# Spherical Aberrations

by Xyius
Tags: aberrations, spherical
 P: 441 1. The problem statement, all variables and given/known data A collimated light beam is incident on the plane side of a of index 1.5, diameter 50mm, and radius 40mm. Find the . 2. Relevant equations Refraction in a plane surface: $$s'=\frac{-n_2}{n_1}s$$ Refraction on a spherical surface: $$\frac{n_1}{s}+\frac{n_2}{s'}=\frac{n_2-n_1}{R}$$ Spherical Wave Aberrations: $$a(Q)= \frac{h^4}{8} \left[ \frac{n_1}{s} \left( \frac{1}{s}+\frac{1}{R}\right)^2 + \frac{n_1}{s'} \left( \frac{1}{s'}-\frac{1}{R}\right)^2 \right]$$ R=Radius of curvature s=Object distance s'=Image distance n=index of refraction 3. The attempt at a solution So the light is traveling in a straight beam incident on the plane surface. By equation 1, the image distance is virtual and at infinity. so $s'=-\infty mm$. I then use this value for the calculation of the image of a spherical surface using equation 2. Plugging everything in, I get a distance of 80mm. So s'=80mm. I then used equation 3 to get the spherical wave aberration and used a height of 25mm since the lens is 50mm in diameter. (Also, R<0 since it is a concave surface from left to right.) I got an aberration of -1.29mm. The book says the correct answer is -0.858mm. What am I doing wrong? :(? EDIT: Figured it out! I used 1.5 for the index in equation 3 instead of 1.