The pressure exerted by electromagnetic wave of intensity I on a non - reflecting surface is ( where c= velocity of light ):

Options

A. IC
B. `Ic^2`
C. `(I )/(c )`
D. `(I )/(c^2)`

Explanation

To find the pressure exerted by an electromagnetic wave of intensity \( I \) on a non-reflecting surface, we can follow these steps:

### Step 1: Understand the relationship between intensity, power, and area
The intensity \( I \) of an electromagnetic wave is defined as the power \( P \) per unit area \( A \):
\[ I = \frac{P}{A} \]

### Step 2: Rearranging the formula for power
From the definition of intensity, we can express power as:
\[ P = I \cdot A \]

### Step 3: Relate power to work done
Power can also be defined in terms of work done \( W \) over time \( T \):
\[ P = \frac{W}{T} \]

### Step 4: Express work done in terms of force and displacement
The work done can be expressed as the product of force \( F \) and displacement \( d \):
\[ W = F \cdot d \]
Thus, we can write:
\[ P = \frac{F \cdot d}{T} \]

### Step 5: Relate displacement to velocity
Displacement over time gives us velocity \( v \):
\[ \frac{d}{T} = v \]
For electromagnetic waves, this velocity is the speed of light \( c \):
\[ P = F \cdot c \]

### Step 6: Equate the two expressions for power
Now, we have two expressions for power:
1. \( P = I \cdot A \)
2. \( P = F \cdot c \)

Setting these equal gives:
\[ I \cdot A = F \cdot c \]

### Step 7: Solve for force per unit area
Dividing both sides by area \( A \):
\[ I = \frac{F \cdot c}{A} \]
This implies:
\[ \frac{F}{A} = \frac{I}{c} \]

### Step 8: Define pressure
The pressure \( P \) is defined as the force per unit area:
\[ P = \frac{F}{A} \]

### Step 9: Substitute to find pressure
From the previous step, we can substitute to find the pressure:
\[ P = \frac{I}{c} \]

### Final Answer
Thus, the pressure exerted by the electromagnetic wave on a non-reflecting surface is:
\[ P = \frac{I}{c} \]

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