Question

The following matrix product is used in discussing two thin lenses in air: $$\mathbf{M}=\left(\begin{array}{cc}1 & -1 / f_{2} \\\0 & 1\end{array}\right)\left(\begin{array}{ll}1 & 0 \\\d & 1\end{array}\right)\left(\begin{array}{cc}1 & -1 / f_{1} \\\0 & 1\end{array}\right)$$ where \(f_{1}\) and \(f_{2}\) are the focal lengths of the lenses and \(d\) is the distance between them. As in Problem 9, element \(M_{12}\) is \(-1 / f\) where \(f\) is the focal length of the combination. Find \(\mathrm{M},\) det \(\mathrm{M},\) and \(1 / f\)

Short answer

The matrix \mathrm{ M} is \left(\begin{array}{cc} 1 & -\left(\frac{1}{f_{1}}+\frac{1}{f_{2}} \right) \ d & 1 -\frac{d}{f_{1}} -\frac{d}{f_{2}} \end{array}\right) .The determinant of \mathrm{ M} is 1. The reciprocal of the focal length is \left(\frac{1}{f_{1}}+\frac{1}{f_{2}}\right).

Step-by-step solution

- Multiply the first two matrices

First, multiply the matrices \(\left(\begin{array}{cc} 1 & 0 \ d & 1 \end{array}\right)\) and \(\left(\begin{array}{cc} 1 & -1 / f_{1} \ 0 & 1 \end{array}\right)\).Using matrix multiplication, we get:\[\left(\begin{array}{cc} 1 & 0 \ d & 1 \end{array}\right) \left(\begin{array}{cc} 1 & -1 / f_{1} \ 0 & 1 \end{array}\right)=\left(\begin{array}{cc} 1 \cdot 1 + 0 \cdot d & 1 \cdot-1 / f_{1} + 0 \cdot 1 \ d \cdot 1 + 1 \cdot 0 & d \cdot-1 / f_{1} + 1 \cdot 1 \end{array}\right)\]\[=\left(\begin{array}{cc} 1 & -1 / f_{1} \ d & 1- d / f_{1} \end{array}\right)\]

- Multiply the result with the remaining matrix

Now, take the result from Step 1 and multiply it with \(\left(\begin{array}{cc} 1 & -1 / f_{2} \ 0 & 1 \end{array}\right)\):\[\left(\begin{array}{cc} 1 & -1 / f_{1} \ d & 1- d / f_{1} \end{array}\right)\left(\begin{array}{cc} 1 & -1 / f_{2} \ 0 & 1 \end{array}\right)=\left(\begin{array}{cc} 1 \cdot 1 + -1 / f_{1} \cdot 0 & 1 \cdot-1 / f_{2} + -1 / f_{1} \cdot 1 \ d \cdot 1 + (1-d / f_{1}) \cdot 0 & d \cdot-1 / f_{2} + (1-d / f_{1}) \cdot 1 \end{array}\right)\]\[=\left(\begin{array}{cc} 1 & -\left(\frac{1}{f_{1}}+\frac{1}{f_{2}} \right) \ d & 1 -\frac{d}{f_{2}} -\frac{d}{f_{1}} \end{array}\right)\]

- Determine the matrix \mathrm{ M}

The matrix \mathrm{ M} after multiplication is \[\text{\(\mathrm{M}\)}=\left(\begin{array}{cc} 1 & -\left(\frac{1}{f_{1}}+\frac{1}{f_{2}} \right) \ d & 1 -\frac{d}{f_{1}} -\frac{d}{f_{2}} \end{array}\right)\]

- Compute the determinant of \mathrm{ M}

The determinant of \mathrm{ M} is computed as follows:\[\text{det(\(\mathrm{M}\))} = \left|\begin{array}{cc} 1 & -\left(\frac{1}{f_{1}}+\frac{1}{f_{2}} \right) \ d & 1 -\frac{d}{f_{2}} -\frac{d}{f_{1}} \end{array}\right| = 1 (1 - \frac{d}{f_{2}} -\frac{d}{f_{1}}) - (- \left(\frac{1}{f_{1}}+\frac{1}{f_{2}} \right) \cdot d)\]\[= 1 - \frac{d}{f_{1}} -\frac{d}{f_{2}} + d \left(\frac{1}{f_{1}}+\frac{1}{f_{2}} \right)\]\[=1\]

- Find \frac{1}{f}

According to the problem, \text{M_{12}} = \frac{-1}{f} .Therefore, \[\left(\begin{array}{cc} 1 & -\left(\frac{1}{f_{1}}+\frac{1}{f_{2}} \right) \ d & 1 - \frac{d}{f_{1}} - \frac{d}{f_{2}} \end{array}\right) \text{M_{12}} = -\left(\frac{1}{f_{1}}+\frac{1}{f_{2}}\right)\]Thus,\[\frac{1}{f}=\left(\frac{1}{f_{1}}+\frac{1}{f_{2}}\right)\]

Key concepts

Matrix Multiplication

Matrix multiplication is an essential concept in linear algebra.
Matrices are arrays of numbers arranged in rows and columns.
When multiplying two matrices, the resulting matrix is formed by taking the dot product of rows and columns.
Consider two matrices, A and B. To find the element in the first row and the first column of the resulting matrix C, you multiply each element in the first row of A by the corresponding element in the first column of B and then add these products.
This process can be expressed mathematically as: \[C_{ij} = \text{sum}( A_{ik} \times B_{kj} )\] where i and j represent the row and column indices respectively.
For our specific problem with lenses, we need to multiply three 2x2 matrices.
This combination allows us to understand the behavior of light passing through optical systems.
It simplifies complex problems into manageable computations.

Optical Lenses

Optical lenses play a crucial role in physics. They bend (refract) light to form images, focusing on objects at various distances.
Thin lenses, the simplest type, follow the lens maker's equation: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] where
\( f \) is the focal length,
\( d_o \) is the object distance, and
\( d_i \) is the image distance.
Different focal lengths lead to different converging or diverging behaviors.
In the matrix provided, \( f_1 \) and \( f_2 \) are the focal lengths of two lenses.
The matrix multiplication captures how these lenses interact with each other.

Determinants in Matrices

Determinants provide information about the properties of a matrix.
For a 2x2 matrix \[ \text{M} = \begin{pmatrix} a & b \ c & d \end{pmatrix}, \]the determinant is calculated as:
\[ \text{det(M)} = ad - bc. \]Determinants are useful for understanding linear transformations.
In our lens system, the determinant tells us if the transformation preserves area.
The determinant of our matrix is 1, indicating that it does.
This is important in optics and other applications of linear algebra.

Focal Length

The focal length \( f \) of a lens is the distance from the lens where parallel light rays converge to a point.
It is a fundamental property in lens design.
Combining lenses changes the overall focal length of the system.
In our problem, this combined focal length is represented as \[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}. \]This suggests that the system's focal length is affected by both lenses' focal lengths.

Linear Algebra Applications

Linear algebra is widely used in many fields including physics, engineering, and computer science.
It provides tools to describe and analyze linear systems.
In optics, matrices model how lenses affect light waves.
This simplifies designing optical systems like microscopes and telescopes.
Understanding these applications helps in solving real-world problems efficiently.
The use of matrices provides a structured way of representing and computing these transformations.
This has broad implications beyond physics, affecting various technological advancements.