Coaxial lens system and its cardinal points | Optics | General theory of image formation

Coaxial lens system and its cardinal points

Let two lenses of focal lengths f₁ and f₂ are placed at a distance d from each other as shown in figure.

Coaxial lens system

If δ₁ and δ₂ are the deviations produced by lens L₁ and L₂, then
Total deviation produced by lens system

δ = δ₁ + δ₂...(1)

Equivalent focal length

The deviation produced by a thin lens, δ = h / f

∴ δ₁ = h₁ / f₁, δ₂ = h₂ / f₂and δ = h₁ / F ...(2)

Here F is the equivalent focal length of the lens system.
From equations (1) and (2), we get

$https://latex.codecogs.com/svg.image?\frac{h_{1}}{F}=\frac{h_{1}}{f_{1}}+\frac{h_{2}}{f_{2}}$

From figure

O₂C = O₂P - CP (∵ δ₁= CP / BP)

h₂ = h₁ - (BP) δ₁

h₂ = h₁ - d (h₁/ f₁) = h₁ (1 - d / f₁)

$https://latex.codecogs.com/svg.image?\frac{h_{1}}{F}=\frac{h_{1}}{f_{1}}+\frac{h_{1}}{f_{2}}\left ( 1-\frac{d}{f_{1}} \right )$

Here Δ is known as optical separation.
If P₁ and P₂ are the power of L₁ and L₂, and P is the power of lens system

P = P₁ + P₂ - d P₁P₂

Position of second focal point (O₂F₂ = β₂)

The real points from where the distances can be measured are O₁ and O₂
The distance of F₂ will be measured from O₂,

O₂F₂ = β₂

From ΔM₂H₂F₂ and ΔCO₂F₂

Position of second principal point (O₂H₂ = α₂)

The distance of second principal point is measured from the second optical centre O₂
From figure,

H₂O₂ = H₂F₂ - O₂F₂

α₂= F - β₂ (∵ H₂F₂ = F and O₂F₂ = β₂)

Since H₂ lies to the left of L₂

$https://latex.codecogs.com/svg.image?\therefore\alpha _{2}=-\frac{Fd}{f_{1}}$

Position of first principal point (O₁H₁ = α₁)

The distance of first principal point is measured from the first optical centre O₁

$https://latex.codecogs.com/svg.image?\alpha _{1}=\frac{Fd}{f_{2}}$

Position of first focal point (O₁F₁ = β₁)

The distance of first focal point can be measured from the first optical centre O₁
From figure

O₁F₁ = H₁F₁ - O₁H₁ (∵ H₁F₁ = F and O₁H₁ = α₁)

β₁ = F - α₁

$https://latex.codecogs.com/svg.image?\beta_{1}=F-\frac{Fd}{f_{2}} =F\left ( 1-\frac{d}{f_{2}} \right )$

Since F₁ lies to the left of L₁

$https://latex.codecogs.com/svg.image?\therefore \beta_{1}=-F\left ( 1-\frac{d}{f_{2}} \right )$
To know more about cardinal points of a lens system please click on the link https://youtu.be/_15xR0-N7ho or https://youtu.be/rq4Yoq9JiQc

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