Thick lens and its cardinal points

Thick lens

The combination of two spherical surfaces, is known as thick lens, if the distance between their poles can not be neglected in comparison to their radii.
It can be treated as the combination of thin lenses.
The thickness of these lenses are comparable to their focal lengths.

Cardinal points of thick lens

Let µ = refractive index of material of lens
t = thickness of lens
R₁ and R₂ = Radii of curvature of first and second curved surface of the given thick lens.
F₁ and F₂ = First and second focal points respectively.
H₁ and H₂ = First and second principal points respectively.
f₁ and f₂ = First and second focal length of thick lens respectively.

Cardinal points of thick lens

Equivalent focal length

If there is same medium on either side of thick lens
H₁F₁ = f₁ = – f , and H₂F₂ = f₂ = f

Refraction from surface PP₁

$\frac{\mu _{2}}{v}-\frac{\mu _{1}}{u}=\frac{\mu _{2}-\mu _{1}}{R}$

∵ u = ∞, v = P₁I, μ₁ = 1, μ₂ = μ and R = R₁

$\therefore \;\;\;\;\;\;\frac{\mu}{P_{1}I}-\frac{1}{\infty }=\frac{\mu-1}{R_{1}}$

$or \;\;\;\;\;\;\frac{1}{P_{1}I}=\frac{\mu-1}{\mu R_{1}}$

Refraction from surface QP₂

This time I works as an object and forms final image is obtained at F₂

$\because\;\;\;\;\;\frac{\mu _{2}}{v}-\frac{\mu _{1}}{u}=\frac{\mu _{2}-\mu _{1}}{R}$

Here u = P₂I, v = P₂F₂, μ₂ = 1, μ₁ = μ and R = R₂

$\therefore \;\;\;\;\;\;\frac{1}{P_{2}F_{2}}-\frac{\mu }{P_{2}I}=\frac{1-\mu}{R_{2}}$

From ΔM₂H₂F₂ and ΔCP₂F₂

$\frac{H_{2}F_{2}}{P_{2}F_{2}}=\frac{M_{2}H_{2}}{CP_{2}}$

$or\;\;\;\;\;\frac{H_{2}F_{2}}{P_{2}F_{2}}=\frac{PP_{1}}{CP_{2}}$

From ΔPP₁I and ΔCP₂I

$\frac{P_{1}I}{P_{2}I}=\frac{PP_{1}}{CP_{2}}$

From above two equations we get

$\frac{H_{2}F_{2}}{P_{2}F_{2}}=\frac{P_{1}I}{P_{2}I}$

$or\;\;\;\;\;\frac{1}{H_{2}F_{2}}=\frac{P_{2}I}{P_{1}I}\times \frac{1}{P_{2}F_{2}$

$or\;\;\;\;\;\frac{1}{f}=\frac{1}{P_{1}I}\times \frac{P_{2}I}{P_{2}F_{2}}\;\;\;\;\;\left ( \because H_{2}F_{2}=f \right )$

$\because \;\;\;\;\;\;\frac{1}{P_{2}F_{2}}-\frac{\mu }{P_{2}I}=\frac{1-\mu}{R_{2}}$

$\therefore \;\;\;\;\;\frac{P_{2}I}{P_{2}F_{2}}-\mu =\left ( \frac{1-\mu}{R_{2}} \right )P_{2}I$

$or \;\;\;\;\;\frac{P_{2}I}{P_{2}F_{2}} =\mu +\left ( \frac{1-\mu}{R_{2}} \right )P_{2}I$

$But\;\;\;\;\;\frac{1}{f}=\frac{1}{P_{1}I}\times \frac{P_{2}I}{P_{2}F_{2}}$

$\therefore \;\;\;\;\;\frac{1}{f} =\frac{1}{P_{1}I}\left [ \mu +\left ( \frac{1-\mu}{R_{2}} \right )P_{2}I \right ]$

∵ P₂I = P₁I − P₁P₂ = P₁I − t

$\therefore \;\;\;\;\;\frac{1}{f} =\frac{1}{P_{1}I}\left [\frac{(1-\mu)}{R_{2}}(P_{1}I-t)+ \mu \right ]$

$or \;\;\;\;\;\frac{1}{f} =\frac{(1-\mu)}{R_{2}}+\frac{(\mu -1)}{R_{2}}\times \frac{t}{P_{1}I}+\frac{\mu}{P_{1}I}$

$But \;\;\;\;\;\;\frac{1}{P_{1}I}=\frac{\mu-1}{\mu R_{1}}$

$\therefore \;\;\;\;\;\frac{1}{f} =\frac{(1-\mu)}{R_{2}}+\frac{(\mu -1)t}{R_{2}}\times \frac{(\mu -1)}{\mu R_{1}}+\frac{(\mu -1)}{R_{1}}$

$or \;\;\;\;\;\frac{1}{f} =(\mu -1)\left [ \frac{1}{R_{1}}-\frac{1}{R_{2}}+ \frac{(\mu -1)t}{\mu R_{1}R_{2}} \right ]$

For thin lens t → 0, and f → f₀

$\therefore \;\;\;\;\;\frac{1}{f_{0}} =(\mu -1)\left [ \frac{1}{R_{1}}-\frac{1}{R_{2}} \right ]$

$or \;\;\;\;\;\frac{1}{f} =\frac{1}{f_{0}}+\frac{(\mu -1)^{2}t}{\mu R_{1}R_{2}}$

Power of thick lens

$P=\frac{1}{f} =\frac{(\mu -1)}{R_{1}}-\frac{(\mu -1)}{R_{2}}+\frac{(\mu -1)^{2}}{R_{1}R_{2}}\times \frac{t}{\mu}$

$\because \;\;\;\;\;P_{1}=\frac{(\mu -1)}{R_{1}}\;\;\;\;\;and\;\;\;\;\;P_{2}=-\frac{(\mu -1)}{R_{2}}$

$\therefore \;\;\;\;\;P=P_{1}+P_{2}-P_{1}P_{2}\times \frac{t}{\mu}$

Position of second focal point (P₂F₂ = β₂)

The distance of second focal point will be measured from pole of second refracting surface.

$\because\;\;\;\;\;\frac{1}{f}=\frac{1}{P_{1}I}\times \frac{P_{2}I}{P_{2}F_{2}}$

$\therefore \;\;\;\;\;P_{2}F_{2}=f\times \frac{P_{2}I}{P_{2}F_{2}}=f\left ( \frac{P_{1}I-t}{P_{1}I} \right )\;\;\;\;\left ( \because P_{2}I=P_{1}I-t \right )$

$or \;\;\;\;\;\beta _{2}=P_{2}F_{2}=f\left ( 1-\frac{t}{P_{1}I} \right )$

$But \;\;\;\;\;\;\frac{1}{P_{1}I}=\frac{\mu-1}{\mu R_{1}}$

$\therefore \;\;\;\;\;\beta _{2}=P_{2}F_{2}=f\left [ 1-\frac{(\mu -1)t}{\mu R_{1}} \right ]$

Position of second principal point (P₂H₂ = α₂)

The distance of second principal point will be measured from pole of second refracting surface.

∵ P₂H₂ = H₂F₂ − P₂F₂

∴ α₂ = P₂H₂ = f − β₂

$But \;\;\;\;\;\beta _{2}=f\left [ 1-\frac{(\mu -1)t}{\mu R_{1}} \right ]$

$\therefore \;\;\;\;\;\alpha _{2}=f-f\left [ 1-\frac{(\mu -1)t}{\mu R_{1}} \right ]=\frac{f(\mu -1)t}{\mu R_{1}}$

Now by using sign convention, we get

$\alpha _{2}=-\frac{f(\mu -1)t}{\mu R_{1}}$

Position of first focal point (P₂F₁ = β₁)

For obtaining it we consider that the ray is incident on lens system from right i.e., O՛Q, this time image will be formed on F₁. So in β₂, R₁ is replaced by -R₂ and f is replaced by −f.

$\beta _{1}=-f\left [ 1+\frac{(\mu -1)t}{\mu R_{2}} \right ]$

Position of first principal point (P₁H₁ = α₁)

For it we use R₁ → –R₂ and f → – f in second principal point α₂

$\alpha _{1}=-\frac{f(\mu -1)t}{\mu (-R_{2})}$

$or\;\;\;\;\;\alpha _{1}=\frac{f(\mu -1)t}{\mu R_{2}}$

To know in detail about thick lens and the cardinal points of thick lens in English click here and in Bilingual (Hindi/English) click here.

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Thick lens and its cardinal points

Thick lens

Cardinal points of thick lens

Equivalent focal length

Refraction from surface PP1

Refraction from surface QP2

Power of thick lens

Position of second focal point (P2F2 = β2)

Position of second principal point (P2H2 = α2)

Position of first focal point (P2F1 = β1)

Position of first principal point (P1H1 = α1)

1 thought on “Thick lens and its cardinal points”

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Refraction from surface PP₁

Refraction from surface QP₂

Position of second focal point (P₂F₂ = β₂)

Position of second principal point (P₂H₂ = α₂)

Position of first focal point (P₂F₁ = β₁)

Position of first principal point (P₁H₁ = α₁)