Light with an energy flux of `18W//cm^(2)` falls on a non-reflecting surface at normal incidence. The pressure exerted on the surface is

Options

A. `2N//m^(2)`
B. `2xx10^(-4)N//m^(2)`
C. `6N//m^(2)`
D. `6xx10^(-4)N//m^(2)`

Explanation

To find the pressure exerted on a non-reflecting surface by light, we can use the formula:

\[ P = \frac{I}{c} \]

where:
- \( P \) is the pressure,
- \( I \) is the intensity of the light (energy flux),
- \( c \) is the speed of light in vacuum.

### Step 1: Identify the given values
- The intensity \( I \) is given as \( 18 \, \text{W/cm}^2 \).
- The speed of light \( c \) is approximately \( 3 \times 10^8 \, \text{m/s} \).

### Step 2: Convert the intensity from \( \text{W/cm}^2 \) to \( \text{W/m}^2 \)
To convert \( 18 \, \text{W/cm}^2 \) to \( \text{W/m}^2 \), we use the conversion factor:
\[ 1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2 \]
Thus,
\[ I = 18 \, \text{W/cm}^2 = 18 \times 10^4 \, \text{W/m}^2 = 1800000 \, \text{W/m}^2 \]

### Step 3: Substitute the values into the pressure formula
Now, we can substitute the values of \( I \) and \( c \) into the pressure formula:
\[ P = \frac{I}{c} = \frac{18 \times 10^4 \, \text{W/m}^2}{3 \times 10^8 \, \text{m/s}} \]

### Step 4: Calculate the pressure
Now we perform the calculation:
\[ P = \frac{18 \times 10^4}{3 \times 10^8} = \frac{18}{3} \times \frac{10^4}{10^8} = 6 \times 10^{-4} \, \text{N/m}^2 \]

### Final Answer
The pressure exerted on the surface is:
\[ P = 6 \times 10^{-4} \, \text{N/m}^2 \]