Problem 56

  A camper is trying to start a fire by focusing sunlight onto a piece of paper. The diameter of the sun is 1.39 x 10⁹ m and its mean distance from the earth is 1.50 x 10¹¹ m. The camper is using a converging lens whose focal length is 10.0 cm. (a) What is the area of the sun's image on the paper? (b) If 0.530 W of sunlight pass through the lens, what is the intensity of the sunlight at the paper?

SOLUTION:
(a) The sun is really, really far away (compared to the focal length of the lens), so using the lens equation with d_o going to $\infty$, we see that d_i is pretty much the same as the focal length. The magnification is, then,

$$m = -\frac{d_i}{d_o} = -\frac{0.1 m}{1.5 \times 10^{11} m} = -6.67 \times 10^{-13}$$

Therefore, the image height is

$$h_i = mh_o = (-6.67 \times 10^{-13}) (1.39 \times 10^9 m) = -9.27 \times 10^{-4} m .$$

The diameter of the sun's image on the paper is h_i. The area of the image is, therefore,

$$A = \pi \left ( \frac{9.27 \times 10^{-4} m}{2} \right )^2$$

$$A = 6.74 \times 10^{-7} m^2$$

(b) The intensity of the light is the power that strikes the paper perpendicularly divided by the illuminated area A

$$I = \frac{P}{A} = \frac{0.530 W}{6.74 \times 10^{-7} m^2}$$

$$I = 7.86 \times 10^5 W/m^2$$