Problem 109 - Physics

Question

A camper is trying to start a fire by focusing sunlight onto a piece of paper. The diameter of the sun is \(1.39 \times 10^{9} \mathrm{~m},\) and its mean distance from the earth is \(1.50 \times 10^{11} \mathrm{~m} .\) The camper is using a converging lens whose focal length is \(10.0 \mathrm{~cm}\) (a) What is the area of the sun's image on the paper? (b) If \(0.530 \mathrm{~W}\) of sunlight pass through the lens, what is the intensity of the sunlight at the paper?

Step-by-Step Solution

Answer

(a) Area is \(6.75 \times 10^{-7} \text{ m}^2\). (b) Intensity is \(7.85 \times 10^5 \text{ W/m}^2\).

Step 1: Understand the problem

We need to find the area of the sun's image formed by the lens on the paper and the intensity of sunlight at that point. The sun is treated as an object, and the lens forms its image. We'll also calculate the light intensity on the paper.

Step 2: Calculate the magnification of the lens

The magnification formula for a lens is given by \[ \text{Magnification} (m) = \frac{-f}{d_o} \]where \( f \) is the focal length of the lens and \( d_o \) is the distance of the object (sun) from the lens. Substitute the given focal length in meters: \[ f = 10.0 \text{ cm} = 0.10 \text{ m}, \, d_o = 1.50 \times 10^{11} \text{ m} \].Thus, \[ m = \frac{-0.10}{1.50 \times 10^{11}} \approx -6.67 \times 10^{-13} \].

Step 3: Calculate the diameter of the sun's image

The diameter of the image (\( d_i \)) is found using the magnification:\[ d_i = |m| \times d_{\text{sun}} \],where \( d_{\text{sun}} = 1.39 \times 10^9 \text{ m} \). So, \( d_i = 6.67 \times 10^{-13} \times 1.39 \times 10^9 \approx 9.27 \times 10^{-4} \text{ m} \).

Step 4: Calculate the area of the sun's image

The area \( A \) of the circular image is given by the formula:\[ A = \pi \left( \frac{d_i}{2} \right)^2 \].Substitute \( d_i \approx 9.27 \times 10^{-4} \text{ m} \):\[ A = \pi \left( \frac{9.27 \times 10^{-4}}{2} \right)^2 \approx 6.75 \times 10^{-7} \text{ m}^2 \].

Step 5: Calculate the intensity of sunlight on the paper

Intensity \( I \) is power per unit area:\[ I = \frac{P}{A} \],where \( P = 0.530 \text{ W} \) is the power through the lens. Using the area calculated:\[ I = \frac{0.530}{6.75 \times 10^{-7}} \approx 7.85 \times 10^5 \text{ W/m}^2 \].

Key Concepts

Lens Formula

The lens formula is a fundamental concept in geometric optics that helps us to determine the relationship between the object distance (\( d_o \)), the image distance (\( d_i \)), and the focal length (\( f \)) of the lens. It is usually represented as:

• \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]

This formula is incredibly useful, especially when dealing with a converging lens like in our problem about lighting a fire with sunlight. Using the focal length and the object's position, one can calculate where the image will form. However, in exercises focusing on celestial objects like the sun, the distance is often treated as infinitely far away, simplifying some calculations.
The important takeaway is that the lens brings together light rays from the object and focuses them at a certain point, forming an image. Understanding this formula helps us explore phenomena such as magnification and image formation.

Magnification

Magnification in the context of lenses describes how much larger or smaller the image is compared to the object. It is an important concept for understanding how the size of an object changes when seen through a lens. Magnification (\( m \)) is defined by the ratio of the image distance (\( d_i \)) to the object distance (\( d_o \)) but can also be expressed as:

• \[ m = \frac{h_i}{h_o} \]
• \( h_i \) is the height of the image, and \( h_o \) is the height of the object.

In our sun and lens problem, we used

• \[ m = \frac{-f}{d_o} \]

to find out how the sun's size changes as a focused image. The value can be negative, indicating that the image is inverted relative to the object. Although the magnitude seems incredibly tiny (like in our example), it highlights how a small image focus crucial in intense light concentration occurs.

Intensity

The concept of intensity refers to the concentration of energy delivered over an area. It is a key aspect in understanding the effects of light on surfaces exposed to a lens. The formula for intensity (\( I \)) is:

• \[ I = \frac{P}{A} \]
• where \( P \) is the power of the light source, and \( A \) is the area over which this power is distributed.

This definition, applied in our exercise, helps see how the power of sunlight passing through a lens gets concentrated into a smaller spot leading to higher intensity. With a high intensity, you can potentially ignite materials like paper as it focuses the light energy more densely compared to when the light is dispersed over a broader area. Understanding intensity is crucial for applications that rely on focused heat or energy, such as solar panels or fire starting.

Light Concentration

Light concentration involves gathering light rays to meet at a point, increasing intensity and energy density in that area. This principle is at the core of many optical applications, from magnifying glasses to solar energy systems. In our camper's scenario, the lens serves to concentrate the generally dispersed sunlight into a small, intense dot on paper.
Light concentration is vital for achieving high temperatures or energy densities without increasing input power. It makes use of the geometry of lenses or mirrors to affect where and how the light is delivered. Here, the effective use of the lens focal length and positioning allows for this concentration, essentially redistributing light energy available over a much smaller area, making materials like paper heat up quickly enough to ignite.