If temperature of Sun=6000K, radius of Sun is `7.2xx10^(5)Km`, radius of earth=6000 Km & distance
between earth and Sun`=15xx10^(7)`Km. find intensity of light on earth.

Options

A. `38.4xx10^(16)`
B. `12.2xx10^(16)`
C. `18.3xx10^(16)`
D. `8.2xx10^(16)`

Explanation

To find the intensity of light on Earth from the Sun, we can use the formula for intensity (I) which is given by:

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

Where:
- \( P \) is the power emitted by the Sun.
- \( A \) is the area over which the power is distributed.

### Step 1: Calculate the Power emitted by the Sun (P)

The power emitted by the Sun can be calculated using the Stefan-Boltzmann Law:

\[ P = \sigma A_s T^4 \]

Where:
- \( \sigma \) is the Stefan-Boltzmann constant, approximately \( 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \).
- \( A_s \) is the surface area of the Sun, given by \( A_s = 4\pi R_s^2 \) where \( R_s \) is the radius of the Sun.
- \( T \) is the temperature of the Sun in Kelvin.

Given:
- Temperature of the Sun, \( T = 6000 \, K \)
- Radius of the Sun, \( R_s = 7.2 \times 10^5 \, km = 7.2 \times 10^8 \, m \) (convert km to m)

First, calculate the surface area of the Sun:

\[ A_s = 4\pi (7.2 \times 10^8)^2 \]

Now calculate \( P \):

\[ P = \sigma A_s T^4 \]

### Step 2: Calculate the area over which the power is distributed (A)

The area \( A \) at the distance of the Earth from the Sun can be calculated as:

\[ A = 4\pi d^2 \]

Where \( d \) is the distance from the Sun to the Earth.

Given:
- Distance from the Sun to the Earth, \( d = 15 \times 10^7 \, km = 1.5 \times 10^{12} \, m \) (convert km to m)

Now calculate the area:

\[ A = 4\pi (1.5 \times 10^{12})^2 \]

### Step 3: Calculate the intensity (I)

Now, substitute \( P \) and \( A \) into the intensity formula:

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

### Final Calculation

1. Calculate \( A_s \) and \( P \).
2. Calculate \( A \).
3. Substitute \( P \) and \( A \) into the intensity formula to find \( I \).