Answer to Question #287126 in Optics for saf

Question #287126

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


1
Expert's answer
2022-01-13T09:20:22-0500

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{\u22125.952}" or 0.02087 m, 0.168 m respectively.

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


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