Lens Ray Calculation

Text Box:

Figure 1: Ray trace showing the incident ray, the refracted rays at both surfaces, as well as the normals to the surfaces where the rays are refracted. The angles that are needed for the refraction calculations are labeled as described below.

In the labels, the first letter t means that it is referring to an angle as in the greek letter theta.

The second letter, L or R, refers to the Left or Right surface.

The third letter can be either r or n. r refers to a refracted angle with respect to the surface normal and n refers to the angle of the surface normal with respect to the x axis. This labeling is needed in order to easily see what are the incident angles and the refracted angles and the final ray angle with respect to the x axis.

Calculations

A double convex lens that is centered on the x axis is defined by the left and right radii of curvature, r_Land r_R. For convenience we will define the signs of these curvatures so that the x_L(y)<0 and x_R(y) >0. If the radius of the lens is r_Lens then the centers of these curves are at:

$\begin{array}{l} x_{L c} = \sqrt{r_{L}^{2} - r_{L e n s}^{2}} \\ x_{R c} = - \sqrt{r_{R}^{2} - r_{L e n s}^{2}} \end{array}$

(1.1)

The x location of any point y<r_Lens on the lens is:

$\begin{array}{l} x_{L} (y) = x_{L c} - \sqrt{r_{L}^{2} - y^{2}} \\ x_{R} (y) = x_{R c} + \sqrt{r_{R}^{2} - y^{2}} \end{array}$

(1.2)

Note that when y=r_Lens, x=0 as expected for these definitions. Also note that when y<>0, x_L is less than zero while x_Ris greater than zero.

If an incident ray that is parallel to the x axis hits the left surface is at $y_{L i}$ then its corresponding x coordinate is at

_{$x_{L i} = x_{L c} - \sqrt{r_{L}^{2} - y_{L i}^{2}}$}

_(1.3)

Also the angle between the surface normal at y_Li and the ray is

$θ_{L i} = \sin^{- 1} (\frac{y_{L i}}{r_{L}})$

(1.4)

Let the index of refraction of the lens be n.

Snell's refraction law is:

$n \sin θ_{L r} = \sin θ_{L i}$

(1.5)

where $θ_{L r}$ is the angle between the surface normal and the refracted ray.

Then, using equations (1.4) and (1.5), the angle between the normal and the refracted ray is

$θ_{L r} = \sin^{- 1} (\frac{y_{L i}}{n r_{L}})$

(1.6)

The angle of the refracted ray with respect to the x axis is

$θ_{L r x} = θ_{L r} - θ_{L i}$

(1.7)

which is a negative angle when y>0 and positive when y<0 so that the refracted ray slopes toward the x axis.

The slope of he refracted ray is then

$m_{r} = \tan θ_{L r x}$

(1.8)

The y coordinate of the refracted ray in the lens can be written as:

$y_{r} (x) = y_{L i} + m_{r} (x - x_{L i})$

(1.9)

In order to compute the intercept of y_r with the right hand surface of the lens we can re-write equation (1.9) using equation (1.2) and obtain

$y_{R r} = y_{L i} + m_{r} (x_{R c} + \sqrt{r_{R}^{2} - y_{R r}^{2}} - x_{L i})$

(1.10)

where y_Rr is the y intercept of y_r(x) with the right surface of the lens and it to be solved for.

We can greatly simplify solving equation (1.10) by separating out the constants by letting

$c = y_{L i} - m_{r} (x_{R c} + x_{L i})$

(1.11)

and then equation (1.10) becomes:

$y_{R r} - c = m_{r} \sqrt{r_{R}^{2} - y_{R r}^{2}}$

(1.12)

We may conveniently square both side of equation (1.12) and solve for y_Rr and obtain:

$y_{R r} = \frac{c \pm m_{r} \sqrt{R_{R}^{2} (1 + m_{r}^{2}) - c^{2}}}{1 + m_{r}^{2}}$

(1.13)

Having obtained the y intercept using the + sign before the radical, we can use equation (1.2) to obtain the x intercept:

$x_{R r} = x_{R c} + \sqrt{r_{R}^{2} - y_{R r}^{2}}$

(1.14)

Finally we may use Snell's law for the refraction of the ray as it exits the right hand surface. The refracted ray's incidence angle with respect to that normal is:

$θ_{R r} = \sin^{- 1} (\frac{y_{R r}}{r_{R}}) + θ_{L r x}$

(1.15)

Since

$n \sin θ_{R r} = \sin θ_{e x i t}$

(1.16)

we have the refracted angle:

$θ_{e x i t} = \sin^{- 1} (n \sin θ_{R r})$

(1.17)

With respect to the x axis this angle is:

$θ_{e x i t X} = θ_{e x i t} - \sin^{- 1} (\frac{y_{R r}}{r_{R}})$

(1.18)

Finally the slope of the this ray is:

$m_{e x i t} = \tan θ_{e x i t X}$

(1.19)

The y coordinate of the exiting ray can be expressed by:

$y_{e x i t} = y_{R r} + m_{e x i t} (x - x_{R r})$

(1.20)

Solving equation (1.20) for x at y_exit=0, the ray will intercept the x axis at location:

$x_{int e r c e p t} = x_{R r} - \frac{y_{R r}}{m_{e x i t}}$

(1.21)