# Taha Azzaoui - 2018.07.14

### Introduction

The principal aim of the field of ray optics is to study the behavior of light and describe how it interacts with its surroundings. It is important to note that in the context of ray optics, we assume the ray model of light. That is, we avoid discussing electromagnetic radiation as a whole and treat light as a ray in the mathematical sense, i.e. a line that starts at some point and extends to infinity in some defined direction. One of the most fundamental equations under this model is Snell’s law, which defines the relationship between a ray’s angle of incidence and its angle of refraction with respect to the material it inhabits. Suppose for example, a ray of light is to penetrate the surface of a body of water. We define θ1 to be the ray’s angle of incidence, (i.e. the angle the ray makes with the norm of the surface upon entrance) and θ2 to be the angle of refraction (i.e. the angle made with the norm of the surface upon exit). We then define the refractive index of a material (denoted n) to be a measure of how light propagates through that material). Intuitively, we would expect the refractive index of air to be smaller than that of a more translucent material like glass. Snell’s law states that the relationship between these two quantities is as follows
$$n_1sin(\theta_1) = n_2sin(\theta_2) \tag{1}$$
Where θ1 and θ2 are the angles of incidence and refraction respectively, and n1 and n2 are the refractive indices of the medium of origin and destination respectively.

### The Thin Lens Equation

An interesting phenomenon arises when light interacts with a converging lens. Lenses such as the one in Figure 2, can produce an inverted image of an object under certain conditions. To dive deeper, we must introduce the concept focal length. The focal length of a lens, is the distance from the lens to the point where light converges onto the surface, the focal point. Simply put, the focal length is a measure of how “zoomed in” a lens is. As it turns out, there is a relationship between the distance between an object and a lens, the distance between that object’s image and the lens, and the lens’s focal point. Using some simple geometry, we can derive this relationship as follows. Turning our attention now to Figure 2, we have an object AB placed beyond the point C, of which an inverted image, AB forms beyond the point G. From similar triangles ΔABG and ΔEDG, we have
$$\frac{A'B'}{ED} = \frac{GA'}{EG} \tag{2}$$
Note that since AB = ED, we also have
$$\frac{A'B'}{AB} = \frac{GA'}{EG} \tag{3}$$
Furthermore, from similar triangles ΔABE and ΔABE, we have
$$\frac{A'B'}{AB} = \frac{EA'}{EA} \tag{4}$$
Combining (3) and (4), we have
$$\frac{GA'}{EG} = \frac{EA'}{EA} \tag{5}$$
We can express GA as EA′ − EG
$$\frac{GA'}{EG} = \frac{EA'}{EA} \tag{6}$$
We can then express these lengths in terms of distance and focal length
$$\frac{d_1 - f}{f} = \frac{d_1}{d_0} \tag{7}$$
Finally, after cross-multiplying and dividing both sides by d0d1f, we arrive at the final expression
$$\frac{1}{f} = \frac{1}{d_1} + \frac{1}{d_0} \tag{8}$$