# Opening angle

The opening angle of a converging lens or objective (generally of a focal optical system ) is the angle that a point on the optical axis forms with the diameter of the entrance or exit pupil . ${\ displaystyle \ omega}$

Specifically, a distinction is made between the object-side and the image-side opening angle

• object-side opening angle
The object-side opening angle is defined as the angle between the object point on the optical axis and the diameter of the entrance pupil .${\ displaystyle \ omega _ {\ mathrm {O}}}$
${\ displaystyle \ omega _ {\ mathrm {O}} = 2 \ cdot \ arctan \ left ({\ frac {\ varnothing _ {\ text {entrance pupil}}} {2 \ cdot {\ text {distance}} _ { \ text {Entry pupil to object point}}}} \ right)}$
• opening angle on the image side
Analogously, the opening angle on the image side is the angle between the image point on the optical axis and the diameter of the exit pupil .${\ displaystyle \ omega _ {\ mathrm {B}}}$
${\ displaystyle \ omega _ {\ mathrm {B}} = 2 \ cdot \ arctan \ left ({\ frac {\ varnothing _ {\ text {exit pupil}}} {2 \ cdot {\ text {distance}} _ { \ text {Exit pupil to image point}}}} \ right)}$

The aperture angle on the image side determines the size of the circles of confusion and thus influences the imaging depth .

## Opening angle of a converging lens

Object- side opening angle and image-side opening angle .
The object distance and the image distance are indicated.${\ displaystyle \ omega _ {\ mathrm {O}}}$${\ displaystyle \ omega _ {\ mathrm {B}}}$ ${\ displaystyle g}$ ${\ displaystyle b}$

In the case of photo lenses, the positions of the entrance and exit pupils generally no longer correspond to the position of the main planes , and their diameter is usually variable. In contrast to lenses that contain additional apertures and are usually made up of several lenses, the entrance and exit pupils of a converging lens coincide in the center of the lens - their diameter simply corresponds to the lens diameter. The distance between the object point and the entrance pupil thus corresponds to the object distance and the distance between the image point and the exit pupil corresponds to the image distance . This simplifies the situation in such a way that the opening angles for a converging lens can be easily calculated: ${\ displaystyle g}$ ${\ displaystyle b}$

${\ displaystyle \ omega _ {\ mathrm {O}} = 2 \ cdot \ arctan \ left ({\ frac {\ varnothing _ {\ text {lens}}} {2 \ cdot {\ text {object distance}}}} \ right)}$

for the object-side opening angle and

${\ displaystyle \ omega _ {\ mathrm {B}} = 2 \ cdot \ arctan \ left ({\ frac {\ varnothing _ {\ text {lens}}} {2 \ cdot {\ text {image width}}}} \ right)}$

for the opening angle on the image side

### Large object range

A further simplification arises if one considers the opening angle for very large object widths (“ infinite ”). If the object range grows beyond all limits , then becomes

${\ displaystyle \ omega _ {\ mathrm {O}} = 0 \ ,.}$

According to the lens equation , the image distance for an infinitely large object distance corresponds to the focal length f , i.e. becomes

${\ displaystyle \ omega _ {\ mathrm {B}} = 2 \ cdot \ arctan \ left ({\ frac {\ varnothing _ {\ text {lens}}} {2 \ cdot f}} \ right) \ ,. }$

## Focal ratio and f-number

For large object widths g , the aperture ratio 1 / k results from the aperture angle on the image side and the focal length f of an optical system${\ displaystyle \ displaystyle \ omega _ {\ mathrm {B}}}$

${\ displaystyle {\ frac {1} {k}} = 2 \ cdot \ tan \ left ({\ frac {\ omega _ {\ mathrm {B}}} {2}} \ right) \,}$

The f-number is the reciprocal of the aperture ratio, so
${\ displaystyle k = {\ frac {1} {2 \ cdot \ tan \ left ({\ frac {\ omega _ {\ mathrm {B}}} {2}} \ right)}}}$

## Numerical aperture

The numerical aperture results from the object-side opening angle and the refractive index n of the medium : ${\ displaystyle \ omega _ {\ mathrm {O}}}$ ${\ displaystyle A_ {N}}$

${\ displaystyle A_ {N} = n \ cdot \ sin \ left ({\ frac {\ omega _ {\ mathrm {O}}} {2}} \ right) \,}$

The refractive index n of air is around 1 (1,000292 near the ground).