91Ó°ÊÓ

Angular diameter is a measure of how large an object appears to an observer, not how large it is physically. This concept is crucial in understanding how the sun is perceived from Earth. For the sun, its angular diameter can be calculated using the formula:

• \( \theta = \frac{D}{d} \)
• Where \( D \) is the actual diameter of the sun, and \( d \) is the distance from the sun to the observer (in this case, Earth).

Given that the sun's diameter is \( 1.39 \times 10^9 \text{ m} \) and the average distance from Earth to the sun is \( 1.50 \times 10^{11} \text{ m} \), the angular diameter \( \theta \) can be calculated as: \[\theta = \frac{1.39 \times 10^9}{1.50 \times 10^{11}} \approx 9.27 \times 10^{-3} \text{ radians}.\] This small angle demonstrates that while the sun is massive, it doesn't appear very large in the sky due to its vast distance from us.

In optics, image formation involves using lenses to project an image of an object onto another surface. A converging lens, like the one in our scenario, focuses parallel rays of light, such as sunlight, to a focal point. The size of the image formed by such a lens is dependent on several factors:

• The focal length of the lens \( f \), which is the distance at which parallel rays converge.
• The angular diameter \( \theta \) of the light source as seen from the observer's position.

To calculate the diameter \( D' \) of the sun's image on the paper, we use:\[D' = f \cdot \theta\]Substituting the given focal length of 10 cm (or 0.1 m) and the previously calculated \( \theta \approx 9.27 \times 10^{-3} \) radians, we find:\[D' = 0.1 \times 9.27 \times 10^{-3} \approx 9.27 \times 10^{-4} \text{ m}.\]This means the sun's image will be about 0.000927 meters, or 0.927 millimeters, in diameter, forming a tiny spot on the paper.

Intensity in optics refers to the power of light per unit area it strikes. It's an important measure, especially when focusing sunlight to start a fire. To find the intensity \( I \), we use the relation:

• \( I = \frac{P}{A} \)
• Where \( P \) is the power of the light source (here, 0.530 W) passing through the lens.
• \( A \) is the area of the image the light is focused onto.

From our previous calculations, the area \( A \) of the sun's image is:\[A = \pi \left(\frac{D'}{2}\right)^2 \approx 6.75 \times 10^{-7} \text{ m}^2.\]By substituting the power and the calculated area into the intensity formula, we get:\[I = \frac{0.530}{6.75 \times 10^{-7}} \approx 7.85 \times 10^{5} \text{ W/m}^2.\]This very high intensity explains why focusing sunlight with a lens can generate enough heat to ignite paper.