# Optics: Focal length of Koenig eyepiece

1. Mar 25, 2014

### xspook

1. The problem statement, all variables and given/known data

A Koenig eyepiece with a focal length of 100mm is constructed in the following way:
The first lens is made of 755276 glass which is 10 mm thick on axis. The radius of curvature of its left hand surface is -225mm, the radius of curvature of it right hand surface is 83.6 mm. The eyepiece is 130mm in diameter.

The second lens shares its left hand surface with the right hand first lens.
It has an axial thickness of 50 mm, and the radius of curvature of its right hand surface is (-102mm). The glass in the second lens is 623569.

The face of the third lens is located 0.7 mm from the right hand surface of the second lens. The gap is filled with air.

The third lens is 33.3 mm thick on axis. The radius of curvature of its left hand surface is 110 mm, the radius of curvature of its right hand surface is -458mm, and it is made of 607567 glass.

2. Relevant equations

$\frac{1}{f}$=($n_{1}$-1)[$\frac{1}{R_{1}}$-$\frac{1}{R_{2}}$+$\frac{(n_{1}-1)d_{1}}{n_{1}R_{1}R_{2}}$]

3. The attempt at a solution

It is stated the eyepiece has a focal length of 100mm, so I assumed that using the equation above and plugging in numbers that I would get the focal length given to us. However I am unable to obtain 100mm. I tried to obtain a focal length for each lens and add them together, but I think the first and second lens need a different equation (since it has glass on one side and air on the other).

2. Mar 25, 2014

### xspook

Attached is the system that I created from the given information

This is the third question on our homework, so the writing on it is from the first two questions.

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