Light with an energy. fiux of `20W/ (cm)^(2)` falls on a nonreflecting surface at normal incidence. If the surface has an area of `30cm^(2)`, the total momentum delivered (for complete absorption) during 30 min is `nxx 10^(-4) kg m s^(-1)`, then find `n`.

Answer

To solve the problem, we need to calculate the total momentum delivered to the non-reflecting surface when light falls on it. Here are the steps to find the value of \( n \):

### Step 1: Calculate the total power incident on the surface
The power \( P \) incident on the surface can be calculated using the formula:
\[ P = \text{Intensity} \times \text{Area} \]
Given:
- Intensity (energy flux) = \( 20 \, \text{W/cm}^2 \)
- Area = \( 30 \, \text{cm}^2 \)

First, convert the area from cm² to m²:
\[ 30 \, \text{cm}^2 = 30 \times 10^{-4} \, \text{m}^2 = 3 \times 10^{-3} \, \text{m}^2 \]

Now calculate the power:
\[ P = 20 \, \text{W/cm}^2 \times 30 \, \text{cm}^2 = 600 \, \text{W} \]

### Step 2: Calculate the total energy delivered in 30 minutes
Energy \( E \) can be calculated using the formula:
\[ E = P \times t \]
where \( t \) is the time in seconds. Convert 30 minutes to seconds:
\[ 30 \, \text{minutes} = 30 \times 60 = 1800 \, \text{seconds} \]

Now calculate the total energy:
\[ E = 600 \, \text{W} \times 1800 \, \text{s} = 1080000 \, \text{J} \]

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

Now calculate the momentum:
\[ p = \frac{1080000 \, \text{J}}{3 \times 10^8 \, \text{m/s}} = 3.6 \times 10^{-3} \, \text{kg m/s} \]

### Step 4: Express the momentum in the required form
The problem states that the momentum delivered is in the form \( n \times 10^{-4} \, \text{kg m/s} \). To find \( n \):
\[ 3.6 \times 10^{-3} \, \text{kg m/s} = 36 \times 10^{-4} \, \text{kg m/s} \]

Thus, \( n = 36 \).

### Final Answer
The value of \( n \) is \( 36 \).

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