A boy is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length `10cm`. The diameter of the sun is `1.39xx10^(9)m` and its mean distance from the earth is `1.5xx10^(11)m`. What is the diameter of the sun's image on the paper ?

Options

A. `9.2xx10^(-4)m`
B. `6.5xx10^(-4)m`
C. `6.5xx10^(-5)m`
D. `12.4xx10^(-4)m`

Explanation

To find the diameter of the sun's image on the paper, we can use the concept of similar triangles and the lens formula. Here’s the step-by-step solution:

### Step 1: Understand the Geometry
The sun can be approximated as a point source of light when considering its distance from the Earth. The rays of sunlight that reach the lens are essentially parallel due to the vast distance.

### Step 2: Use the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( f \) is the focal length of the lens (10 cm or 0.1 m),
- \( v \) is the image distance from the lens,
- \( u \) is the object distance (distance from the sun to the lens).

Since the sun is very far away compared to the focal length of the lens, we can approximate \( u \) as infinity. Thus, \( \frac{1}{u} \) approaches 0, leading to:
\[ \frac{1}{f} = \frac{1}{v} \implies v = f = 0.1 \text{ m} \]

### Step 3: Calculate the Angular Size of the Sun
The angular size \( \theta \) of the sun can be calculated using the formula:
\[ \theta = \frac{D}{d} \]
Where:
- \( D \) is the diameter of the sun (1.39 x \( 10^9 \) m),
- \( d \) is the distance from the sun to the Earth (1.5 x \( 10^{11} \) m).

Substituting the values:
\[ \theta = \frac{1.39 \times 10^9}{1.5 \times 10^{11}} \approx 0.00927 \text{ radians} \]

### Step 4: Calculate the Diameter of the Image
The diameter of the image \( d_i \) formed on the paper can be calculated using the formula:
\[ d_i = v \cdot \theta \]
Substituting the values:
\[ d_i = 0.1 \cdot 0.00927 \approx 0.000927 \text{ m} = 0.927 \text{ mm} \]

### Final Answer
The diameter of the sun's image on the paper is approximately **0.927 mm**.

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