Answer to Question #287126 in Optics for saf

Question

Calculate the position of the six cardinal points (focal points, principal points, and nodal points) in the human cornea. Assume the cornea is a thick lens separating air (n = 1) and aqueous humour (n = 1.336). The refractive powers of the first and second surfaces are +47.898 D and −5.952 D, respectively. The cornea is 565 µm thick and has a refractive index of 1.376. Draw a diagram showing where all the cardinal points are located

Answer 1

Solution:

The refractive powers of the first and second surfaces are +47.898 D and −5.952 D.

So, radii of curvature are $\dfrac 1{+47.898}, \dfrac 1{−5.952}$ or 0.02087 m, 0.168 m respectively.

$\begin{aligned} &\text { focalpoints. }\\ &f=\frac{-n_{1}}{D}, \quad f^{\prime}=\frac{n_{3}}{D} .\\ &n_{1}=\text { air }, \quad n_{3}=\text { aqueous humour }=1.337\\ &R_{a} \text { : radius of } 1 \text {st focal pant }=0.02087 \mathrm{~m}\\ &R_{p}=\text { radics of } 2{\text {nd }} =0.168 \mathrm{~mm}\\ &D=\text { Total power } d=565 \mu \mathrm{m} \text {. }\\ &n_{2}=\text { lens }=1.376 \text {. }\\ &D=\frac{n_{2}-n_{1}}{R_{a}}+\frac{n_{3}-n_{2}}{R_{p}}+\frac{\left(n_{2}-n_{1}\right)\left(n_{3}-n_{2}\right) d}{n_{2}\left(R_{a} R_{p}\right)} .\\ &\Rightarrow D=-0.25\ \ [\text{By putting\ all\ values}]\\ &f_{1}=\frac{-1}{-0.25}=4 ,\quad f_{2}=\frac{1.337}{-0.25}=-5.348 .\\ &\text { Princple point } h_{1}=-\frac{f_{1}\left(n_{2}-1\right) d}{R_{p} n_{2}}=-0.092; h_{2}=\frac{-f_{2}\left(n_{2}-1\right) d}{R_{a} n_{2}}=0.047 . \end{aligned}$