A simple magnifying glass can focus sunlight enough to heat...

A simple magnifying glass can focus sunlight enough to heat up paper or dry grass and start a fire. A magnifying glass with a diameter of $4.0 \mathrm{~cm}$ has a focal length of $6.0 \mathrm{~cm} .$ (a) Using the information found on the inside back cover of the book, estimate the size of the image of the Sun when the magnifying glass focuses the image to its smallest size. (b) If the intensity of the Sun falling on the magnifying glass is $0.85 \mathrm{~kW} / \mathrm{m}^{2}$, what is the intensity of the image of the Sun?

Key Concepts

Image Formation by Convex Lenses

A convex lens takes parallel rays from a distant object (such as the Sun) and brings them together at its focal point, forming a real image. This principle, which is central to geometrical optics, shows that when an object is effectively at infinity, its image is formed at the focal plane of the lens, and the size of that image can be estimated using the lens’s focal length and the object’s angular size.

Angular Size of Celestial Objects

Celestial objects, like the Sun, subtend a small angle as seen from Earth. This angular size is key in determining the dimensions of the image produced by a lens, because when an object at infinity is imaged by a converging lens, the linear size of the image is given by the product of the object's angular size (in radians) and the focal length of the lens. This approximation is widely used in various branches of astronomy and optics.

Optical Intensity Concentration

When a lens focuses light, it concentrates the power from a larger aperture into a smaller image area. The intensity of light (or power per unit area) in the image increases relative to the original source by the ratio of the area of the collecting aperture (the lens) to the area of the resultant image. This concept is useful in understanding how simple optical devices can concentrate sunlight sufficiently to produce effects like heating or ignition.

Transcript

00:02 In this problem we have a circular glass that is used to focus the light from the sun.

00:12 So if these are sun rays that hit the glass, they emerge focusing on a smaller region.

00:27 So in the first part of the problem we want to compute what is the size of the image that is produced.

00:35 When it gets its smaller size.

00:42 So when this happens, this distance that we call q satisfies the length equation.

00:52 So we can say that for the class 1 over f is equal 1 over p plus 1 over q.

01:04 So from here we can get the value of q.

01:08 So first we're going to put 1 over q equal 1 over f minus 1 over p.

01:18 And from here we have that q is just f times p over p over p minus f.

01:29 So this relation here is useful because if we want to compute what is the final size of the image, we can use this relation.

01:40 That is related with a magnification.

01:46 So from here, we have the value of h prime, it's just minus q times h over p.

01:55 And we already have the value of q, that is this one.

01:59 So this is minus f, p divided by p minus f, and all of this divided by p also.

02:11 And we need an h here, this one.

02:15 So from here we can get the value of h prime.

02:21 So we just need to replace the values here.

02:25 So f is the focal length of the glass.

02:29 It has this value here, these six centimeters.

02:34 H is the initial object size.

02:40 So it's the size of the object.

02:42 In this case, the object is the sun.

02:45 So this is the diameter of the sun.

02:49 That is this value here...

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