Lens Assembly Structural Analysis

Data Recording

In our lab experiments, we must measure and record a collimated light beam. This is done by shining the beam into a camera (the Casio Exilim EX-F1) and photographing the results. In order to properly interpret the information, it is necessary to understand the effects of the lenses on the image. This is incredibly difficult because of a number of factors. Most importantly, we have no way to directly measure the camera and acquire most of the information that we need. The lens assembly on the Exilim EX-F1 cannot be removed, so we cannot replace it with a custom- built lens assembly of known dimensions. All we can measure are a limited number of lengths, and we know the ranges of F-numbers and focal lengths. In order to solve this problem, a number of significant approximations and generalizations must be made. While the actual lens assembly contains twelve disparate lenses, our approximation has only four, and unlike the actual assembly, our approximation contains only perfect, idealized spherical lenses. These approximations would certainly affect our final results, but most likely not enough to severely skew our results.

Inside the Lens Assembly

This leads to a significant problem: How can we analyze the structure of something we cannot access? We contacted Casio, but the company was unwilling to divulge any detailed information about the lens assembly. Actually dismantling the lens assembly is out of the question unless a second camera could be acquired, and acquiring a second camera to dismantle would be prohibitively expensive. Instead, we chose to geometrically approximate the light paths through the lens apparatus and compare these results with data we were able to acquire. This would prove difficult, but it would also prove to be a generally successful approximation.

A Sliding Scale

By representing the lens assembly numerically, we would be able to calculate the effects of the lens experimentally. This would be done by taking photographs of the light beam and comparing the magnification of the final image with the size of the initial object. These values could be plugged into an Excel spreadsheet, and various values for the spacing of the lenses and their focal lengths could be entered appropriately. In addition, the few known and measurable values could be added as well.

First Iteration

The top and bottom light paths for an object (on the left), traveling through arbitrary lenses and resolving an image (on the right).

Calculating the height of an image. In this diagram, the image is formed (unfocused) before the focal point because of the lens spacing. Only one light path is shown. This math, while invalid, led to more accurate calculations in the third iteration.

With this known, we were able to create a basic spreadsheet. While many details of this sheet would later be proven wrong or simply be abandoned, the concepts are valid, and it is a sufficient starting point. To make the math easier, the lens asembly would be broken down into the four lenses, all of which would be approximated as thin lenses. For each lens, an image would be generated from the object, and that image would then be treated as the object for the next lens. This means, for instance, that the light rays traveling from the object would pass through the first lens and generate an image at some distance within the lens assembly. This image would then be treated as the object for the second lens, which would generate an image used as the object for the fourth lens, and so on. This was fairly simply done. For a thin lens,

${\frac {1}{D}}+{\frac {1}{L}}={\frac {1}{F}}$

which can be expressed as

$L={\frac {D*F}{D-F}}$

where D is the distance from the object to the lens, L is the distance from the lens to the object, and F is the focal length of the lens. This equation was used to generate each image. The magnification of each image can be calculated by

${\frac {M_{1}}{D}}={\frac {M_{2}}{L}}$

which can be expressed as

$M_{2}={\frac {S_{2}}{S_{1}}}M_{1}$

where $M_{1}$ is the size of the object and $M_{2}$ is the size of the image. By repeating these calculations for all lenses in the lens assembly, a final image magnification could be calculated. Unfortunately, there were problems with this approximation, namely that it uses the wrong type of light. This approximation assumes that the object is a physical thing, which would allow light rays to leave it from any angle, allowing the required light rays shown in the illustration to pass through the lens. However, the light used is collimnated laser light, meaning that all the light rays are parallel when they enter the lens assembly. Therefore, this approximation is invalid.

Second Iteration

Two light paths, from an object off the left side of the frame. The vertical lines are the lenses. Notice how the light rays cross through the focal point and continue through to the next lens.

If the light entering the lens assembly is colimnated and enters the , then the top and bottom beams (the only two beams needed to locate the image) will travel through the first lens parallel, and both will cross through the focal point. The size of the image will then be

${\frac {M_{1}}{f}}={\frac {M_{2}}{L-F}}$

which can be expressed as

$M_{2}={\frac {L-F}{F}}M_{1}$

Using the same logic as the first iteration, this magnified image can be treated as the object for the next lens. This approximation has the advantage of being very simple, mathematically, while also being more accurate for collimnated light. Unfortunately, that approximation cannot be used. Because the light is collimnated, we must instead follow the paths of the beams as they are deflected through the lenses rather than simply treat the image as a new object. Worse yet, the light paths will almost certainly not pass through the focal point; rather, they will pass through some point aligned with the focal point. The rays will all pass through one certain specific point on the focal plane. This mst be approximated, and the math must be rectified to correct for this.

Third Iteration

A ray entering the lens assembly at an angle such that it is normal to the lens. If we treat this ray is the topmost ray in a collimnated beam and its reflection around the center of the lenses (the horizontal line) as the lowermost ray, we can find the image height. Values are as explained in the text.

Still assuming that the collimnated light will enter the lens at a zero angle, any given light path will pass through the focal point of the first lens and will then pass through a particular point in the focal plane of the other lenses. This point can be calculated by tracing the trajectory of a theoretical beam of light passing through the center of the first lens at the same angle that the actual beams traced before entering the lens.

$\tan \theta ={\frac {m}{f_{1}}}$

From this and simple geometry, we find that

$\tan \theta ={\frac {c}{f_{2}}}$

and therefore

${\frac {m}{f_{1}}}={\frac {c}{f_{2}}}$

and therefore

$c={\frac {f_{2}}{f_{1}}}m$

where c is the horizontal displacement from the focal point on the focal plane.

From this, we can use a bit of linear regression to chart the size of the image.

$y={\frac {lf_{2}-s_{1}c_{1}+s_{1}l+cx-lx}{f_{2}}}$

By inputting

$x=s_{2}$

we can calculate the magnification of the image.

This approximation is correct, accurate, and valid. It works perfectly, except that it is only correct, accurate, and valid when the collimnated light enters the lens at a zero angle.

Fourth Iteration

The fourth iteration is elementary. The light is now allowed to enter the lens assembly at a selected angle $\theta$ which is input by the user. Since

$\tan {\theta }={\frac {c}{f_{2}}}$

the equation is very simple to appropriately modify to take this into account.

This equation forms the basis of the fifth iteration.

Fifth Iteration

The fifth iteration must generate a proper two-dimensional image through a three-dimensional camera. This model must allow for the light to enter the camera at any angle and offset. To simplify this, the two-dimensional model is used for each axis. The user will input an entry angle in both the X and Y directions, and he will input an entry offset in the same fashion.

We know logically that the image will not be compressed or stretched by passing through the lenses, as this cannot be seen in photography. First, therefore, we will track the center of the image as it passes through the lenses. This is done by calculating the offsets using the equation

$O=F_{p}*\tan {(\theta _{p})}+{(S_{p}-F_{p})}*\tan {(\theta )}$

where the entry angle is defined by

$\theta =\arctan {(\tan({\theta _{p}})-{\frac {O_{p}}{F}}})$

where $\theta$ is the entry angle at a given lens, $\theta _{p}$ is the entry angle at the previous lens, $O_{p}$ is the directional offset at the previous lens, and F is the focal length of the given lens.

Aperture

Math for the aperture has been tried. For the third iteration, an aperture can be calculated by finding the magnification of an image at the location of the aperture, and if it is larger than the aperture allows, the amount of image remaining can be calculated by simple ratio. An aperture would be more difficult to add for the fourth iteration, but it should be possible.

An aperture for the fourth iteration is being calculated and will be posted alongside the fourth iteration.

Future Improvements

• Add diagrams to this page
• Add Excel spreadsheets to this page, particularly for the third iteration
• Solve the fourth iteration
• Solve for the aperture in the fourth iteration