A beam of light of intensity 12 W/`cm^(2)` in incident on a totally reflecting plane mirror of area 1.5 `cm^(2)` . The force in Newton's acting on the mirror will be

Options

A. `2.4xx10^(6)`
B. `1.2xx10^(-9) `
C. `3.6xx10^(-8)`
D. `5.6xx10^(-5)`

Answer

To find the force acting on the mirror due to the incident beam of light, we can follow these steps:

### Step 1: Understand the relationship between intensity, area, and energy
The intensity \( I \) of the light is given as \( 12 \, \text{W/cm}^2 \) and the area \( A \) of the mirror is \( 1.5 \, \text{cm}^2 \). The energy \( E \) incident on the mirror can be calculated using the formula:
\[ E = I \times A \]

### Step 2: Calculate the energy incident on the mirror
Substituting the values:
\[ E = 12 \, \text{W/cm}^2 \times 1.5 \, \text{cm}^2 = 18 \, \text{W} \]
Since \( 1 \, \text{W} = 1 \, \text{J/s} \), we can say that \( 18 \, \text{W} \) means \( 18 \, \text{J} \) of energy is incident on the mirror per second.

### Step 3: Calculate the momentum of the light
The momentum \( p \) associated with the energy can be found using the relationship:
\[ p = \frac{E}{c} \]
where \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \).

Substituting the values:
\[ p = \frac{18 \, \text{J}}{3 \times 10^8 \, \text{m/s}} = 6 \times 10^{-8} \, \text{kg m/s} \]

### Step 4: Calculate the change in momentum
Since the light is totally reflected, the change in momentum \( \Delta p \) is twice the momentum:
\[ \Delta p = 2p = 2 \times 6 \times 10^{-8} \, \text{kg m/s} = 1.2 \times 10^{-7} \, \text{kg m/s} \]

### Step 5: Calculate the force acting on the mirror
The force \( F \) can be calculated using the formula:
\[ F = \frac{\Delta p}{t} \]
Assuming \( t = 1 \, \text{s} \):
\[ F = \Delta p = 1.2 \times 10^{-7} \, \text{N} \]

### Final Answer
The force acting on the mirror is:
\[ F = 1.2 \times 10^{-7} \, \text{N} \]

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