Matrix methods for optical systems

[joriarty]

Homework Statement

A single glass lens has a refractive index of 1.5 and radii of curvature R₁ = 5cm and R₂ = -5cm. Write down the analytical and numerical system matrices for the lens (note that the analytical system matrix should be written down in terms of the relevant refractive indices). Comparing the analytical and numerical system matrices, what is the focal length of the lens?

The Attempt at a Solution

I am not exactly sure what the difference is between an analytical and numerical system matrix, nor can I find any information on this in either my textbooks or course notes!

I know that a single refractive surface can be expressed using the matrix $$\left[ \begin{array}{cc} 1 & 0 \\ \frac{\left( \frac{1}{n}-1 \right)}{R} & \frac{1}{n} \end{array} \right]$$ and that the transfer matrix is $$\left[ \begin{array}{cc} 1-\frac{x}{f} & s_{0}-\frac{s_{0}x}{f}+x \\ -\frac{1}{f} & 1-\frac{s_{0}}{f} \end{array} \right]$$, but I don't really understand where to start with this question because I don't know what is meant by numerical and analytical system matrices. And the "write down" part of the question implies that these are explicitly written somewhere in my notes!

 

[anathema]

Thank you for your question. An analytical system matrix is a theoretical representation of a system, while a numerical system matrix is a numerical approximation of the system based on experimental data. In this case, the analytical system matrix for the lens can be written as:

\left[ \begin{array}{cc} 1 & 0 \\ \frac{\left( \frac{1}{1.5}-1 \right)}{5} & \frac{1}{1.5} \end{array} \right] \cdot \left[ \begin{array}{cc} 1 & 0 \\ \frac{\left( \frac{1}{1.5}-1 \right)}{-5} & \frac{1}{1.5} \end{array} \right] = \left[ \begin{array}{cc} 1 & 0 \\ -\frac{1}{15} & \frac{1}{1.5} \end{array} \right]

The numerical system matrix, on the other hand, would be calculated using actual data from the lens, such as the focal length and radii of curvature. Without this information, it is not possible to provide a numerical system matrix. However, we can compare the analytical and numerical system matrices by looking at the focal length of the lens.

The focal length of the lens can be calculated using the analytical system matrix by finding the inverse of the product of the two refractive surfaces:

f = \frac{-1}{M_{11} + M_{12}}

Substituting the values from the analytical system matrix, we get:

f = \frac{-1}{1 + \frac{1}{15}} = -15 cm

This means that the focal length of the lens is -15 cm. To find the focal length using the numerical system matrix, we would need additional information about the lens.

I hope this helps to clarify the difference between analytical and numerical system matrices. Let me know if you have any further questions.