What is the Correct Calculation for Spherical Aberrations in a Lens?

[Xyius]

Homework Statement

A collimated light beam is incident on the plane side of a of index 1.5, diameter 50mm, and radius 40mm. Find the .

Homework 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

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.