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`s
D. `3.0xx10^(-6)N`

Answer

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

### Step 1: Understand the given values
- Energy flux (intensity of light) \( I = 25 \times 10^4 \, \text{W/m}^2 \)
- Surface area \( A = 15 \, \text{cm}^2 \)

### Step 2: Convert surface area to square meters
Since the area is given in square centimeters, we need to convert it to square meters:
\[ A = 15 \, \text{cm}^2 = 15 \times 10^{-4} \, \text{m}^2 \]

### Step 3: Use the formula for average force
The average force \( F \) exerted on a perfectly reflecting surface by light is given by the formula:
\[ F = \frac{2 \times I \times A}{c} \]
where \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \).

### Step 4: Substitute the values into the formula
Now, we can substitute the values into the formula:
\[ F = \frac{2 \times (25 \times 10^4) \times (15 \times 10^{-4})}{3 \times 10^8} \]

### Step 5: Calculate the numerator
Calculating the numerator:
\[ 2 \times (25 \times 10^4) \times (15 \times 10^{-4}) = 2 \times 25 \times 15 \times 10^{4-4} = 750 \]

### Step 6: Calculate the force
Now substituting back into the force equation:
\[ F = \frac{750}{3 \times 10^8} = 2.5 \times 10^{-6} \, \text{N} \]

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