the wave (in W/m²),
- \( A \) is the area of the mirror (in m²),
- \( c \) is the speed of light (approximately \( 3 \times 10^8 \) m/s).

### Step 2: Convert the area from cm² to m²
The area of the mirror is given as \( 40 \, \text{cm}^2 \). We need to convert this to square meters:
\[ A = 40 \, \text{cm}^2 = 40 \times 10^{-4} \, \text{m}^2 = 0.004 \, \text{m}^2 \]

### Step 3: Substitute the values into the formula
Now, we can substitute the values into the momentum formula. The intensity \( S \) is given as \( 6 \, \text{W/m}^2 \), and the speed of light \( c \) is \( 3 \times 10^8 \, \text{m/s} \):
\[ P = \frac{2 \cdot 6 \, \text{W/m}^2 \cdot 0.004 \, \text{m}^2}{3 \times 10^8 \, \text{m/s}} \]

### Step 4: Calculate the numerator
Calculating the numerator:
\[ 2 \cdot 6 \cdot 0.004 = 0.048 \]

### Step 5: Calculate the momentum
Now substituting back into the equation:
\[ P = \frac{0.048}{3 \times 10^8} \]
\[ P = 1.6 \times 10^{-10} \, \text{kg m/s} \]

### Final Answer
Thus, the momentum transferred by the wave to the mirror each second is:
\[ P = 1.6 \times 10^{-10} \, \text{kg m/s} \]

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