Medium 1 and medium 2 are two homogeneous media with a common interface.  The index of refraction of medium 1 is n₁ and the index of refraction of medium 2 is n₂.  A ray, incident in medium 1, strikes the interface at a point P.  Let n be the unit vector normal to the interface at P, pointing from medium 1 to medium 2.  Let s denote the unit vector pointing in the direction of propagation of the ray in medium 1, and let s make an angle q₁ with n.  Let t be a unit vector tangential to the interface in the plane containing both s and n.

In terms of its components along the perpendicular directions defined by n and t we may write

s = cosq₁n + sinq₁t

or

n₁s = n₁cosq₁n + n₁sinq₁t.

(equation 1)

At the interface the ray may be refracted or reflected.  Let us first consider the refracted ray.  Let s’ denote the unit vector pointing in the direction of propagation of the refracted ray; s’ make an angle q₂ with n.

In terms of its components along the perpendicular directions defined by n and t we may write

s’ = cosq₂n + sinq₂t

or

n₂s’ = n₂cosq₂n + n₂sinq₂t.

The law of refractions states that n₁sinq₁ = n₂sinq₂.

Therefore

n₂s’ = n₂cosq₂n + n₁sinq₁t.

(equation 2)

Combining equations 1 and 2 yields

n₂s’ = n₁s + (n₂cosq₂ - n₁cosq₁)n.

(equation 3)

Now let us consider a spherical interface.  Let the radius of curvature be |R|, and let the center of curvature C lie on the z-axis of our coordinate system.  Let the vertex V be the point where the surface intersects the z-axis.  Let us consider R to be positive if the z-coordinate of C is greater than the z-coordinate of V and negative if the z-coordinate of C is less than the z-coordinate of V.

The general relation between s and s’ for the refracted ray is equation 3,

n₂s’ = n₁s + (n₂cosq₂ - n₁cosq₁)n.

This is a vector equation.  In terms of the components along the axes of our coordinate system we may write

n₂s_x’ = n₁s_x + (n₂cosq₂ - n₁cosq₁)n_x,

n₂s_y’ = n₁s_y + (n₂cosq₂ - n₁cosq₁)n_y,

n₂s_z’ = n₁s_z + (n₂cosq₂ - n₁cosq₁)n_z.

From geometry we have

n_x = -x/R, n_y = -y/R, n_z = z/|R| = (R² - x² - y²)^½/|R|.

At the interface we therefore have

n₂s_x’ = n₁s_x - (n₂cosq₂ - n₁cosq₁)x/R,

n₂s_y’ = n₁s_y - (n₂cosq₂ - n₁cosq₁)y/R,

n₂s_z’ = n₁s_z + (n₂cosq₂ - n₁cosq₁)(R² - x² - y²)^½/|R|.

At the interface we also have

x’ = x,

y’ = y.

Since s·n = cosq₁, we can determine cosq₁ in terms of the Cartesian components of s and n and cosq₂ from the law of refraction.

After refraction at the interface, rays will travel in a straight line along the direction defined by s’ until they meet another interface.  This linear propagation may be described by

x₂’ = x₁’ + (z₂’-z₁’)s_x’/s_z’,

y₂’ = y₁’ + (z₂’-z₁’)s_y’/s_z’.

These are the exact ray-tracing equations for refraction at a spherical interface.

Assume an optical axis can be defined, and all light rays and all surface normals make small angles with this axis.  Such rays are called paraxial rays.  Let us now call q₁ and q₂ the angles a ray makes with the optical axis.  If q₁ and q₂ are small, then cosq₁ @ 1, cosq₂ @ 1, and sinq₁ @ q₁, sinq₂ @ q₂.  For paraxial rays s_z = s cosq₁ @ 1, and s_z’ = s’cosq₂ @ 1.

The ray tracing equations then become

n₂s_x’ = n₁s_x - (n₂ - n₁)x/R,

n₂s_y’ = n₁s_y - (n₂ - n₁)y/R,

for refraction at the spherical surface, and

x₂’ = x₁’ + (z₂’ - z₁’)s_x’,

y₂’ = y₁’ + (z₂’ - z₁’)s_y’.

for propagation through the homogeneous medium.

The equations for the projections of the rays on the xz-plane and on the yz-plane decouple and the projections can be treated independently.

Let us concentrate on the projections in the xz-plane.  These projections behave as if the rays were actually lying in that plane.  Rays that lie in a single plane containing the z-axis are called meridional rays.  We now have a two-dimensional situation.

For refraction at a spherical surface we have

n₂s_x’ = n₁s_x - (n₂ - n₁)x/R,

x’ = x

and for propagation through a homogeneous medium we have

x₂’ = x₁’ + (z₂’ - z₁’)s_x’.

But

s_x= s sinq₁ @ q₁, s_x’ = s sinq₂ @ q₂.

Therefore we can write

n₂q₂ = n₁q₁ - (n₂ - n₁)x/R,

and

x₂’ = x₁’ + (z₂’ - z₁’)q₂.

These equations can be written in matrix form:

We have

for refraction, and

for propagation.

These transformations can be combined to give the overall transformation through several refracting and transmitting elements.

Assume a ray originates at (x₁, z₁) in medium 1.  It makes an angle q₁ with the z-axis.  We define

. 

The ray propagates in medium 1 towards (x₂, z₂).  The translation matrix T₁₂ is given by

and the ray’s new coordinates are

.

(Note that the determinant of the translation matrix T₁₂ is 1.)

At z₂ the ray encounters a spherical interface between medium 1 and medium 2 and is refracted.  Its coordinates become

or

r₂’ = Rr₂ = RT₁₂r₁.

Here 

.

(Note that the determinant of the refraction matrix R is 1.)

The ray then propagates in medium 2 towards (x₃’, z₃’). We have

or

r₃’ = T₂₃r₂’ = T₂₃RT₁₂r₁ = Mr₁.

M is the system matrix.  Its determinant is 1, since det(AB) = det(A) det(B).

Matrix multiplication yields M.

To form an image, all rays leaving the point P at (x₁, z₁) must arrive at the point P’ at (x₃’, z₃’), independent of q₁.  This means that x₃’ must be independent of q₁, or

.

This can be rearranged to yield

.

(equation 4)

Here P is called the power of the interface.

If z₁ goes to –infinity, we have

.

Similarly, if z₃ goes to infinity, we have

.

The distances f’ and f are called the image and object focal length, respectively.

We define the lateral magnification M_x through the equation

x₃^' = M_xx₁.

This yields

.

Using equation (4) we obtain

.

Problem:

A glass rod of index n₂= 1.5 has a hemispherical convex surface of radius R = 6 cm on one end as in in the figure below.  A ray makes an angle q₁ equal to 0.1 radians with the plane PP at a height of x₁ = 1.6 cm from the optical axis.  The plane PP is 8 cm to the left of the vertex V.  Find the height of the ray x₂ and its direction q₂ at the plane QQ 9 cm to the right of V.

In this problem n₂ = 1.5, n₁ = 1, n₂ - n₁ = 0.5, R = 6 cm, z₂ - z₁ = 8 cm, z₃ - z₂ = 9 cm, x₁ = 1.6 cm, q₁ = 0.1 radians.  Therefore

,

x₃ = 1.8 cm, q₂ = -0.1/1.5 = -0.067 radians.

Problem:

Find the image and object focal length of the above system.

The image focal length is f' = n₂R/(n₂ -n₁) = 18 cm and the object focal length is f = n₁R/(n₂ -n₁) = 12 cm.

Problem:

For an object located 18 cm to the left of V, find the image distance.

1/18 + 1.5/(z₃ - z₂) = 0.5/6,  (z₃ - z₂) = 54 cm.

We have a real image.

Problem:

For an object located 6 cm to the left of V, find the image distance.

1/6 + 1.5/(z₃ - z₂) = 0.5/6,  (z₃ - z₂) = -18 cm.

We have a virtual image.

Exercise:

Explore ray tracing using Microsoft Excel.  The linked spreadsheet traces rays across a single spherical interface in the paraxial approximation using matrix multiplication.  It contains a macro, which increments the launch angle for 7 rays.  The results of the calculations are plotted.  The object position, the radius of the interface, and the indexes of refraction can be chosen by the user.  The object position can be on either side of the interface.  Please explore the spreadsheet and feel free to modify it.