A man is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length `10 cm`. The diameter of the sun is `1.39 xx 10^(9) m` and the diameter of the sun's image on the paper is

Options

A. `3.1 xx 10^(-4) cm`
B. `6.5 xx 10^(-5) cm`
C. `6.5 xx 10^(-4) m`
D. `9.2 xx 10^(-4) m`

Explanation

To find the diameter of the Sun's image on the paper when focused by an equiconvex lens of focal length 10 cm, we can follow these steps:

### Step 1: Understand the setup
The Sun is very far away, so the rays coming from it can be considered parallel. When these parallel rays pass through the lens, they will converge at the focal point of the lens.

### Step 2: Identify the given values
- Focal length of the lens (f) = 10 cm = 0.1 m
- Diameter of the Sun (H₀) = 1.39 x 10^9 m
- Distance from the Earth to the Sun (U) = 1.5 x 10^11 m (this is the object distance)

### Step 3: Use the lens formula
Since the rays from the Sun are parallel, we can use the lens formula to find the image distance (V):
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Here, since the object is very far away, we can approximate:
\[ \frac{1}{u} \approx 0 \]
Thus, the formula simplifies to:
\[ \frac{1}{v} = \frac{1}{f} \]
So, \( v = f = 0.1 \) m.

### Step 4: Calculate the magnification
The magnification (M) of the lens can be calculated using the formula:
\[ M = \frac{h_i}{h_o} = \frac{v}{u} \]
Where:
- \( h_i \) = height of the image (or diameter of the image)
- \( h_o \) = height of the object (or diameter of the Sun)

### Step 5: Substitute the values into the magnification formula
We can rearrange the magnification formula to find \( h_i \):
\[ h_i = M \cdot h_o \]
First, we calculate the magnification:
\[ M = \frac{v}{u} = \frac{0.1}{1.5 \times 10^{11}} \]

### Step 6: Calculate the height of the image
Now substituting the values:
\[ h_i = \left(\frac{0.1}{1.5 \times 10^{11}}\right) \cdot (1.39 \times 10^9) \]

### Step 7: Perform the calculation
Calculating \( h_i \):
\[ h_i = \frac{0.1 \times 1.39 \times 10^9}{1.5 \times 10^{11}} \]
\[ h_i = \frac{1.39 \times 10^8}{1.5} \]
\[ h_i \approx 9.27 \times 10^{-4} \text{ m} \]

### Step 8: Convert to centimeters
To express the diameter in centimeters:
\[ h_i \approx 9.27 \times 10^{-4} \text{ m} = 0.0927 \text{ cm} \]

### Final Answer
The diameter of the Sun's image on the paper is approximately **0.0927 cm**.

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