# Using Tube Lenses with Infinity Corrected Objectives

This is Section 9.2 of the Imaging Resource Guide.

In order to create an image with an infinity-corrected objective, a tube lens must be used to focus the image. One advantage to using an infinity-corrected objective with a tube lens is that there can be a space between the objective and tube lens. The space allows additional optical components to be inserted into the system, such as optical filters or beamsplitters. The distance between an infinity-corrected objective and the tube lens (L) can be varied from the recommended or optimal, but this will affect the image field diameter ($\varnothing$). Equations 1 and 2 are approximation formulas to determine the relation between ($\varnothing$) and L.

(1)$$\varnothing_1 =2 \times f_1 \times \text{NA}$$
(1)
$$\varnothing_1 =2 \times f_1 \times \text{NA}$$

where $\small{\varnothing _1}$ is exit pupil of the objective, $\small{f_1}$ is focal length of the objective, and NA is the numerical aperture of the objective.

(2)$$L = \frac{ \left( \varnothing_2 - \varnothing_1 \right) \times f_2 }{\varnothing}$$
(2)
$$L = \frac{ \left( \varnothing_2 - \varnothing_1 \right) \times f_2 }{\varnothing}$$

Where L is the distance between the objective and the tube lens, $\small{\varnothing _2}$ is entrance pupil of the tube lens, $\small{f_2}$ is focal length of the tube lens, and $\small{\varnothing }$ is the image field diameter.

## Application Example:

Using an M Plan APO 10X objective (#46-144), MT-1 tube lens (#54-774), and a 2/3” sensor camera, what is the maximum spacing between the tube lens and objective without vignetting? The focal length of the objective ($\small{f_1}$) is 20mm and NA is 0.28, so the exit pupil diameter can be calculated:

(3)$$\varnothing_1 = 2 \times 20 \text{mm} \times 0.28 = 11.2 \text{mm}$$
(3)
$$\varnothing_1 = 2 \times 20 \text{mm} \times 0.28 = 11.2 \text{mm}$$

A 2/3” image sensor features an 11mm diagonal, therefore $\varnothing$ needs to be at least 11mm. The focal length of the MT-1 tube lens is 200mm and the entrance pupil diameter is 24mm. Therefore,

(4)$$L = \frac{ \left( 24 \text{mm} - 11.2 \text{mm} \right) \times 200 \text{mm} }{11 \text{mm}} = 232.7 \text{mm}$$
(4)
$$L = \frac{ \left( 24 \text{mm} - 11.2 \text{mm} \right) \times 200 \text{mm} }{11 \text{mm}} = 232.7 \text{mm}$$

As long as the spacing between the tube lens and objective is less than 232.7mm, there will be no vignetting.

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