Light with an enargy flux of `25xx10^(4) Wm^(-2)` falls on a perfectly reflecting surface at normal incidence. If the surface area is `15 cm^(2)`, the average force exerted on the surface is

Options

A. `1.25xx10^(-6) N`
B. `2.50xx10^(-6) N`
C. `1.20xx10^(-6) N`
D. `3.0xx10^(-6) N`

Explanation

To find the average force exerted on a perfectly reflecting surface by light, we can follow these steps:

### Step 1: Understand the relationship between energy flux and momentum change
The average force exerted on the surface can be calculated using the formula:
\[ F = \frac{\Delta p}{\Delta t} \]
where \( \Delta p \) is the change in momentum and \( \Delta t \) is the time interval.

### Step 2: Relate momentum change to energy flux
For light reflecting off a surface, the change in momentum \( \Delta p \) can be expressed as:
\[ \Delta p = \frac{2I}{c} \times A \]
where:
- \( I \) is the energy flux (intensity) of the light,
- \( c \) is the speed of light (approximately \( 3 \times 10^8 \, \text{m/s} \)),
- \( A \) is the area of the surface.

### Step 3: Convert the area from cm² to m²
Given the surface area \( A = 15 \, \text{cm}^2 \):
\[ A = 15 \, \text{cm}^2 = 15 \times 10^{-4} \, \text{m}^2 \]

### Step 4: Substitute values into the momentum change formula
Substituting the values into the momentum change formula:
- Energy flux \( I = 25 \times 10^4 \, \text{W/m}^2 \)
- Area \( A = 15 \times 10^{-4} \, \text{m}^2 \)
- Speed of light \( c = 3 \times 10^8 \, \text{m/s} \)

The change in momentum becomes:
\[ \Delta p = \frac{2 \times (25 \times 10^4)}{3 \times 10^8} \times (15 \times 10^{-4}) \]

### Step 5: Calculate the average force
Now, we can calculate the average force:
\[ F = \frac{\Delta p}{\Delta t} = \Delta p \text{ (for } \Delta t = 1 \text{ second)} \]
Thus,
\[ F = \frac{2 \times (25 \times 10^4) \times (15 \times 10^{-4})}{3 \times 10^8} \]

### Step 6: Simplify the expression
Calculating the above expression:
\[ F = \frac{2 \times 25 \times 15 \times 10^4 \times 10^{-4}}{3 \times 10^8} \]
\[ = \frac{750}{3 \times 10^8} \]
\[ = 250 \times 10^{-8} \]
\[ = 2.5 \times 10^{-6} \, \text{N} \]

### Final Answer
The average force exerted on the surface is:
\[ F = 2.5 \times 10^{-6} \, \text{N} \]

---