Файл: Ersoy O.K. Diffraction, Fourier optics, and imaging (Wiley, 2006)(ISBN 0471238163)(427s) PEo .pdf

Figure 8.2. Minimum path between A and B after reflection from a mirror.

In a homogeneous medium, n is constant. Then, the path of minimum optical path length between two points is the straight line between these two points. This is often stated as ‘‘rays travel in straight lines.’’

Fermat’s principle can be used to prove what happens when a ray is reflected from a mirror, or is reflected and refracted at the boundary between two optical media, as illustrated in the following examples.

EXAMPLE 8.1 Show that when a ray reflects from a mirror, the angle of reflection should be equal to the angle of incidence in a medium of constant n.

Solution: The geometry involved is shown in Figure 8.2. y1 and y2 are the angles of incidence and of reflection, respectively. The path between A and C after reflection from the mirror should be a minimum.

The path length is given by

Lp ¼ nðAB þ BCÞ

which can be written as

Lp ¼ nð½h21 þ ðd1 d2Þ2&1=2 þ ½h22 þ d22&1=2Þ

Lp depends on d2 as the independent variable. Setting dLp=dd2 ¼ 0 yields

sinðy1Þ ¼ sinðy2Þ

or

y1 ¼ y2

116 GEOMETRICAL OPTICS

(b) When the rays travel close to the optical axis, sin y1 y1 and sin y2 y2 so that

n1y1 n1y2

EXAMPLE 8.3 A thin lens has the property shown in Figure 8.3.

where y is the distance from the optical axis, and f is the focal length of the lens given by

n is the index of refraction of the lens material, R1 and R2 are the radii of curvature of the two surfaces of the lens.

Using the information given above, show the lens law of image formation. Solution: Consider Figure 8.4(b). As a consequence of Eq. (8.2-7), and paraxial approximation such that sin y y, the ray starting at P1 and parallel to the optical axis passes through the focal point F since y1 equals zero; the ray starting at P1 and passing through the center point of the lens on the optical axis goes through the lens without change in direction since y equals 0. The two rays intersect at P2, which determines the image.

(a)

P1

F

P2

f

z1 z2

(b)

Figure 8.4. (a) Ray bending by a thin lens, (b) Image formation by a thin lens.

Figure 8.5. The trajectory of a ray in a GRIN medium.

8.3THE RAY EQUATIONS

In a graded-index (GRIN) material, the refractive index is a function n(r) of position r. In such a medium, the rays follow curvilinear paths in accordance with Fermat’s principle.

The trajectory of a ray can be represented in terms of the coordinate functions x(s), y(s) and z(s), s being the length of the trajectory. This is shown in Figure 8.5.

The trajectory is such that Fermat’s principle is satisfied along the trajectory between the points A and B:

r being the position vector. Using calculus of variations, Eq. (8.3-1) can be converted to the ray equation given by

where r is the gradient operator. The ray equation is derived rigorously in Section 8.4 by using the Eikonal equation.

The ray equation given above is expressed in vector form. In Cartesian coordinates, the component equations can be written as

Solution of Eq. (8.3-2) is not trivial. It can be considerably simplified by using the paraxial approximation in which the rays are assumed to be traveling close to the optical axis so that ds dz. Equation (8.3-2) can then be written in terms of two