(a) Draw rays to form the image of a vertical object on the optical axis and farther than the focal point from a converging lens. (b) Use plane geometry in your figure and prove that the magnification \(m\) is given by $$m=\frac{h_{\mathrm{i}}}{h_{\mathrm{o}}}=-\frac{d_{\mathrm{i}}}{d_{\mathrm{o}}}$$

A converging lens focuses parallel rays of light to a single point, called the focal point. When an object is placed farther than the focal point, we can use three principal rays to locate the image: 1. A ray parallel to the optical axis, which will pass through the focal point after the lens. 2. A ray passing through the center of the lens, which will be undeviated. 3. A ray passing through the focal point before the lens, which will become parallel to the optical axis after the lens.

Place the object to the left of the lens, farther than the focal point. Let the height of the object, \(h_\text{o}\), be positive above the optical axis. Draw the three principal rays mentioned in Step 1 to find the image's position and orientation: 1. Draw a ray parallel to the optical axis from the object's top; it will pass through the right-side focal point after the lens. 2. Draw a ray from the top of the object through the center of the lens; it will continue undeviated. 3. Draw a ray from the top of the object through the left-side focal point; it will become parallel to the optical axis after the lens. The intersection of these rays will give the position of the top of the image. The image will appear inverted compared to the object and lie on the opposite side of the lens.

Using the ray diagram, we can form two sets of similar triangles in the planes of the object and the image: - Triangle A: formed by the height of the object \(h_\text{o}\), the distance of the object from the lens \(d_\text{o}\), and the distance between the object and the left-side focal point. - Triangle B: formed by the height of the image \(h_\text{i}\), the distance of the image from the lens \(d_\text{i}\), and the distance between the image and the right-side focal point. Notice that the triangles are similar by the AA (angle-angle) similarity criterion, as they share a common angle at the center of the lens, and their adjacent angles are vertically opposite. Using the corresponding side ratios of these similar triangles, we have: \(\frac{h_\text{i}}{h_\text{o}} = \frac{d_\text{i} - f}{d_\text{o} - f}\) Next, let's rearrange this equation and factor out the focal length term (f), giving: \(h_\text{i} (d_\text{o} - f) = h_\text{o} (d_\text{i} - f)\) Now, since the object and image heights have opposite signs (due to the image being inverted), we can put a minus sign on one side of the equation: \(h_\text{i} (d_\text{o} - f) = -h_\text{o} (d_\text{i} - f)\) Finally, we can cancel out the focal length terms (f) and get the desired magnification formula: \(m = \frac{h_\text{i}}{h_\text{o}} = -\frac{d_\text{i}}{d_\text{o}}\) This derivation proves that the magnification of an object placed on the optical axis and farther than the focal point of a converging lens is given by \(m = \frac{h_\text{i}}{h_\text{o}} = -\frac{d_\text{i}}{d_\text{o}}\).