Assume a ray starts in a medium with index of refraction n₁ at position (x₁, y₁, z₁) and propagate in this medium to a position (x₂, y₂, z₂).  If we use the paraxial approximation, then, for propagation in a single medium we have
x₂ = x₁ + (z₂ - z₁)θ_1x,   θ_2x = θ_1x,
y₂ = y₁ + (z₂ - z₁)θ_1y,   θ_2y = θ_1y.

Assume a ray crosses a spherical interface with radius of curvature R.  The ray crosses at a position with coordinates (x₁, y₁, z₁).  Assume that the medium in front of the interface has index of refraction n₁ and the medium behind the interface has index of refraction n₂.

If we use the paraxial approximation, then ths position (x₁, y₁, z₁) lies in the plane perpendicular to the z-axis containing the vertex, and for refraction at an interface for a ray going from a medium with n₁ into a medium with n₂ we have
n₂θ_2x = n₁θ_1x - (n₂ - n₁)x₁/R,  x₂ = x₁,
n₂θ_2y = n₁θ_1y - (n₂ - n₁)y₁/R,  y₂ = y₁.

The equations can be written in matrix form. 

Let us concentrate on a ray in the xz-plane.  We can then drop the index x on the angle, and we have for translation in a medium with index n

nθ₂ = nθ₁ + 0*x₁,
x₂ = [(z₂ - z₁)/n]nθ₁ + x₁,

or

.

For refraction at an interface we have

or

.

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

Tracing a ray across an interface

Assume a ray propagates in a medium with index of refraction n.  Let the optical axis be the z axis.  At any position z along this axis, a meridional ray is completely specified by giving its perpendicular distance from the axis and the angle θ it makes with the optical axis.  Let us orient our coordinate system so that the ray lies in the xz-plane.

Assume a ray originates at (x₁, z₁) in medium 1.  It makes an angle θ₁ with the z-axis.  We define the coordinates of the ray as

. 

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.)
P is called the power of the interface.

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.

Image formation

To form an image, all rays leaving the point P at (x₁, z₁) must arrive at the point P’ at (x₃’, z₃’), independent of θ₁.  This means that x₃’ must be independent of θ₁, or

.

This can be rearranged to yield

.
 

In terms of the object distance d_o and the image distance d_i from the interface we write
n₁/d_o + n₂/d_i = (n₂ - n₁)/R.

• If the image distance d_i is positive, then the image position lies behind the interface.  We have a real image.
• If the image distance d_i is negative, then the the image position lies in front of the interface.  We have a virtual image.
• The radius of curvature R is positive, if the center of curvature lies behind the interface.
• The radius of curvature R is negative, if the center of curvature lies in front of the interface.

A real image of an object produces the same pattern of light as the object does somewhere in space.  Some portion of the light from the real image reaches our detector along a straight-line path.  The detector cannot distinguish between light coming from the object and light coming from the image.  We interpret the same patterns in the same way.

A virtual image is the apparent position from which a pattern of light reaches our detector, if we make the assumption that it has traveled from its source to the detector along a straight-line path.  Virtual images are formed when light from an object or from an image is reflected or bend on the way to the detector.

Let us revisit

.

If z₁ goes to –infinity, we have

.

Similarly, if z₃ goes to infinity, we have

.

The distances f’ = n₂/P and f = n₁/P are called the image and object focal length, respectively.

Note: If n₂ < n₁, then f and f' are negative.

Lateral magnification

We define the lateral magnification M_x through the equation
x₃^' = M_xx₁.
This yields

.

Using from above

we obtain

.

In terms of object and image distance we have
M  = -(n₁d_i)/(n₂d_o).
If M is negative, the image is inverted.

AI Study Tip:

Example prompt:  'I am confused about the sign conventions for a single spherical interface. Create a summary table for me based on the 'Light Travels Left-to-Right' convention. Include columns for object distance, image distance, and radius of curvature. Then, give me a practice problem involving a virtual object behind a concave surface and walk me through the solution step-by-step to ensure I don't drop a negative sign.'