Optical Engineering Science. Stephen Rolt

spherical refractive surface. If this surface is deemed to comprise ‘the optical system’ in its entirety, then one can use Eq. (1.14) to calculate the location of all Cardinal Points, expressed as a displacement, z along the optical axis. Positive z is to the right and the origin lies at the intersection of the optical axis and the surface. The Cardinal points are listed below. Cardinal points for a spherical refractive surface

Both Principal Points: z = 0
Both Nodal Points: z = R

      In this instance, the two focal lengths, f₁ and f₂ are different since the object and image spaces are in different media. If we take the first focal length as the distance from the first focal point to the first principal point, then the first focal length is positive. Similarly, the second focal length, the distance from the second principal point to the second focal point, is also positive. The principal points are both located at the surface vertex and the nodal points at the centre of curvature of the sphere. It is important to note that, in this instance, the principal and nodal points do not coincide. Again, this is because the refractive indices of object and image space differ.

      1.4.4 Refraction at Two Spherical Surfaces (Lenses)

Figure 1.14 shows a lens made up of two spherical surfaces, of radius, R₁ and R₂. Once again, the convention is that the spherical radius is positive if the centre of curvature lies to the right of the relevant vertex.

So, in the biconvex lens illustrated in Figure 1.14, the first surface has a positive radius of curvature and the second surface has a negative radius of curvature. The lens is made from a material of refractive index n₂ and is bounded by two surfaces with radius of curvature R₁ and R₂ respectively. It is immersed totally in a medium of refractive index, n₁ (e.g. air). In addition, it is assumed that the lens has negligible thickness (the thin lens approximation). Of course, as for the treatment of the single curved surface, we assume all angles are small and θ ∼ sinθ. First, we might calculate the angle of refraction, φ₁, produced by the first curved surface, R₁. This can be calculated using Eq. (1.14):

Figure 1.14 Refraction by two spherical surfaces (lens).

      Of course, the final angle, φ, can be calculated from φ₁ by another application of Eq. (1.14):

      Substituting for φ₁ we get:

(1.15)

As for Eq. (1.14) there are two parts to Eq. (1.15). First, there is an angular term that is equal to the incident angle. Second, there is a focusing contribution that produces a deflection proportional to ray height. Equation (1.15) allows the tracing of all rays in a system containing the single lens and it is straightforward to calculate the Cardinal points of the thin lens: Cardinal points for a thin lens

Both Principal Points: At centre of lens
Both Nodal Points: At centre of lens

      Since both object and image spaces are in the same media, then both focal lengths are equal and the principal and nodal points are co-located. One can take the above expressions for focal length and cast it in a more conventional form as a single focal length, f. This gives the so-called Lensmaker's Equation, where it is assumed that the surrounding medium (air) has a refractive index of one (i.e. n₁ = 1) and we substitute n for n₂.

      (1.16)

      1.4.5 Reflection by a Plane Surface

Figure 1.15 shows the process of reflection at a plane surface. As in the previous case of refraction, the reflected ray lies in the same plane as the incident ray and the angle of reflection is equal and opposite to the angle of incidence.

Figure 1.15 Reflection at a plane surface.

The virtual projected ray shown in Figure 1.15 illustrates an important point about reflection. If one considers the process as analogous to refraction, then a mirror behaves as a refractive material with an index of −1. This, in itself has an important consequence. The image produced is inverted in space. As such, there is no combination of positive magnification and pure rotation that will map the image onto the object. That is to say, a right handed object will be converted into a left handed image. More generally, if an optical system contains an odd number of reflective elements, the parity of the image will be reversed. So, for example, if a complex optical system were to contain nine reflective elements in the optical path, then the resultant